GroupBlog

G - Coding Contest

A coding contest will be held in this university, in a huge playground. The whole playground would be divided into N blocks, and there would be M directed paths linking these blocks. The i-th path goes from the uiui-th block to the vivi-th block. Your task is to solve the lunch issue. According to the arrangement, there are sisi competitors in the i-th block. Limited to the size of table, bibi bags of lunch including breads, sausages and milk would be put in the i-th block. As a result, some competitors need to move to another block to access lunch. However, the playground is temporary, as a result there would be so many wires on the path.
For the i-th path, the wires have been stabilized at first and the first competitor who walker through it would not break the wires. Since then, however, when a person go through the i - th path, there is a chance of pipi to touch
the wires and affect the whole networks. Moreover, to protect these wires, no more than cici competitors are allowed to walk through the i-th path.
Now you need to find a way for all competitors to get their lunch, and minimize the possibility of network crashing.

InputThe first line of input contains an integer t which is the number of test cases. Then t test cases follow.
For each test case, the first line consists of two integers N (N ≤ 100) and M (M ≤ 5000). Each of the next N lines contains two integers si and bibi ( sisi , bibi ≤ 200).
Each of the next M lines contains three integers uiui , vivi and ci(cici(ci ≤ 100) and a float-point number pipi(0 < pipi < 1).
It is guaranteed that there is at least one way to let every competitor has lunch. OutputFor each turn of each case, output the minimum possibility that the networks would break down. Round it to 2 digits. Sample Input

Sample Output

0.50

此题一眼网络流,不过不同的是,其是求乘积最大.

由于其是求最大值,且最大值不为0,可以求其倒数最小

这样就变成了最小费用最大流

把加减的成分变为乘除负号用倒数代替

就OK了

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MAXM=100000;
const int INF=1e9;
const double eps = 1e-8;
struct Edge
{
    int to,next,cap,flow;
    double cost;
    Edge(int _to=0,int _next=0,int _cap=0,int _flow=0,double _cost=1.0) :
        to(_to),next(_next),cap(_cap),flow(_flow),cost(_cost) {}
}edge[MAXM];


struct MinCostMaxFlow
{
    int head[MAXM],tot;
    int cur[MAXM];
    double dis[MAXM];
    bool vis[MAXM];
    int ss,tt,N;
    int max_flow;
    double min_cost;

    void init()
    {
        tot=0;
        memset(head,-1,sizeof(head));
    }
    void addedge(int u,int v,int cap,double cost)
    {
//        cout << cost << 1.0/cost <<endl;
        edge[tot]=Edge(v,head[u],cap,0,cost);
        head[u]=tot++;
        edge[tot]=Edge(u,head[v],0,0,1.0/cost);
        head[v]=tot++;
    }
    int aug(int u,int flow)
    {
        if(u==tt) return flow;
        vis[u]=true;
        for(int i=cur[u];i!=-1;i=edge[i].next)
        {
            int v=edge[i].to;
            if(edge[i].cap>edge[i].flow&&!vis[v]&&fabs(dis[u]-dis[v]*edge[i].cost)<eps)
            {
                int tmp=aug(v,min(flow,edge[i].cap-edge[i].flow));
                edge[i].flow+=tmp;
                edge[i^1].flow-=tmp;
                cur[u]=i;
                if(tmp) return tmp;
            }
        }
        return 0;
    }

    bool modify_label()
    {
        double d=1e9;
        for(int u=0;u<N;u++)
            if(vis[u])
            for(int i=head[u];i!=-1;i=edge[i].next)
            {
                int v=edge[i].to;
                if(edge[i].cap>edge[i].flow&&!vis[v])
                {
                    d=min(d,dis[v]*edge[i].cost/dis[u]);

                }

            }
        if(fabs(d-1e9)<eps) return false;
        for(int i=0;i<N;i++)
            if(vis[i])
            {
                vis[i]=false;
                dis[i]*=d;
            }
        return true;
    }

    double mincostmaxflow(int start,int end,int n)
    {
        ss=start,tt=end,N=n;
        max_flow = 0;
        min_cost = 1.0;
        for(int i=0;i<=n;i++) dis[i]=1.0;
        while(1)
        {
            for(int i=0;i<n;i++) cur[i]=head[i];
            while(1)
            {
                for(int i=0;i<n;i++) vis[i]=false;
                int tmp=aug(ss,INF);
                if(tmp==0) break;
                max_flow+=tmp;

                for(int ii=0;ii<tmp;ii++)
                {
                    min_cost *=dis[ss];
                }

            }
            if(!modify_label()) break;
        }
        return min_cost;
    }
}sol;


int n,m;
int T;

int main()
{
//    freopen("data.txt","r",stdin);
//    ios_base::sync_with_stdio(false);
    int u,v,c;
    double p;
    int s,b;
    int ss,tt;

    scanf("%d",&T);
    while(T--)
    {
        sol.init();
        scanf("%d%d",&n,&m);
        ss = 0 ;
        tt = n+1;
        for(int i=1;i<=n;i++)
        {
            scanf("%d%d",&s,&b);
            if(s>b)
            {
                sol.addedge(ss,i,s-b,1.0);
            }
            else if(b>s)
            {
                sol.addedge(i,tt,b-s,1.0);
            }
        }
        for(int i=0;i<m;i++)
        {
            scanf("%d%d%d%lf",&u,&v,&c,&p);
            if(c==0) continue;
            sol.addedge(u,v,1,1.0);
            if(c==1) continue;
            sol.addedge(u,v,c-1,1.0/(1-p));
        }
        printf("%.2f\n",1.0-1/sol.mincostmaxflow(ss,tt,n+3));
    }

    return 0;
}