SRM 578 DIV 2
250:
简单题目：
500：
题意：
给定一个矩形，里面要么是"v"表示，要么是"."，v表示可能是g,也可能是d，如果是g的话，那么它的哈弗曼距离dis之内如果是v的话，一定是g。求有多少种满足条件的可能数。
思路：
将每一个块分出来，自这一联通块里面，所有的v要么是g，要么d,bfs把所有的快求出来，假设为n，则最后的总数为2^n - 1
View Code
#line 5 "GooseInZooDivTwo.cpp" #include <iostream> #include <cstdio> #include <cmath> #include <vector> #include <cstring> #include <algorithm> #include <string> #include <set> #include <functional> #include <numeric> #include <sstream> #include <stack> #include <map> #include <queue> #define CL(arr, val) memset(arr, val, sizeof(arr)) #define lc l,m,rt<<1 #define rc m + 1,r,rt<<1|1 #define pi acos(-1.0) #define ll long long #define L(x) (x) << 1 #define R(x) (x) << 1 | 1 #define MID(l, r) (l + r) >> 1 #define Min(x, y) (x) < (y) ? (x) : (y) #define Max(x, y) (x) < (y) ? (y) : (x) #define E(x) (1 << (x)) #define iabs(x) (x) < 0 ? -(x) : (x) #define OUT(x) printf("%I64d\n", x) #define lowbit(x) (x)&(-x) #define Read() freopen("din.txt", "r", stdin) #define Write() freopen("dout.txt", "w", stdout); #define tr(container, it) for(typeof(container.begin()) it = container.begin(); it != container.end(); it++) #define M 107 #define N 107 using namespace std; const ll mod = 1000000007; struct node { int x,y; }nd; int n,m; vector<string> tmp; int Abs(int x) { if (x >= 0) return x; else return (-x); } void bfs(int x,int y,int k) { queue<node>q; nd.x = x; nd.y = y; q.push(nd); int i,j; while (!q.empty()) { node u = q.front(); q.pop(); for (i = max(0,u.x - k); i <= min(n - 1, u.x + k); ++i) { for (j = max(0,u.y - k); j <= min(m - 1,u.y + k); ++j) { if (Abs(u.x - i) + Abs(u.y - j) <= k && tmp[i][j] == 'v') { tmp[i][j] = 'g'; nd.x = i,nd.y = j; q.push(nd); } } } } } class GooseInZooDivTwo { public: int count(vector <string> fd, int dis) { int i,j; n = fd.size(); m = fd[0].size(); tmp.clear(); for (i = 0; i < n; ++i) { tmp.push_back(fd[i]); } n = fd.size(); m = fd[0].size(); ll cnt = 0; for (i = 0; i < n; ++i) { for (j = 0; j < m; ++j) { if (tmp[i][j] == 'v') { cnt++; tmp[i][j] = 'g'; bfs(i,j,dis); } } } // cout<<cnt<<endl; ll ans = 1; for (i = 0; i < cnt; ++i) { ans = ans*2%mod; } return ans - 1; } }; // Powered by FileEdit // Powered by TZTester 1.01 [25-Feb-2003] // Powered by CodeProcessor // Powered by FileEdit // Powered by TZTester 1.01 [25-Feb-2003] // Powered by CodeProcessor // Powered by FileEdit
1000:
题意：
个顶n个段，每个段里面要么有狼，要么没狼，然后给出m个区域间，[l[i],r[i]]表示在区间内至少有一只狼，问满足条件的一共有多少种可能数？
思路：
记忆化搜索DP , dp[i][j]表示到i个段，有j个区间还没有处理， 对于第i个段我们有两种选择要么方狼，要么不放狼。 如果方狼的话，dp[i][j] = dp[i - 1][k] k表示必须多少个区间被处理了。
如果不放的话，要必须满足区间条件。 dp[i][j] += dp[i - 1][j]
View Code
#line 5 "WolfInZooDivTwo.cpp" #include <iostream> #include <cstdio> #include <cmath> #include <vector> #include <cstring> #include <algorithm> #include <string> #include <set> #include <functional> #include <numeric> #include <sstream> #include <stack> #include <map> #include <queue> #define CL(arr, val) memset(arr, val, sizeof(arr)) #define lc l,m,rt<<1 #define rc m + 1,r,rt<<1|1 #define pi acos(-1.0) #define ll long long #define L(x) (x) << 1 #define R(x) (x) << 1 | 1 #define MID(l, r) (l + r) >> 1 #define Min(x, y) (x) < (y) ? (x) : (y) #define Max(x, y) (x) < (y) ? (y) : (x) #define E(x) (1 << (x)) #define iabs(x) (x) < 0 ? -(x) : (x) #define OUT(x) printf("%I64d\n", x) #define lowbit(x) (x)&(-x) #define Read() freopen("din.txt", "r", stdin) #define Write() freopen("dout.txt", "w", stdout); #define tr(container, it) for(typeof(container.begin()) it = container.begin(); it != container.end(); it++) #define M 107 #define N 307 using namespace std; const ll mod = 1000000007; struct node { int l,r; }nd[N]; ll pow2[N]; ll dp[N][N]; int off[N],cnt; ll DP(int x,int y) { int i; if (x < 0) return 1; if (dp[x][y] != -1) return dp[x][y]; if (y == 0) return (dp[x][y] = pow2[x + 1]); dp[x][y] = 0; for (i = y; i >= 1; --i) { if (x > nd[i].r) break; } dp[x][y] += DP(x - 1,i); dp[x][y] %= mod; for (i = y; i >= 1; --i) { if (x == nd[i].l) break; } if (i == 0) dp[x][y] += DP(x - 1, y); dp[x][y] %= mod; return dp[x][y]; } vector<int> getR(string s) { istringstream is(s); int no; vector<int> rs; while (is >> no) { rs.push_back(no); } return rs; } class WolfInZooDivTwo { public: void init() { pow2[0] = 1; for (int i = 1; i <= 300; ++i) { pow2[i] = pow2[i - 1]*2%mod; } } int count(int n, vector <string> L, vector <string> R) { int i; init(); string lstr = "",rstr = ""; for (size_t i = 0; i < L.size(); ++i) lstr += L[i]; for (size_t i = 0; i < R.size(); ++i) rstr += R[i]; vector<int> Ls = getR(lstr); vector<int> Rs = getR(rstr); if (Ls.size() != Rs.size()) puts("BUBIUBYUGB"); set<int> sr(Rs.begin(),Rs.end()); CL(off,-1); int m = Ls.size(); for (i = 0; i < m; ++i) off[Rs[i]] = max(off[Rs[i]],Ls[i]); cnt = 0; set<int>::iterator it = sr.begin(); for (; it != sr.end(); ++it) { nd[++cnt].l = off[*it]; nd[cnt].r = *it; } CL(dp,-1); return DP(n - 1,cnt); } };
