斜率优化DP
hdu 2993 MAX Average Problem
http://acm.hdu.edu.cn/showproblem.php?pid=2993
这题都整死我了，代码基本上和别人AC的都快一样了还是不对。郁闷死了快。。最后终于发现了错误点自判断斜率时y1*x2 <= y2*x1 如果用整型就会超数据类型肯定会错，而上边的用整型可以过，不过最好用实型因为毕竟k的大小不确定。。。这题的详细解释http://www.docin.com/p-47950655.html 例2 这里在转到0-n个点求斜率的时候不好理解。
数形结合以形住树。
View Code
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <queue> #include <stack> #include <set> #include <map> #include <string> #define CL(a,num) memset((a),(num),sizeof(a)) #define iabs(x) ((x) > 0 ? (x) : -(x)) #define Min(a,b) (a) > (b)? (b):(a) #define Max(a,b) (a) > (b)? (a):(b) #define ll long long #define inf 0x7f7f7f7f #define MOD 100000007 #define lc l,m,rt<<1 #define rc m + 1,r,rt<<1|1 #define pi acos(-1.0) #define test puts("<------------------->") #define maxn 100007 #define M 100007 #define N 100007 using namespace std; int q[N],sum[N]; double max(double a,double b) { return a > b ? a:b; } int GetInt() { char ch = getchar(); while (ch < '0' || ch > '9') ch = getchar(); int num = 0; while (ch >= '0' && ch <= '9') { num = num*10 + ch - '0'; ch = getchar(); } return num; } int main() { //freopen("data.in","r",stdin); int n,i,k; while (~scanf("%d%d",&n,&k)) { sum[0] = 0; for (i = 1; i <= n; ++i) { int x = GetInt(); sum[i] = sum[i - 1] + x; } int front = 0,tail = 0; q[0] = 0; double ans = 0; for (i = k; i <= n; ++i) { int now = i - k; while (tail > front) { int y1 = sum[q[tail]] - sum[q[tail - 1]]; int x1 = q[tail] - q[tail - 1]; int y2 = sum[now] - sum[q[tail]]; int x2 = now - q[tail]; if (y1*x2 >= y2*x1) tail--; else break; } q[++tail] = now; while(front<tail){ double y1=sum[i] - sum[q[front]]; double x1=i - q[front]; double y2=sum[i] - sum[q[front+1]]; double x2=i - q[front+1]; if(y1*x2<=y2*x1)front++; else break; } double tmp = (sum[i] - sum[q[front]])*1.0/(i - q[front]); if (tmp > ans) ans = tmp; } printf("%.2lf\n",ans); } return 0; }
hdu 3507 Print Article
http://acm.hdu.edu.cn/showproblem.php?pid=3507
题意：
给定长度为n的整数序列，
对他进行划分，每一段的花费为上述公式:假设从i到j为一段划分则花费为 (∑Ck)^2 + m ， i<=k<=j。求一个划分使花费最小。
思路：
这里参考别人的思路，给出个人感觉讲解比较好的连接：
http://www.cnblogs.com/ka200812/archive/2012/08/03/2621345.html
代码参考：
http://blog.csdn.net/woshi250hua/article/details/7901433
View Code
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <queue> #include <stack> #include <set> #include <map> #include <string> #define CL(a,num) memset((a),(num),sizeof(a)) #define iabs(x) ((x) > 0 ? (x) : -(x)) #define Min(a,b) (a) > (b)? (b):(a) #define Max(a,b) (a) > (b)? (a):(b) #define ll long long #define inf 0x7f7f7f7f #define MOD 100000007 #define lc l,m,rt<<1 #define rc m + 1,r,rt<<1|1 #define pi acos(-1.0) #define test puts("<------------------->") #define maxn 100007 #define M 100007 #define N 500007 using namespace std; //freopen("din.txt","r",stdin); struct point { int x,y; }pot[N]; int sum[N],q[N]; ll dp[N]; int n,m; int cross(int x1,int y1,int x2,int y2) { return x1*y2 - x2*y1; } int det(point a,point b,point c) { return cross(b.x - a.x,b.y - a.y,c.x - a.x,c.y - a.y); } bool CheckIt(int x,int y,int z)//叉积判断点在线段的方向 { int mk = det(pot[x],pot[y],pot[z]); if (mk <= 0) return true; else return false; } bool NotBest(int a,int b,int k) { point p0 = pot[a], p1 = pot[b]; if (pot[a].y - k*pot[a].x >= pot[b].y - k*pot[b].x) return true; else return false; } int GetInt()//开挂输入 { char ch = getchar(); while (ch < '0' || ch > '9') ch = getchar(); int num = 0; while (ch >= '0' && ch <= '9') { num = num*10 + ch - '0'; ch = getchar(); } return num; } int main() { //freopen("din.txt","r",stdin); int i; while (~scanf("%d%d",&n,&m)) { sum[0] = 0; for (i = 1; i <= n; ++i) { //scanf("%d",&sum[i]); sum[i] = GetInt(); sum[i] += sum[i - 1];//累加和 } int front = 0,tail = 0; pot[0].x = pot[0].y = 0;//出事化原点 q[0] = 0; for (i = 1; i <= n; ++i) { pot[i].x = sum[i - 1]; pot[i].y = dp[i - 1] + sum[i - 1]*sum[i - 1];//记录每个状态的x,y; while (front + 1 <= tail && CheckIt(q[tail - 1],q[tail],i)) tail--;//删除不可能为最优的上凸点 q[++tail] = i; while (front + 1 <= tail && NotBest(q[front],q[front + 1],2*sum[i])) front++;//将在下凸折线不是最有的点删除 int k = q[front]; dp[i] = pot[k].y - 2*sum[i]*pot[k].x + sum[i]*sum[i] + m;//获取最优值 } printf("%I64d\n",dp[n]); } return 0; }
hdu 3408 Division
http://acm.hdu.edu.cn/showproblem.php?pid=3480
题意：
给你一个含有n个数的集合，让你划分成M个子集，每个集合里面的(MAX - MIN)^2为这个集合的花费，求总的花费最小。
思路：
首先我们要经整体排序
我们可以得到状态转移方程dp[i][j]表示以j结尾将序列划分成i块的最小花费则有dp[i][j] = min(dp[i - 1][k] + (a[j] - a[k + 1])*(a[j] - a[k + 1]));这样求的话复杂度O(n*m*m)肯定会超时。所以要优化，我们把方程拆开得到dp[i][j] = dp[i - 1][k] + a[j]^2 + a[k + 1]^ - 2*a[j]*a[k + 1];令y = dp[i - 1][k] + a[k + 1]^2 ; x = a[k + 1].
得到 dp[i][j] = y - 2*a[j]*x + a[j]^2;
则套用斜率优化即可。
View Code
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <queue> #include <stack> #include <set> #include <map> #include <string> #define CL(a,num) memset((a),(num),sizeof(a)) #define iabs(x) ((x) > 0 ? (x) : -(x)) #define Min(a,b) (a) > (b)? (b):(a) #define Max(a,b) (a) > (b)? (a):(b) #define ll long long #define inf 0x7f7f7f7f #define MOD 100000007 #define lc l,m,rt<<1 #define rc m + 1,r,rt<<1|1 #define pi acos(-1.0) #define test puts("<------------------->") #define maxn 100007 #define M 5007 #define N 10007 using namespace std; //freopen("din.txt","r",stdin); struct point { int x; int y; }pot[N]; int dp[M][N]; int n,m,a[N],q[N]; int cmp(int x,int y) { return x < y; } int cross(int x1,int y1,int x2,int y2) { return x1*y2 - x2*y1; } int det(point a,point b,point c) { return cross(b.x - a.x,b.y - a.y,c.x - a.x,c.y - a.y); } int GetInt() { char ch = getchar(); while (ch < '0' || ch > '9') ch = getchar(); int num = 0; while (ch >= '0' && ch <= '9') { num = num*10 + ch - '0'; ch = getchar(); } return num; } bool CheckIt(int x,int y,int z) { int mk = det(pot[x],pot[y],pot[z]); if (mk <= 0) return true; else return false; } bool NotBest(int a,int b,int k) { if (pot[a].y - k*pot[a].x <= pot[b].y - k*pot[b].x) return true; else return false; } int main() { //freopen("din.txt","r",stdin); int i,j,t,k; int cas = 1; scanf("%d",&t); while (t--) { scanf("%d%d",&n,&m); for (i = 1; i <= n; ++i) scanf("%d",&a[i]); sort(a + 1,a + 1 + n,cmp); for (i = 1; i <= n; ++i)//出事划分成一个集合的 dp[1][i] = (a[i] - a[1])*(a[i] - a[1]); pot[0].x = pot[0].y = 0; q[0] = 0; for (i = 2; i <= m; ++i)//枚举划分的大小 { int front = 0,tail = 0; for (j = i; j <= n; ++j)//枚举可能的结束点 { pot[j].x = a[j]; pot[j].y = dp[i - 1][j - 1] + a[j]*a[j]; while (front + 1 <= tail && CheckIt(q[tail - 1],q[tail],j)) tail--; q[++tail] = j; while (front + 1 <= tail && NotBest(q[front + 1],q[front],2*a[j])) front++; k = q[front]; dp[i][j] = pot[k].y - 2*a[j]*pot[k].x + a[j]*a[j]; } } printf("Case %d: %d\n",cas++,dp[m][n]); } return 0; }
hdu 4258 Covered Walkway
http://acm.hdu.edu.cn/showproblem.php?pid=4258
和上边的两道题目基本上一样，都是划分的题目。
题意：
给你x坐标上的n个点，让你让你两点连线将所有点覆盖，假设我们在a[i]到a[j]连线，则i-j的点都已经覆盖并且这一段的费用为(j - i)^2,求把所有点覆盖后的最小花费。
思路：
由题可得状态转移方程dp[i] = min(dp[j] + (a[i] - a[j + 1])^2 + c) (1<=j < i) 将方程展开后记得
dp[i] = dp[j] + a[j + 1]^2 + a[i]^2 - 2*a[i]*a[j + 1] + c; 令y = dp[j] + a[j + 1]^2 ; x = a[j + 1]; 则有dp[i] = y - 2*a[i]*x + a[i]^2 + c;
套用斜率优化即可。注意数据类型的处理，这里因为超了数据类型错了一次。
View Code
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <queue> #include <stack> #include <set> #include <map> #include <string> #define CL(a,num) memset((a),(num),sizeof(a)) #define iabs(x) ((x) > 0 ? (x) : -(x)) #define Min(a,b) (a) > (b)? (b):(a) #define Max(a,b) (a) > (b)? (a):(b) #define ll long long #define inf 0x7f7f7f7f #define MOD 100000007 #define lc l,m,rt<<1 #define rc m + 1,r,rt<<1|1 #define pi acos(-1.0) #define test puts("<------------------->") #define maxn 100007 #define M 5007 #define N 1000007 using namespace std; //freopen("din.txt","r",stdin); struct point { ll x,y; }pot[N]; ll dp[N]; ll a[N],c; int q[N]; int n; ll cross(ll x1,ll y1,ll x2,ll y2) { return x1*y2 - x2*y1; } ll det(point a,point b,point c) { return cross(b.x - a.x,b.y - a.y,c.x - a.x,c.y - a.y); } bool CheckIt(int x,int y,int z) { ll mk = det(pot[x],pot[y],pot[z]); if (mk <= 0) return true; else return false; } bool NotBest(int a,int b,ll k) { if (pot[a].y - k*pot[a].x <= pot[b].y - k*pot[b].x) return true; else return false; } ll GetInt() { char ch = getchar(); while (ch < '0' || ch > '9') ch = getchar(); ll num = 0; while (ch >= '0' && ch <= '9') { num = num*10 + ch - '0'; ch = getchar(); } return num; } int main() { //freopen("din.txt","r",stdin); int i; while (~scanf("%d%I64d",&n,&c)) { if (!n && !c) break; for (i = 1; i <= n; ++i) a[i] = GetInt(); //scanf("%I64d",&a[i]); int front = 0,tail = 0; pot[0].x = pot[0].y = 0; q[0] = 0; for (i = 1; i <= n; ++i) { pot[i].x = a[i]; pot[i].y = dp[i - 1] + a[i]*a[i]; while (front + 1 <= tail && CheckIt(q[tail - 1],q[tail],i)) tail--; q[++tail] = i; while (front + 1 <= tail && NotBest(q[front + 1],q[front],2*a[i])) front++; int k = q[front]; dp[i] = pot[k].y - 2*a[i]*pot[k].x + a[i]*a[i] + c; //if (i == 1) printf("%I64d %I64d %I64d %I64d %d\n",pot[i].x,pot[i].y,dp[1],a[i],c); } printf("%I64d\n",dp[n]); } return 0; }
hdu 2829 Lawrence
http://acm.hdu.edu.cn/showproblem.php?pid=2829
基本上和上几个题类似：
题意：
Lawrence要建一条铁路，铁路包括一些收费站（这里描述为一些点，对应着有自己的vale），还有连接两点的线，这条铁路的花费计算方法为：
4*5 + 4*1 + 4*2 + 5*1 + 5*2 + 1*2 = 49.
由于资金材料的限制只能毁掉一些线，使的耗费减小.给定n个点，以及要炸掉m个边求如何炸才能得到最小耗费，输出最小耗费。
思路：
首先我们得到状态转移方程dp[i][j]表示以i结尾炸掉j条路线的最小耗费。则有
dp[i][j] = min(dp[k][j - 1] + cost[k + 1][i]) (1<=k<i). cost[i][j]表示i到j这串数字的乘积和,sum[i]表示1到i为止这些数字的和
cost[1][i] = cost[1][k] + cost[k + 1][i] + sum[k]*(sum[i] - sum[k]);
==>cost[k+1][i]=cost[1][i]-cost[1][k]-sum[k]*(sum[i]-sum[k])
==>dp[i][j]=dp[k][j-1]+cost[1][i]-cost[1][k]-sum[k]*(sum[i]-sum[k])
==>dp[k][j-1]-cost[1][k]+sum[k]^2-sum[i]*sum[k]+cost[1][i]
==>由于sum[i]是与i有关的常量，所以很明显，斜率k=sum[i]，
设y=dp[k][j-1]-cost[1][k]+sum[k]^2
那么dp[i][j]=y-k*x+cost[1][i]
到这一步就可以套用斜率优化了。
View Code
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <queue> #include <stack> #include <set> #include <map> #include <string> #define CL(a,num) memset((a),(num),sizeof(a)) #define iabs(x) ((x) > 0 ? (x) : -(x)) #define Min(a,b) (a) > (b)? (b):(a) #define Max(a,b) (a) > (b)? (a):(b) #define ll long long #define inf 0x7f7f7f7f #define MOD 100000007 #define lc l,m,rt<<1 #define rc m + 1,r,rt<<1|1 #define pi acos(-1.0) #define test puts("<------------------->") #define maxn 100007 #define M 5007 #define N 1007 using namespace std; //freopen("din.txt","r",stdin); int n,m; int q[N]; struct point { ll x,y; }pot[N]; ll cost[N],sum[N],a[N],dp[N][N]; ll cross(ll x1,ll y1,ll x2,ll y2) { return x1*y2 - x2*y1; } ll det(point a,point b,point c) { return cross(b.x - a.x,b.y - a.y,c.x - a.x,c.y - a.y); } bool CheckIt(int x,int y,int z) { ll mk = det(pot[x],pot[y],pot[z]); if (mk <= 0) return true; else return false; } bool NotBest(int a,int b,ll k) { if (pot[a].y - k*pot[a].x <= pot[b].y - k*pot[b].x) return true; else return false; } ll solve_DP() { int i,j; int front,tail; pot[0].x = pot[0].y = 0; q[0] = 0; for (j = 1; j <= m; ++j) { front = tail = 0; for (i = j + 1; i <= n; ++i) { pot[i].x = sum[i - 1]; pot[i].y = dp[i - 1][j - 1] - cost[i - 1] + sum[i - 1]*sum[i -1]; while (front + 1 <= tail && CheckIt(q[tail - 1],q[tail],i)) tail--; q[++tail] = i; while (front + 1 <= tail && NotBest(q[front + 1],q[front],sum[i])) front++; int k = q[front]; dp[i][j] = pot[k].y - sum[i]*pot[k].x + cost[i]; } } return dp[n][m]; } ll GetInt() { char ch = getchar(); while (ch < '0' || ch > '9') ch = getchar(); ll num = 0; while (ch >= '0' && ch <= '9') { num = num*10 + ch - '0'; ch = getchar(); } return num; } int main() { //freopen("din.txt","r",stdin); int i; while (~scanf("%d%d",&n,&m)) { if (!n && !m) break; sum[0] = 0; for (i = 1; i <= n; ++i) { //scanf("%I64d",&a[i]); a[i] = GetInt(); sum[i] = sum[i - 1] + a[i]; } cost[0] = 0; for (i = 1; i <= n; ++i) { cost[i] = cost[i - 1] + sum[i - 1]*a[i]; dp[i][0] = cost[i];//初始化 } /* for (i = 1; i <= n; ++i) printf("%I64d ",cost[i]); puts("");*/ ll ans = solve_DP(); printf("%I64d\n",ans); } return 0; }
