链接：
http://acm.split.hdu.edu.cn/showproblem.php?pid=5988
题意：
n个点，每个点有a个人和b包饭，m条边，第一次经过这条边没有问题，以后每一个经过的有p概率出问题
每条边有限制，问使得所有人都有饭吃且出问题的概率最小
题解：
无向图的费用流
把相乘的取下对数变成相加的就可以用最小费用流了
代码：
30 #include <iomanip> 31 const double eps = 1e-8; 32 const int MAXM = 1e5 + 7; 33 34 struct node{ 35 int to, next, cap; 36 double cost; 37 }edge[MAXM]; 38 int head[MAXN]; 39 int pre[MAXN]; 40 int vis[MAXN]; 41 double dist[MAXN]; 42 int n, m, tot; 43 44 void init() { 45 memset(head, -1, sizeof(head)); 46 tot = 0; 47 } 48 49 void add_edge(int u, int v, int cap, double cost) { 50 edge[tot].to = v; 51 edge[tot].cap = cap; 52 edge[tot].cost = cost; 53 edge[tot].next = head[u]; 54 head[u] = tot++; 55 edge[tot].to = u; 56 edge[tot].cap = 0; 57 edge[tot].cost = -cost; 58 edge[tot].next = head[v]; 59 head[v] = tot++; 60 } 61 62 double mcf(int s, int t, int f) { 63 double res = 0; 64 while (f > 0) { 65 rep(i, 0, n + 2) { 66 dist[i] = 1e18; 67 vis[i] = 0; 68 pre[i] = -1; 69 } 70 queue<int> q; 71 q.push(s); 72 dist[s] = 0; 73 vis[s] = 1; 74 while (!q.empty()) { 75 int u = q.front(); q.pop(); 76 vis[u] = 0; 77 for (int i = head[u]; i != -1; i = edge[i].next) { 78 int v = edge[i].to; 79 if (edge[i].cap > 0 && 80 dist[v] > dist[u] + edge[i].cost + eps){ 81 dist[v] = dist[u] + edge[i].cost; 82 pre[v] = i; 83 if (!vis[v]) { 84 vis[v] = 1; 85 q.push(v); 86 } 87 } 88 } 89 } 90 if (dist[t] == 1e18) return -1; 91 int d = f; 92 for (int v = pre[t]; v != -1; v = pre[edge[v ^ 1].to]) 93 d = min(d, edge[v].cap); 94 f -= d; 95 res += d * dist[t]; 96 for (int v = pre[t]; v != -1; v = pre[edge[v ^ 1].to]) { 97 edge[v].cap -= d; 98 edge[v ^ 1].cap += d; 99 } 100 } 101 return res; 102 } 103 104 int main() { 105 int T; 106 scanf("%d", &T); 107 while (T--) { 108 init(); 109 cin >> n >> m; 110 int s = 0, t = n + 1, f = 0; 111 rep(i, 1, n + 1) { 112 int a, b; 113 scanf("%d%d", &a, &b); 114 add_edge(s, i, a, 0); 115 add_edge(i, t, b, 0); 116 f += a; 117 } 118 while (m--) { 119 int u, v, c; 120 double p; 121 scanf("%d%d%d%lf", &u, &v, &c, &p); 122 p = -log2(1 - p); 123 if (c > 0) add_edge(u, v, 1, 0); 124 if (c > 1) add_edge(u, v, c - 1, p); 125 } 126 double p = mcf(s, t, f); 127 printf("%.2f ", 1 - pow(2, -p)); 128 } 129 return 0; 130 }