标签:blog os io for div ar amp type
不懂这个建模是什么原理,以后把二分图相关的东西看完再补上把= =
#include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #include <climits> #include <string> #include <iostream> #include <map> #include <cstdlib> #include <list> #include <set> #include <queue> #include <stack> using namespace std; typedef long long LL; const int maxn = 105; const double INF = 1e60; const double eps = 1e-9; int level[maxn],n,m,c,q[maxn],qs,qe; double cap[maxn][maxn]; int s,t; bool bfs() { qs = qe = 0; memset(level,0,sizeof(level)); level[s] = 1; q[qe++] = s; while(qs < qe) { int now = q[qs++]; if(now == t) break; for(int i = s;i <= t;i++) if(cap[now][i] >= eps) { if(level[i] == 0) { q[qe++] = i; level[i] = level[now] + 1; } } } return level[t]; } double dfs(int now,double alpha) { if(now == t) return alpha; double sum = 0; for(int i = s;i <= t && alpha >= eps;i++) { if(cap[now][i] >= eps && level[i] == level[now] + 1) { double ret = dfs(i,min(alpha,cap[now][i])); sum += ret; alpha -= ret; cap[now][i] -= ret; cap[i][now] += ret; } } if(sum < eps) level[now] = -1; return sum; } void solve() { double ans = 0; while(bfs()) ans += dfs(s,INF); printf("%.4f\n",exp(ans)); } int main() { int T; scanf("%d",&T); while(T--) { memset(cap,0,sizeof(cap)); scanf("%d%d%d",&n,&m,&c); s = 0; t = n + m + 1; for(int i = 1;i <= n;i++) { double tmp; scanf("%lf",&tmp); cap[s][i] = log(tmp); } for(int i = 1;i <= m;i++) { double tmp; scanf("%lf",&tmp); cap[i + n][t] = log(tmp); } for(int i = 1;i <= c;i++) { int a,b; scanf("%d%d",&a,&b); cap[a][b + n] = INF; } solve(); } return 0; }
POJ 3308 Paratroopers 最小点权覆盖 求最小割,布布扣,bubuko.com
POJ 3308 Paratroopers 最小点权覆盖 求最小割
标签:blog os io for div ar amp type
原文地址:http://www.cnblogs.com/rolight/p/3872739.html