CodeVs天梯之Gold

CodeVs天梯之Gold

2018.01.04 By gwj233

0x01贪心

1. 均分纸牌

   
   #include<iostream>
   
   using namespace std;
   const int maxn = 110;
   int a[maxn], sum, ans;
   int main(){
      int n;  cin>>n;
      for(int i = 1; i <= n; i++){
          cin>>a[i];  sum += a[i];
      }
      sum /= n;
      for(int i = 1; i <= n; i++){
          if(a[i] != sum){
              ans++;
              a[i+1] -= sum-a[i];
          }
      }
      cout<<ans<<'\n';
      return 0;
   }

2. 线段覆盖

   
   #include<iostream>
   
   
   #include<algorithm>
   
   using namespace std;
   const int maxn = 110;
   int a[maxn], b[maxn], c[maxn];
   bool cmp(int x, int y){
      return b[x]==b[y] ? a[x]<a[y] : b[x]<b[y];
   }
   int main(){
      int n;  cin>>n;
      for(int i = 1; i <= n; i++){
          int x, y;  cin>>x>>y;  //输入有坑,可能先大再小
          a[i]=min(x,y);  b[i]=max(x,y);  c[i]=i;
      }
      sort(c+1,c+n+1,cmp);
      int t = -999, ans = 0;
      for(int i = 1; i <= n; i++){
          if(t <= a[c[i]]){
              t = b[c[i]];  ans++;
          }
      }
      cout<<ans<<"\n";
      return 0;
   }

0x02高精度入门

1. 高精度练习之减法

   
   #include<iostream>
   
   
   #include<algorithm>
   
   
   #include<string>
   
   using namespace std;
   const int maxn = 510;
   int a[maxn], b[maxn], c[maxn];
   int main(){
      string s1, s2;
      cin>>s1>>s2;
      if(s1.size()<s2.size()||(s1.size()==s2.size()&&s1<s2)){ cout<<"-"; swap(s1,s2); }
      a[0] = s1.size();
      b[0] = s2.size();
      for(int i = 1; i <= a[0]; i++)a[i] = s1[a[0]-i]-'0';
      for(int i = 1; i <= b[0]; i++)b[i] = s2[b[0]-i]-'0';
      c[0] = max(a[0], b[0]);
      for(int i = 1; i <= c[0]; i++){
          if(a[i]<b[i]){ a[i+1]--; a[i]+=10; }
          c[i] = a[i]-b[i];
      }
      while(c[0]>1 && c[c[0]]==0)c[0]--;
      for(int i = c[0]; i >= 1; i--)cout<<c[i];
      return 0;
   }

2. 高精度练习之加法

   
   #include<iostream>
   
   
   #include<algorithm>
   
   
   #include<string>
   
   using namespace std;
   const int maxn = 510;
   int a[maxn], b[maxn], c[maxn];
   int main(){
      string s1, s2;
      cin>>s1>>s2;
      a[0] = s1.size();
      b[0] = s2.size();
      for(int i = 1; i <= a[0]; i++)a[i] = s1[a[0]-i]-'0';
      for(int i = 1; i <= b[0]; i++)b[i] = s2[b[0]-i]-'0';
      c[0] = max(a[0], b[0])+1;
      for(int i = 1; i <= c[0]; i++){
          c[i] += a[i]+b[i];
          if(c[i]>=10){ c[i+1]++; c[i]%=10; }
      }
      while(c[0]>1 && c[c[0]]==0)c[0]--;
      for(int i = c[0]; i >= 1; i--)cout<<c[i];
      return 0;
   }

3. 高精度练习之乘法

   
   #include<iostream>
   
   
   #include<algorithm>
   
   
   #include<string>
   
   using namespace std;
   const int maxn = 510;
   int a[maxn], b[maxn], c[maxn];
   int main(){
      string s1, s2;
      cin>>s1>>s2;
      a[0] = s1.size();
      b[0] = s2.size();
      for(int i = 1; i <= a[0]; i++)a[i] = s1[a[0]-i]-'0';
      for(int i = 1; i <= b[0]; i++)b[i] = s2[b[0]-i]-'0';
      c[0] = a[0]+b[0];
      for(int i = 1; i <= a[0]; i++){
          for(int j =1; j <= b[0]; j++){
              c[i+j-1] += a[i]*b[j];
              if(c[i+j-1]>=10){
                  c[i+j] += c[i+j-1]/10;
                  c[i+j-1] %= 10;
              }
          }
      }
      while(c[0]>1 && c[c[0]]==0)c[0]--;
      for(int i = c[0]; i >= 1; i--)cout<<c[i];
      return 0;
   }

0x03背包型动态规划

1. 装箱问题

   //数据太水，搜索就好
   
   #include<iostream>
   
   
   #include<algorithm>
   
   using namespace std;
   int n, m, a[110];
   int search(int i, int r){
      if(i == n)return r;
      if(r < a[i])return search(i+1, r);
      return min(search(i+1,r), search(i+1,r-a[i]));
   }
   int main(){
      cin>>m>>n;
      for(int i = 0; i < n; i++)cin>>a[i];
      cout<<search(0,m);
      return 0;
   }

   //f[i][j],表示在选第i个物品时, 大小为j的体积能否取到
   
   #include<iostream>
   
   using namespace std;
   const int maxn = 20010;
   int n, m, a[35], f[35][maxn];
   int main(){
      cin>>m>>n;
      for(int i = 1; i <= n; i++)cin>>a[i];
      f[0][0] = 1;
      for(int i = 1; i <= n; i++)
          for(int j = 0; j <= m; j++)
              if(f[i-1][j]){
                  f[i][j] = f[i-1][j];
                  if(j+a[i]<=m)f[i][j+a[i]] = 1;
              }
      for(int i = m; i >= 0; i--)
          if(f[n][i]){ cout<<m-i<<"\n"; return 0;}
      return 0;
   }

   //f[i]表示大小为i的体积能否取到
   
   #include<iostream>
   
   using namespace std;
   const int maxn = 20010;
   int n, m, a[maxn], f[maxn];
   int main(){
      cin>>m>>n;
      for(int i = 1; i <= n; i++)cin>>a[i];
      f[0] = 1;
      for(int i = 1; i <= n; i++)
          for(int j = m; j >= 0; j--)
              if(f[j] && j+a[i]<=m)
                  f[j+a[i]] = 1;
      for(int i = m; i >= 0; i--)
          if(f[i]){ cout<<m-i<<"\n"; return 0;}
      return 0;
   }

   //01背包, f[i]表示体积为i时得到的最大价值
   
   #include<iostream>
   
   using namespace std;
   const int maxn = 20010;
   int n, m, w[35], v[35], f[maxn];
   int main(){
      cin>>m>>n;
      for(int i = 1; i <= n; i++){
          cin>>w[i];  v[i] = w[i];
      }
      for(int i = 1; i <= n; i++)
          for(int j = m; j >= w[i]; j--)
              f[j] = max(f[j],f[j-w[i]]+v[i]);
      cout<<m-f[m]<<"\n";
      return 0;
   }

2. 乌龟棋

   
   #include<iostream>
   
   
   #include<algorithm>
   
   using namespace std;
   int n, m, a[350], num[5], f[50][50][50][50];
   int main(){
      cin>>n>>m;
      for(int i = 1; i <= n; i++)cin>>a[i];
      for(int i = 1; i <= m; i++){
          int x;  cin>>x;  num[x]++;
      }
      for(int i = 0; i <= num[1]; i++){
          for(int j = 0; j <= num[2]; j++){
              for(int k = 0; k <= num[3]; k++){
                  for(int l = 0; l <= num[4]; l++){
                      int t = 0;
                      if(i)t = max(t, f[i-1][j][k][l]);
                      if(j)t = max(t, f[i][j-1][k][l]);
                      if(k)t = max(t, f[i][j][k-1][l]);
                      if(l)t = max(t, f[i][j][k][l-1]);
                      f[i][j][k][l] = t+a[i+j*2+k*3+l*4+1];
                  }
              }
          }
      }
      cout<<f[num[1]][num[2]][num[3]][num[4]]<<"\n";
      return 0;
   }

0x04序列型动态规划

1. 拦截导弹

   //LDS的个数等于LIS的长度
   
   #include<iostream>
   
   
   #include<algorithm>
   
   using namespace std;
   int n, a[30], f[30], ans;
   int main(){
      while(cin>>a[n])n++;
      for(int i = 0; i < n; i++)f[i] = 1;
      for(int i = 0; i < n; i++)
          for(int j = 0; j < i; j++)
              if(a[j]>=a[i])f[i] = max(f[i], f[j]+1);
      for(int i = 0; i < n; i++)ans = max(ans, f[i]);
      cout<<ans<<"\n";
      for(int i = 0; i < n; i++)f[i] = 1;
      for(int i = 0; i < n; i++)
          for(int j = 0; j < i; j++)
              if(a[j]<a[i])f[i] = max(f[i], f[j]+1);
      ans = 0;
      for(int i = 0; i < n; i++)ans = max(ans, f[i]);
      cout<<ans<<"\n";
      return 0;
   }

2. 最长严格上升子序列

   //f[i]:到i为止的LIS的长度。
   //f[i]=max{1,f[j]+1|j<i&&aj<ai}
   
   #include<iostream>
   
   
   #include<algorithm>
   
   using namespace std;
   int n, a[510], f[510], ans;
   int main(){
      cin>>n;
      for(int i = 1; i <= n; i++)cin>>a[i];
      for(int i = 1; i <= n; i++){
          int t = 0;
          for(int j = 1; j < i; j++)
              if(a[j]<a[i])t = max(t, f[j]);
          f[i] = t+1;
          ans = max(ans, f[i]);
      }
      cout<<ans<<"\n";
      return 0;
   }

3. 线段覆盖 2

   //f[i]:到第i条线段为止能获得的最大价值
   //f[i]=max{s[i].c,f[j]+s[i].c|j<i&&s[j].b<=s[i].a}
   
   #include<iostream>
   
   
   #include<algorithm>
   
   using namespace std;
   const int maxn = 1010;
   struct seg{ int a, b, c; }s[maxn];
   bool cmp(seg a, seg b){ return a.a<b.a; }
   int n, f[maxn], ans;
   int main(){
      cin>>n;
      for(int i = 1; i <= n; i++)
          cin>>s[i].a>>s[i].b>>s[i].c;
      sort(s+1,s+n+1,cmp);
      for(int i = 1; i <= n; i++){
          f[i] = s[i].c;
          for(int j = 1; j < i; j++)
              if(s[j].b<=s[i].a)f[i] = max(f[i], f[j]+s[i].c);
          ans = max(ans, f[i]);
      }
      cout<<ans<<"\n";
      return 0;
   }

0x05区间型动态规划

1. 石子归并

   
   #include<iostream>
   
   
   #include<algorithm>
   
   using namespace std;
   const int maxn = 1010, inf=0xffffff;
   int w[maxn], f[maxn][maxn];
   int main(){
      int n;  cin>>n;
      for(int i = 1; i <= n; i++)cin>>w[i];
      for(int i = 1; i <= n; i++)
          for(int j = 1; j <= n; j++)
              f[i][j] = i==j ? 0 : inf;
      for(int i = 1; i <= n; i++)w[i] += w[i-1];
      //转移顺序长度从小到大覆盖即可
      for(int i = n; i >= 1; i--)//枚举起点
          for(int j = i; j <= n; j++)//枚举终点
              for(int k = i; k <= j; k++)//枚举断点
                  f[i][j] = min(f[i][j],f[i][k]+f[k+1][j]+w[j]-w[i-1]);
      cout<<f[1][n]<<"\n";
      return 0;
   }

   //解释一下, DP题写几个版本主要看心情。。。
   
   #include<iostream>
   
   
   #include<algorithm>
   
   using namespace std;
   const int maxn = 1010, inf=0xffffff;
   int w[maxn], f[maxn][maxn];
   int dp(int i, int j){
      if(f[i][j] != inf)return f[i][j];
      for(int k = i; k <= j; k++)
          f[i][j] = min(f[i][j], dp(i,k)+dp(k+1,j)+w[j]-w[i-1]);
      return f[i][j];
   }
   int main(){
      int n;  cin>>n;
      for(int i = 1; i <= n; i++)cin>>w[i];
      for(int i = 1; i <= n; i++)
          for(int j = 1; j <= n; j++)
              f[i][j] = i==j ? 0 : inf;
      for(int i = 1; i <= n; i++)w[i] += w[i-1];
      cout<<dp(1,n)<<"\n";
      return 0;
   }

2. 能量项链

   //环形DP, 复制一份, 拆环成链，然后做区间DP(即对每条链做一遍区间DP，取最大值。)
   //解释一下， 注释详细度主要看心情。。。
   
   #include<iostream>
   
   
   #include<algorithm>
   
   using namespace std;
   const int maxn = 1010;
   int w[maxn], f[maxn][maxn];
   int dp(int i, int j){
      if(f[i][j])return f[i][j];
      //注意边界, [i~i,i+1~j] and [i~j-1,j~j];
      for(int k = i; k <= j-1; k++)
          //通过k合并后, 两颗珠子分别为[i,k],[k+1,j]其中计算能量时headi*taili*headj;
          f[i][j] = max(f[i][j], dp(i,k)+dp(k+1,j)+w[i]*w[k+1]*w[j+1]);
      return f[i][j];
   }
   int main(){
      int n;  cin>>n;
      for(int i = 1; i <= n; i++){ cin>>w[i];  w[i+n]=w[i]; }
      dp(1, 2*n);
      int ans = 0;
      for(int i = 1; i <= n; i++)//扫描每条长度为n的链, 最大值为答案。
          ans = max(ans, f[i][i+n-1]);
      cout<<ans<<'\n';
      return 0;
   }

3. 矩阵取数游戏

   //每行独立区间DP, 贪心反例->某行像这样，4 1 1 1 1 1 233 3 3
   //2^80数据, 所以记得高精.
   
   #include<iostream>
   
   
   #include<algorithm>
   
   
   #include<cstring>
   
   
   #include<cstdlib>
   
   
   #include<cstdio>
   
   using namespace std;
   typedef long long LL;
   typedef __int128 LLL;
   const int maxn = 110;
   int n, m, a[maxn];
   LLL t[maxn], f[maxn][maxn], _max, ans;
   void print(LLL ans){
      if(ans == 0)return ;
      else print(ans/10);
      putchar(ans%10+'0');
   }
   int main(){
      cin>>n>>m;
      t[0] = 1;
      for(int i = 1; i <= m; i++)t[i] = t[i-1]*2;
      while(n--){
          for(int i = 1; i <= m; i++)cin>>a[i];
          memset(f, 0, sizeof f);
          //f[i][j]:这行还剩下[i,j]时能得到的最高分
          //转移加上分别取了左边的和右边的数的时候的得分
           //转移顺序，区间从大到小。
          for(int i = 1; i <= m; i++)
              for(int j = m; j >= i; j--)
                  f[i][j]=max(f[i-1][j]+t[m-(j-i+1)]*a[i-1], f[i][j+1]+t[m-(j-i+1)]*a[j+1]);
          //枚举最后一个取的是哪个数，得到这一行的最高分
          _max = 0;
          for(int i = 1; i <= m; i++)_max = max(_max, f[i][i]+t[m]*a[i]);
          ans += _max;
      }
      if(ans == 0)cout<<"0\n";
      else print(ans);
      return 0;
   }

0x06棋盘型动态规划

1. 过河卒

   //填表法
   
   #include<iostream>
   
   using namespace std;
   int n, m, x, y, a[20][20], f[20][20];
   int main(){
      cin>>n>>m>>x>>y; x++;y++;n++;m++;//+1方便赋初始值
      //9 points could'n be find
      a[x][y] = 1;
      a[x-1][y-2] = a[x-1][y+2] = a[x+1][y-2] = a[x+1][y+2] = 1;
      a[x-2][y-1] = a[x-2][y+1] = a[x+2][y-1] = a[x+2][y+1] = 1;
      f[0][1] = 1; //边界
      for(int i = 1; i <= n; i++)
          for(int j = 1; j <= m; j++)
              if(!a[i][j])f[i][j] = f[i-1][j]+f[i][j-1];
      cout<<f[n][m]<<"\n";
      return 0;
   }

   //刷表法
   
   #include<iostream>
   
   using namespace std;
   int n, m, x, y, a[20][20], f[20][20];
   int main(){
      cin>>n>>m>>x>>y;
      a[x][y] = 1;
      a[x-1][y-2] = a[x-1][y+2] = a[x+1][y-2] = a[x+1][y+2] = 1;
      a[x-2][y-1] = a[x-2][y+1] = a[x+2][y-1] = a[x+2][y+1] = 1;
      f[0][0] = 1;//初始值, 刷表时不会覆盖所以可以直接放
      for(int i = 0; i <= n; i++){
          for(int j = 0; j <= m; j++){
              if(!a[i+1][j])f[i+1][j] += f[i][j];
              if(!a[i][j+1])f[i][j+1] += f[i][j];
          }
      }
      cout<<f[n][m]<<"\n";
      return 0;
   }

2. 传纸条

   //考虑题设，找到两条不重复的路径，所以从上到下直接DP，状态四维(上往下,下往上分别DP,没办法考虑路径重叠)
   //f[i][j][k][l]表示分别到(i,j),(k,l)时候的最大好心值
   
   #include<iostream>
   
   
   #include<algorithm>
   
   using namespace std;
   int n, m, a[55][55], f[55][55][55][55];
   int main(){
      cin>>n>>m;
      for(int i = 1; i <= n; i++)
          for(int j = 1; j <= m; j++)
              cin>>a[i][j];
      for(int i = 1; i <= n; i++)
          for(int j = 1; j <= m; j++)
              for(int k = 1; k <= n; k++)
                  for(int l = 1; l <= m; l++)
                      if(!(i==k&&j==l) || (i==n&&j==m&&k==n&&l==m))
                          f[i][j][k][l] = max(max(f[i-1][j][k-1][l], f[i][j-1][k-1][l]), max(f[i-1][j][k][l-1],f[i][j-1][k][l-1]))+a[i][j]+a[k][l];
      cout<<f[n][m][n][m]<<"\n";
      return 0;
   }

3. 骑士游历

   //bugs:行列弄反(x,y是坐标轴...)+longlong
   
   #include<iostream>
   
   using namespace std;
   typedef long long LL;
   LL n, m, x1, y1, x2, y2, f[55][55];
   int main(){
      cin>>n>>m>>x1>>y1>>x2>>y2;
      f[x1][y1] = 1;
      for(int i = x1+1; i <= x2; i++)//避免覆盖掉x1,y1时候的1种方案
          for(int j = 1; j <= n; j++)//因为日字，所以要全
              f[i][j] = f[i-1][j-2]+f[i-1][j+2]+f[i-2][j-1]+f[i-2][j+1];
      cout<<f[x2][y2]<<"\n";
      return 0;
   }

   //行列式反的，，，
   
   #include<iostream>
   
   using namespace std;
   typedef long long LL;
   LL n, m, x1, y1, x2, y2, f[55][55];
   int main(){
      cin>>n>>m>>x1>>y1>>x2>>y2;
      f[y1][x1] = 1;
      //转移的时候按照列每层去取(因为只能往右走)
      for(int i = x1; i <= x2; i++){//枚举y
          for(int j = 1; j <= m; j++){//枚举x
              //之前的状态无法到达那么当前状态也无法到达
              if(!f[j][i])continue;
              //刷表状态转移
              f[j+1][i+2] += f[j][i];
              f[j-1][i+2] += f[j][i];
              f[j+2][i+1] += f[j][i];
              f[j-2][i+1] += f[j][i];
          }
      }
      cout<<f[y2][x2]<<"\n";
      return 0;
   }

4. 数字三角形

   //f[i][j]:从(i,j)出发能获得的最大值 _裸DFS
   
   #include<iostream>
   
   
   #include<algorithm>
   
   using namespace std;
   int n, a[110][110], f[110][110];
   int dfs(int i, int j){
      if(f[i][j])return f[i][j];
      if(i>n || j>n)return 0;
      return f[i][j] = max(dfs(i+1,j),dfs(i+1,j+1))+a[i][j];
   }
   int main(){
      cin>>n;
      for(int i = 1; i <= n; i++)
          for(int j = 1; j <= i; j++)
              cin>>a[i][j];
      cout<<dfs(1,1)<<"\n";
      return 0;
   }

   //f[i][j]:从(i,j)出发能获得的最大值 _裸递推
   
   #include<iostream>
   
   
   #include<algorithm>
   
   using namespace std;
   int n, a[110][110], f[110][110];
   int main(){
      cin>>n;
      for(int i = 1; i <= n; i++)
          for(int j = 1; j <= i; j++)
              cin>>a[i][j];
      for(int i = n; i >= 1; i--)
          for(int j = 1; j <= i; j++)
              f[i][j] = max(f[i+1][j], f[i+1][j+1])+a[i][j];
      cout<<f[1][1]<<"\n";
      return 0;
   }

   //f[i][j]:从(1,1)到(i,j)能获得的最大值 _裸递推
   
   #include<iostream>
   
   
   #include<algorithm>
   
   using namespace std;
   int n, a[110][110], f[110][110];
   int main(){
      cin>>n;
      for(int i = 1; i <= n; i++)
          for(int j = 1; j <= i; j++)
              cin>>a[i][j];
      for(int i = 1; i <= n; i++)
          for(int j = 1; j <= i; j++)
              f[i][j] = max(f[i-1][j], f[i-1][j-1])+a[i][j];
      int ans = -0xffffff;
      for(int i = 1; i <= n; i++)ans = max(ans, f[n][i]);
      cout<<ans<<'\n';
      return 0;
   }

   //f[i][j]:从(i,j)出发能获得的最大值 _滚动数组
   
   #include<iostream>
   
   
   #include<algorithm>
   
   using namespace std;
   int n, a[110][110], f[2][110];
   int main(){
      cin>>n;
      for(int i = 1; i <= n; i++)
          for(int j = 1; j <= i; j++)
              cin>>a[i][j];
      for(int i = n; i >= 1; i--)
          for(int j = 1; j <= i; j++)
              f[i%2][j] = max(f[(i+1)%2][j], f[(i+1)%2][j+1])+a[i][j];
      cout<<f[1][1]<<"\n";
      return 0;
   }

   //f[i][j]:从(1,1)到(i,j)能获得的最大值 _滚动数组2
   
   #include<iostream>
   
   
   #include<algorithm>
   
   using namespace std;
   int n, a, f[2][110];
   int main(){
      cin>>n;
      for(int i = 1; i <= n; i++){
          for(int j = 1; j <= i; j++){
              cin>>a;
              f[i%2][j] = max(f[(i-1)%2][j], f[(i-1)%2][j-1])+a;
          }
      }
      int ans = -0xffffff;
      for(int i = 1; i <= n; i++)ans = max(ans, f[n%2][i]);
      cout<<ans<<"\n";
      return 0;
   }

0x07划分型动态规划

1. 乘积最大

   //f[i][j]:前i位数包含j个乘号时能获得的最大值
   //转移，枚举每个乘号的位置即可，O(n^3)可过。
   
   #include<iostream>
   
   
   #include<algorithm>
   
   
   #include<string>
   
   using namespace std;
   typedef long long LL;
   int n, m;
   LL f[110][110];
   string s;
   LL mid(int l, int r){
      LL t = 0;
      for(int i = l; i <= r; i++)
          t = t*10+s[i-1]-'0';//第i位在s[i-1];
      return t;
   }
   int main(){
      cin>>n>>m>>s;
      for(int i = 1; i <= n; i++)f[i][0] = mid(1,i);//边界条件，没有乘号
      for(int i = 1; i <= n; i++) //枚举前i位数
          for(int j = 1; j <= min(m,i-1); j++)//枚举每个乘号(即子状态)
              for(int k = j; k < i; k++)//枚举该乘号的位置,乘号放后面(保证第j个乘号时, 前j-1个乘号的最优状态已经算出来了）
                  f[i][j] = max(f[i][j], f[k][j-1]*mid(k+1,i));
      cout<<f[n][m]<<"\n";
      return 0;
   }

2. 数的划分

   //step1:把n个苹果放到m个盘子里，不允许有空盘。等价于每个盘子放一个苹果先允许有空盘
   //step2:f[i][j]表示i个苹果j个盘子的放法数目
   //step3:转移,j>i时,去掉空盘不影响结果; j<=i时,对盘子是否空着分类讨论;
   
   #include<iostream>
   
   using namespace std;
   int n, m, f[210][10];
   int main(){
      cin>>n>>m;
      n-=m;//每个盘子先放一个苹果就不会有空盘了。。。
      for(int i = 0; i <= n; i++)f[0][i]=f[i][1]=1;
      for(int i = 1; i <= n; i++)
          for(int j = 2; j <= m; j++)
              f[i][j] = j>i?f[i][i]:f[i][j-1]+f[i-j][j];//所有盘子有苹果时每个盘子都去掉一个苹果不影响结果
      cout<<f[n][m]<<"\n";
      return 0;
   }

   //Step1:f[i][j]:将i这个整数划分成j份且不重复的方法数
   //Step2:因为划分成的每一份至少为1，所以我们把它每份减去1
   //Step3:将i这个数划分成j份等价于将i-j这个数划分成1份、2份、3份。。。j份的和
   //Step4:f[i-1][j-1]=f[(i-1)-(j-1)][1]+...+f[(i-1)-(j-1)][j-1]; 代入化简
   
   #include<iostream>
   
   
   #include<algorithm>
   
   using namespace std;
   int n, k, f[210][10];
   int main(){
   cin>>n>>k;
   f[0][0] = 1;
   for(int i = 1; i <= n; i++)
       for(int j = 1; j <= min(i,k); j++)
           f[i][j] = f[i-j][j]+f[i-1][j-1];
   cout<<f[n][k];
   return 0;
   }

3. 统计单词个数

   //just for test4
   
   #include<iostream>
   
   
   #include<algorithm>
   
   
   #include<cstring>
   
   
   #include<string>
   
   using namespace std;
   int n, m, x, d[210], f[210][50];//c[210][210];
   string s, w[20];
   void pre(){
      memset(d, 0x3f, sizeof d);
      for(int i = 1; i <= s.size(); i++){//枚举每个字母
          for(int j = 1; j <= x; j++){//枚举每个单词
              if(s.substr(i).find(w[j])==0){//如果存在i字母开头的单词
                  d[i] = min(d[i], i+(int)w[j].size()-1);
              }
          }
      }
   }
   int main(){
      int T;  cin>>T;
      while(T--){
          //input
          cin>>n>>m;
          s = " ";
          for(int i = 1; i <= n; i++){
              string t;  cin>>t;  s += t;
          }
          n *= 20;
          cin>>x;
          for(int i = 1; i <= x; i++)cin>>w[i];
          //预处理1：得到c[i][j]为[i,j]中的最大单词数
          //预处理2：得到d[i]为字母i开头的最短单词的结束位置(小贪心,每个字母只能按照第一个字母取一次)
          pre();
          //dp:f[i][j]为前i个字母划分成j段能得到的最大单词数
          //转移:f[i][j] = max{ f[k][j-1]+c[k+1][i] | k>=j&&k<i}
          for(int i = 1; i <= n; i++){
              for(int j = 1; j <= m; j++){
                  int w = 0;
                  for(int k = i; k >= j; k--){ //逆序覆盖
                      if(d[k]<=i)w++; //w=[k,i]
                      f[i][j] = max(f[i][j],f[k-1][j-1]+w);
                      //f[i][j] = max(f[i][j], f[k][j-1]+w);
                      //if(d[k]<=i)w++; //w=[k+1,i]
                  }
              }
          }
          cout<<f[n][m]<<"\n";
      }
      return 0;
   }

0x08宽度优先搜索

1. 四子连棋

   //思路：把空白当棋，交替黑白走。
   //实现：BFS, 打表判断是否成立
   
   #include<iostream>
   
   
   #include<algorithm>
   
   
   #include<string>
   
   
   #include<queue>
   
   using namespace std;
   string s;
   struct node{
   string ma;  int step;  char next;
   node(string x, int y, char ch):ma(x),step(y),next(ch){}
   };
   queue<node>q;
   int dz[] = {4,-4,1,-1};
   char change(char ch){
   if(ch == 'B')return 'W';
   if(ch == 'W')return 'B';
   }
   int check(string s){
   //check diagonal 1
   if(s[0]==s[5] && s[5]==s[10] && s[10]==s[15])return 1;
   //check diagonal 2
   if(s[3]==s[6] && s[6]==s[9] && s[9]==s[12])return 1;
   //check row
   for(int i = 0; i < 4; i++){
       int ok = 1, t = 4*i;
       for(int j = 0; j < 4; j++)
           if(s[t] != s[t+j])ok = 0;
       if(ok)return 1;
   }
   //check col
   for(int i = 0; i < 4; i++){
       int ok = 1, t = i;
       for(int j = 0; j < 4; j++)
           if(s[t] != s[t+j*4])ok = 0;
       if(ok)return 1;
   }
   return 0;
   }
   int bfs(){
   while(q.size()){
       string t = q.front().ma;
       int st = q.front().step;
       char ch = q.front().next;
       q.pop();
       //check
       if(check(t))return st;
       //find O
       int o1=-1, o2;
       for(int i = 0; i < 16; i++){
           if(t[i]=='O'){
               if(o1==-1)o1 = i;
               else o2 = i;
           }
       }
       //o1go
       for(int i = 0; i < 4; i++){
           if(dz[i]==1 && o1%4==3)continue;
           if(dz[i]==-1 && o1%4==0)continue;
           int nz = o1+dz[i];
           if(nz>=0 && nz<16 && t[nz]==ch){
               string nt = t;
               swap(nt[o1],nt[nz]);
               q.push(node(nt,st+1,change(ch)));
           }
       }
       //o2go
       for(int i = 0; i < 4; i++){
           if(dz[i]==1 && o2%4==3)continue;
           if(dz[i]==-1 && o2%4==0)continue;
           int nz = o2+dz[i];
           if(nz>=0 && nz<16 && t[nz]==ch){
               string nt = t;
               swap(nt[o2],nt[nz]);
               q.push(node(nt,st+1,change(ch)));
           }
       }
   }
   }
   int main(){
   ios::sync_with_stdio(false);
   for(int i = 0; i < 4; i++){
       string t;  cin>>t;  s += t;
   }
   if(check(s)){ cout<<"0"; return 0;}
   int ans = 0xffffff;
   q.push(node(s,0,'W'));
   ans = min(ans, bfs());
   while(q.size())q.pop();
   q.push(node(s,0,'B'));
   ans = min(ans, bfs());
   cout<<ans<<"\n";
   return 0;
   }

2. 逃跑的拉尔夫

   
   #include<cstdio>
   
   
   #include<queue>
   
   
   #include<set>//set判重防MLE
   
   using namespace std;
   
   const int dx[4] = {-1, 0, 1, 0};
   const int dy[4] = {0, 1, 0, -1};
   
   int r, c, c1, r1, n, go[1010];
   char a[55][55];
   struct node{ 
   int x, y, step; 
   node(int x, int y, int step):x(x),y(y),step(step){}
   };
   queue<node>q;
   set<int>s;
   
   bool inside(int x, int y){ return (x<r && x>=0 && y<c && y>=0 && a[x][y]!='X'); }
   
   int togo(char ch){
   if(ch == 'N')return 0;
   if(ch == 'E')return 1;
   if(ch == 'S')return 2;
   if(ch == 'W')return 3;
   }
   
   void bfs(){
   q.push(node(r1, c1, 0));
   while(q.size()){
       node t = q.front();  q.pop();
       if(t.step == n)a[t.x][t.y] = '*';
       int tt=go[t.step], nx=t.x+dx[tt], ny=t.y+dy[tt];
       while(inside(nx,ny)){
           int ok = nx*1000000+ny*10000+t.step+1;
           if(!s.count(ok)){
               q.push(node(nx,ny,t.step+1));
               s.insert(ok);
           }
           nx += dx[tt], ny += dy[tt];
       }
   }
   return ;
   }
   
   int main(){
   scanf("%d%d", &r, &c);
   for(int i = 0; i < r; i++)scanf("%s", a[i]);
   for(int i = 0; i < r; i++)
       for(int j = 0; j < c; j++)
           if(a[i][j] == '*'){ r1 = i; c1 = j; break;}
   a[r1][c1] = '.';
   scanf("%d",&n);
   for(int i= 0; i < n; i++){
       char t[10];  scanf("%s", t);  go[i] = togo(t[0]);
   }
   bfs();
   for(int i = 0; i < r; i++)printf("%s\n", a[i]);
   return 0;
   }

3. 字串变换

   //思路就是对于每个状态下的字符串，枚举可以替换的部分替换作为下一个新的状态。
   
   #include<iostream>
   
   
   #include<queue>
   
   
   #include<string>
   
   
   #include<map>
   
   using namespace std;
   int n = 1, flag;
   string a, b, ai[1010], bi[1010];
   queue<string>q;
   map<string, int>ma;//map判重防MLE
   int main(){
   cin>>a>>b;
   while(cin>>ai[n]>>bi[n])n++;
   q.push(a);
   ma[a] = 0;
   while(q.size()){
       string t = q.front();  q.pop();
       if(t == b){ flag = 1; break; }
       if(ma[t]>10)break;
       //如果没有这层循环的话，就只能找到第一个子串，后面的会被忽略,如abaaaba abcdaba
       for(int j = 0; j < t.size(); j++){ 
           string nt = t.substr(j);
           for(int i = 1; i < n; i++){
               int tt = nt.find(ai[i]);
               if(tt == string::npos)continue;
               tt += j; //边界条件调起来很麻烦，以及最后直接+j就好了
               string ttt = t.substr(0,tt)+bi[i]+t.substr(tt+ai[i].size());
               if(!ma.count(ttt)){
                   ma[ttt] = ma[t]+1;
                   q.push(ttt);
               }
           }
       }
   }
   if(flag)cout<<ma[b]<<"\n";
   else cout<<"NO ANSWER!\n";
   return 0;
   }

   
   #include<iostream>
   
   
   #include<string>
   
   
   #include<queue>
   
   
   #include<set> //set判重
   
   
   #define maxn 1010
   
   using namespace std;
   string ai[maxn],bi[maxn];
   struct node{
      string str;
      int st;
      node(string a, int b):str(a),st(b){}
   };
   set<string>s;
   int ans;
   int main(){
      string a, b;
   cin>>a>>b;
      int n = 1; 
      while(cin>>ai[n]>>bi[n])n++;
      queue<node>q;
      q.push(node(a,0));
      while(q.size()){
          node t = q.front(); q.pop();
       if(t.str == b){ ans = t.st; break;}
       if(t.st > 10){ ans = 20; break; }
       for(int j = 0; j < t.str.size(); j++){
           string nt = t.str.substr(j);
           for(int i = 1; i < n; i++){
               int tt = nt.find(ai[i]);
               if(tt==string::npos)continue;
               tt += j;
               string ttt = t.str.substr(0,tt)+bi[i]+t.str.substr(tt+ai[i].size());
               if(!s.count(ttt)){
                   q.push(node(ttt,t.st+1));
                   s.insert(ttt);
               }
           }
       }
      }
   //如果<10就因为找不到出来也是不成立的, 即ans没有被赋过值的话
      if(ans == 20 || ans == 0)cout<<"NO ANSWER!\n";
      else cout<<ans<<"\n";
      return 0;
   }

0x09深度优先搜索

1. 单词接龙

   //直接搜索就好啦，几乎没什么技巧，就是状态建模会有点难想到(应该有多种)
   //包含的情况可以证明是不需要考虑的,因为包含后一定不会比不包含要来的长
   
   #include<iostream>
   
   
   #include<algorithm>
   
   
   #include<string>
   
   using namespace std;
   const int maxn = 30;
   int n, ans, used[maxn];
   string w[maxn];
   //找到a的末尾与b的前端重复的子串并返回其长度
   int find(string a, string b){
      int mm = min(a.size(), b.size());
      for(int i = 1; i <= mm; i++)
          if(a.substr(a.size()-i)==b.substr(0,i))
              return i;
      return 0;
   }
   //深度优先搜索寻找解, 状态：s为当前字符串
   void dfs(string s){
      ans = max(ans, (int)s.size());
      for(int i = 1; i <= n; i++)if(used[i]<2){
          int t = find(s, w[i]);
          if(t == s.size() && s!=w[0])continue;//包含关系
          if(t){
              used[i]++;
              dfs(s.substr(0,s.size()-t)+w[i]);
              used[i]--;
          }
      }
   }
   int main(){
      cin>>n;
      for(int i = 1; i <= n; i++)cin>>w[i];
      cin>>w[0];
      dfs(w[0]);
      cout<<ans<<"\n";
      return 0;
   }

2. 四色问题

   //尝试填每个点每种颜色填过去就好啦
   
   #include<iostream>
   
   using namespace std;
   int n, e[10][10];
   int c[10], ans;
   void dfs(int cur){
   if(cur == n)ans++;
   else for(int i = 0; i < 4; i++){
       c[cur] = i;
       bool ok = true;
       for(int j = 0; j < cur; j++)//判断和前面的点有没有冲突
           if(e[j][cur] && c[j]==c[cur])//如果联通且同色那就翻车
               { ok = false; break; }
       if(ok){
           dfs(cur+1);
       }
   }
   }
   int main(){
   cin>>n;
   for(int i = 0; i < n; i++)
       for(int j = 0; j < n; j++)
           cin>>e[i][j];
   dfs(0);
   cout<<ans<<"\n";
   return 0;
   }

3. 全排列

   
   #include<iostream>
   
   using namespace std;
   int n, c[20];
   void dfs(int cur){
   if(cur == n){
       for(int i = 0; i < n; i++)cout<<c[i]<<" ";
       cout<<"\n";
   }else for(int i = 1; i <= n; i++){
       int ok = 1;
       for(int j = 0; j < cur; j++)
           if(c[j]==i)ok = 0;
       if(ok){
           c[cur] = i;
           dfs(cur+1);
       }
   }
   }
   int main(){
   cin>>n;
   dfs(0);
   return 0;
   }

4. N皇后问题

   //c[i]:第i行的皇后放在第几列
   
   #include<iostream>
   
   using namespace std;
   int n, c[20], ans;
   void dfs(int cur){
   if(cur > n)ans++;
   else for(int i = 1; i <= n; i++){
       int ok = 1;
       for(int j = 1; j < cur; j++)
           if(c[j]==i || c[j]-j==i-cur || c[j]+j==i+cur)
               { ok = 0; break; }
       if(ok){
           c[cur] = i;
           dfs(cur+1);
       }
   }
   }
   int main(){
   cin>>n;
   dfs(1);
   cout<<ans<<"\n";
   return 0;
   }