hdu 5988 Coding Contest 费用流

题目：

http://acm.split.hdu.edu.cn/showproblem.php?pid=5988

题意：

有n个区域，有m条有向边连接它们，每条边都有一个被破环的几率，但第一个人通过不会造成任何影响，之后的人通过才会有影响。现在每个区域内有一定的队员和背包，要求每个队员都拿到一个背包，且使道路奔溃的几率最小，求这个最小几率

思路：

被破坏的几率最小就是不被破坏的几率最大。可以用费用流去做，因为是乘法，那么先取log再去跑费用流，再对结果去exp即可

#include <bits/stdc++.h>using namespace std;const int N = 100 + 10, INF = 0x3f3f3f3f;const double eps = 1e-8;struct edge{    int to, cap, next;    double cost;} g[N*N*10];int cnt, head[N];int pre[N];double dis[N];bool vis[N];int nv;void init(){    cnt = 0;    memset(head, -1, sizeof head);}void add_edge(int v, int u, int cap, double cost){    g[cnt].to = u, g[cnt].cap = cap, g[cnt].cost = cost, g[cnt].next = head[v], head[v] = cnt++;    g[cnt].to = v, g[cnt].cap = 0, g[cnt].cost = -cost, g[cnt].next = head[u], head[u] = cnt++;}bool spfa(int s, int t){    for(int i = 0; i < nv; i++) dis[i] = INF;    memset(vis, 0, sizeof vis);    memset(pre, -1, sizeof pre);    queue<int> que;    que.push(s);    dis[s] = 0.0, vis[s] = true;    while(! que.empty())    {        int v = que.front(); que.pop();        vis[v] = false;        for(int i = head[v]; i != -1; i = g[i].next)        {            int u = g[i].to;            if(g[i].cap > 0 && dis[u] - dis[v] - g[i].cost > eps)//这里不取eps会tle            {                dis[u] = dis[v] + g[i].cost;                pre[u] = i;                if(! vis[u]) que.push(u), vis[u] = true;            }        }    }    return !(abs(dis[t] - INF) < eps);}double cost_flow(int s, int t, int flow){    double ans = 0;    while(flow > 0)    {        if(! spfa(s, t)) return -1;        int d = INF;        for(int i = pre[t]; i != -1; i = pre[g[i^1].to])            d = min(d, g[i].cap);        for(int i = pre[t]; i != -1; i = pre[g[i^1].to])            g[i].cap -= d, g[i^1].cap += d;        ans += d * dis[t];        flow -= d;    }    return ans;}int main(){    int t, n, m;    scanf("%d", &t);    while(t--)    {        scanf("%d%d", &n, &m);        init();        int ss = 0, tt = n + 1;        int a, b, c;        double p;        int sum = 0;        for(int i = 1; i <= n; i++)        {            scanf("%d%d", &a, &b);            if(a - b > 0) add_edge(ss, i, a-b, 0), sum += a-b;            if(a - b < 0) add_edge(i, tt, -(a-b), 0);        }        for(int i = 1; i <= m; i++)        {            scanf("%d%d%d%lf", &a, &b, &c, &p);            if(c >= 1) add_edge(a, b, 1, 0);            if(c > 1) add_edge(a, b, c-1, -log(1-p));        }        nv = tt + 1;        double ans = cost_flow(ss, tt, sum);        printf("%.2f\n", 1.0 - exp(-ans));    }    return 0;}