hdu5988（网络流）

链接

题意：一个网络流的图，有n个点，从1~n，然后m条边，每个点有两个值，一个是人的数量一个是饭的数量。每条边有容量，还有走上去可能踩断电线的几率。问让所有人吃到饭的前提下断电线的最小概率是多少。
题解：
踩断电线的最小概率就是1-没踩断电线的最大概率。那就可以把一条路上的所有可能性乘起来就好，最小费用流没法处理乘法，就要取对数。要求最大概率就可以取相反数求费用流。求最短路的时候松弛要用eps。
（我的代码是可以卡时间过的。。。具体可以过得关键在111行）

#include<bits/stdc++.h>
using namespace std;
#define MAXN 110
#define MAXM 25000
#define INF 2e9
const double eps = 1e-8;
struct Edge
{
    int to, next, cap, flow;
    double cost;
}edge[MAXM];
int head[MAXN], tol;
int pre[MAXN];
double dis[MAXN];
bool vis[MAXN];
int N;//节点总个数，节点编号从0~N-1
void init(int n)
{
    N = n;
    tol = 0;
    memset(head, -1, sizeof(head));
}
void addedge(int u, int v, int cap, double cost)
{
    edge[tol].to = v;
    edge[tol].cap = cap;
    edge[tol].cost = cost;
    edge[tol].flow = 0;
    edge[tol].next = head[u];
    head[u] = tol++;
    edge[tol].to = u;
    edge[tol].cap = 0;
    edge[tol].cost = -cost;
    edge[tol].flow = 0;
    edge[tol].next = head[v];
    head[v] = tol++;
}
bool spfa(int s, int t)
{
    queue<int>q;
    for (int i = 0; i <= N; i++)
    {
        dis[i] = INF;
        vis[i] = false;
        pre[i] = -1;
    }
    dis[s] = 0;
    vis[s] = true;
    q.push(s);
    while (!q.empty())
    {
        //cout<<1<<endl;
        int u = q.front();
        q.pop();
        vis[u] = false;
        for (int i = head[u]; i != -1; i = edge[i].next)
        {
            int v = edge[i].to;
            if (edge[i].cap > edge[i].flow &&
                dis[v]-dis[u]-edge[i].cost>eps)
            {
                dis[v] = dis[u] + edge[i].cost;
                pre[v] = i;
                if (!vis[v])
                {
                    vis[v] = true;
                    q.push(v);
                }
            }
        }
    }
    if (pre[t] == -1)return false;
    else return true;
}
//返回的是最大流，cost存的是最小费用
int minCostMaxflow(int s, int t, double &cost)
{
    int flow = 0;
    cost = 0;
    while (spfa(s, t))
    {
        //cout<<1<<endl;
        int Min = INF;
        for (int i = pre[t]; i != -1; i = pre[edge[i ^ 1].to])
        {
            if (Min > edge[i].cap - edge[i].flow)
                Min = edge[i].cap - edge[i].flow;
        }
        for (int i = pre[t]; i != -1; i = pre[edge[i ^ 1].to])
        {
            edge[i].flow += Min;
            edge[i ^ 1].flow -= Min;
            cost += edge[i].cost * Min;
        }
        flow += Min;
    }
    return flow;
}
int main()
{
    int t,n,m,a,b;
    scanf("%d",&t);

    while (t--)
    {
        scanf("%d %d",&n,&m);
        init(n + 1);
        for (int i = 1; i <= n; i++)
        {
            scanf("%d %d",&a,&b);
            int temp = a - b;//如果不用temp而用a-b做，会超时。这样会快20ms
            if(temp > 0)
            {
                addedge(0, i, temp, 0);
            }
            else if(temp < 0)
            {
                addedge(i, n + 1, -temp, 0);
            }
        }
        int u,v,f;
        double w;
        for (int i = 0; i < m; i++)
        {
            scanf("%d %d %d %lf",&u,&v,&f,&w);
            w = -log2(1 - w);
            if(f > 0)
                addedge(u,v,1,0);
            if(f - 1 > 0)
                addedge(u,v,f - 1, w);
        }
        double cost = 0;
        minCostMaxflow(0,n + 1,cost);
        cost = 1 - pow(2,-cost);
        printf("%.2f\n",cost);
    }
}