驚了,TLE半天發現浮點運算是要加esp,真是學到了。

題目鏈接https://www.nowcoder.com/acm/contest/207/G

題意就是給你N個點，每個點有si個人和bi個食物背包,然後是M條邊,每條邊有個觸電概率pi,第一個人經過這條邊不會觸電,問你讓所有人吃完飯的最小觸電概率.

方法:如果si>bi 連接0到i 容量為si-bi,費用為0,否則就連接 i到n+1,容量為bi-si,費用為0

然後把每線路拆成兩條,一條容量為1,費用為0,一條容量為c-1,費用為-log(1-pi)

因為最後要輸出所有概率的乘積的最小值,取對數後變成負數,加上負號後最小變最大,所以再把概率取反變成(1-pi)

跑一邊最短路輸出1-exp(-ans)即可

#include<bits/stdc++.h>
#define esp 1e-7
using namespace std;
const int maxn=205;
const int INF=0x3fffffff;
struct Edge
{
    int from,to,cap,flow;
    double cost;
};
struct MCMF
{
    int n,m,s,t;
    vector<Edge>edges;
    vector<int> G[maxn];
    
 
int inq[maxn];
    double d[maxn];
    int p[maxn];
    int a[maxn];
    void init(int n)
    {
        this->n=n;
        for(int i=0; i<=n; i++)G[i].clear();
        edges.clear();
    }
    void AddEdge(int from,int to,int cap,double cost)
    {
        Edge e;
        e.from=from
 
,e.to=to,e.cap=cap,e.flow=0,e.cost=cost;
        edges.push_back(e);
        e.from=to,e.to=from,e.cap=0,e.flow=0,e.cost=-cost;
        edges.push_back(e);
        m=edges.size();
        G[from].push_back(m-2);
        G[to].push_back(m-1);
    }
    bool BellmanFord(int s,int t,int& flow,double& cost)
    {
        for(int i=0; i<=n; i++)d[i]=INF;
        memset(inq,0,sizeof(inq));
        d[s]=0;
        inq[s]=1;
        p[s]=0;
        a[s]=INF;

        queue<int>Q;
        Q.push(s);
        while(!Q.empty())
        {
            int u=Q.front();
            Q.pop();
            inq[u]=0;
            for(int i=0; i<(int)G[u].size(); i++)
            {
                Edge& e=edges[G[u][i]];
                if(e.cap>e.flow&&d[e.to]-d[u]-e.cost>esp)
                {
                    d[e.to]=d[u]+e.cost;
                    p[e.to]=G[u][i];
                    a[e.to]=min(a[u],e.cap-e.flow);
                    if(!inq[e.to])
                    {
                        Q.push(e.to);
                        inq[e.to]=1;
                    }
                }
            }
        }
        if(d[t]==INF)return false;
        flow+=a[t];
        cost+=d[t]*a[t];
        int u=t;
        while(u!=s)
        {
            edges[p[u]].flow+=a[t];
            edges[p[u]^1].flow-=a[t];
            u=edges[p[u]].from;
        }
        return true;
    }
    double Mincost(int s,int t)
    {
        int flow=0;
        double cost=0;
        while(BellmanFord(s,t,flow,cost));
        return cost;
    }
} mcmf;

int main()
{
    int t,n,m,s,b,u,v,c;
    double pi;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        mcmf.init(n+2);
        for (int i=1; i<=n ; i++ )
        {
            scanf("%d%d",&s,&b);
            if(s>b)
            {
                mcmf.AddEdge(0,i,s-b,0);
            }
            else if(b>s)
            {
                mcmf.AddEdge(i,n+1,b-s,0);
            }
        }
        for (int i=1; i<=m ; i++ )
        {
            scanf("%d%d%d%lf",&u,&v,&c,&pi);
            if(c>0)mcmf.AddEdge(u,v,1,0);
            if(c-1>0)
            {
                mcmf.AddEdge(u,v,c-1,-log(1.0-pi));
            }
        }
        printf("%.2lf\n",1.0-exp(-mcmf.Mincost(0,n+1)));
    }
    return 0;
}

hdu5988最小費用最大流浮點