【HDU 5988】 Coding Contest 【费用流】

题意：给你一张图n个点m条边，每个点有s个人，但只能容纳b个人，然后如果第一个经过一条边，不会有任何事情发生，下面经过这条边的人都有p的概率损坏这条边，问你使每个点的人小于等于容量的最小损坏概率是多少

题解：

最小损坏概率=最大不损坏概率

也就是说(1-p1)*(1-p2)..

然后乘法可以用取对数的形式换成加法，就可以用费用流写了

1.原点向每个i连一条容量为si,代价为0的边

2.每个i向汇点连一条容量为bi,代价为0的边

3.对于每条边如果c>=1,连一条容量为1,代价为0的边,再连一条容量为c-1,代价为log(1-p)的边

#include<cmath>
#include<stack>
#include<queue>
#include<bitset>
#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define PB push_back
#define MP make_pair
#define ll long long
#define MS(a,b) memset(a,b,sizeof(a))
#define LL (rt<<1)
#define RR (rt<<1|1)
#define lson l,mid,LL
#define rson mid+1,r,RR
#define pii pair<int,int>
#define pll pair<ll,ll>
#define lb(x) (x&(-x))
void In(){freopen("in.in","r",stdin);}
void Out(){freopen("out.out","w",stdout);}
const int N=1e2+10;
const int M=3e4+10;
const int Mbit=1e6+10;
const int inf=0x3f3f3f3f;
const ll mod=1e9+7;
const double eps=1e-8;
struct Edge
{
    int u,v,cap,next;
    double cost;
}edge[M<<2];
int NE;
int head[N],pp[N],Q[N];
double dist[N];
bool vis[N];
void init()
{
    NE=0;
    memset(head,-1,sizeof(head));
}
void add(int u,int v,int cap,double cost)
{
    edge[NE].u=u;edge[NE].v=v;edge[NE].cap=cap;edge[NE].cost=cost;
    edge[NE].next=head[u];head[u]=NE++;

    edge[NE].u=v;edge[NE].v=u;edge[NE].cap=0;edge[NE].cost=-cost;
    edge[NE].next=head[v];head[v]=NE++;
}
bool SPFA(int s,int t,int n)
{
    int i,u,v,l=0,r=1;
    memset(vis,0,sizeof(vis));
    memset(pp,-1,sizeof(pp));
    for(i=0;i<n;i++) dist[i]=-inf;
    vis[s]=1;dist[s]=0;
    Q[0]=s;
    while(l!=r)
    {
        u=Q[l++];vis[u]=0;
        if(l==N)l=0;
        for(i=head[u];i!=-1;i=edge[i].next)
        {
            v=edge[i].v;
            if(edge[i].cap&&dist[v]<dist[u]+edge[i].cost-eps)
            {
                dist[v]=dist[u]+edge[i].cost;
                pp[v]=i;
                if(!vis[v])
                {
                    Q[r++]=v;
                    vis[v]=1;
                    if(r==N)r=0;
                }
            }
        }
    }
    return dist[t]>-inf;
}
pair<int,double> MCMF(int s,int t,int n)
{
    int flow=0,i,minflow;
    double mincost=0;
    while(SPFA(s,t,n))
    {
        minflow=inf;
        for(i=pp[t];i!=-1;i=pp[edge[i].u])
            minflow=min(edge[i].cap,minflow);
        flow+=minflow;
        for(i=pp[t];i!=-1;i=pp[edge[i].u])
            edge[i].cap-=minflow,edge[i^1].cap+=minflow;
        mincost+=dist[t]*minflow;
    }
    return make_pair(flow,mincost);
}
int s[N],b[N];
int main()
{
    //In();
    int T,n,m,u,v,c;
    double p;
    scanf("%d",&T);
    while(T--){
        scanf("%d%d",&n,&m);
        init();
        int vs=0,vt=n+1,NV=vt+1;
        for(int i=1;i<=n;i++){
            scanf("%d%d",&s[i],&b[i]);
            add(vs,i,s[i],0);
            add(i,vt,b[i],0);
        }
        for(int i=1;i<=m;i++){
            scanf("%d%d%d%lf",&u,&v,&c,&p);
            if(c){
                add(u,v,1,0);
                if(c-1)add(u,v,c-1,log(1-p));
            }
        }
        printf("%.2f\n",1-pow(10,MCMF(vs,vt,NV).second/log(10)));
    }
    return 0;
}