HDU - 5988 Coding Contest （最小费用最大流）

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.

Input

The 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.

Output

For each turn of each case, output the minimum possibility that the networks would break down. Round it to 2 digits.

Sample Input

1
4 4
2 0
0 3
3 0
0 3
1 2 5 0.5
3 2 5 0.5
1 4 5 0.5
3 4 5 0.5

Sample Output

0.50

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5988

题目大意：t组输入，操场被分成n块，有m条路，每一块开始有ai个参赛人员，有bi份午饭， 每条路从ui到vi 最多能通过ci个人，每个人通过导致道路网络瘫痪的概率是pi，其中第一个通过这条路的人不会导致道路瘫痪。问每个人都吃到午饭道路瘫痪的最小概率。

把概率取log，把乘法变成加法，然后就是最小费用最大流模板

代码：

#include<bits/stdc++.h>
using namespace std;
#define mem(a,b) memset(a,b,sizeof(a))
const int N=105;
const int M=50005;
const double eps=1e-7;
const int inf=0x3f3f3f3f;
double dis[N];
int pre[N],vis[N];
int maxflow,first[N],tot;
struct node
{
    int v,nex,cap,flow;
    double cost;
    node(){};
    node(int _v,int _c,int _nex,double _cost)
    {
        v=_v,cap=_c,nex=_nex,cost=_cost,flow=0;
    }
}e[M];
void init()
{
    mem(first,-1);
    tot=maxflow=0;
}
void add(int u,int v,int c,double cost)
{
    e[tot]=node(v,c,first[u],cost);
    first[u]=tot++;
    e[tot]=node(u,0,first[v],-cost);
    first[v]=tot++;
}
double spfa(int s,int t,int n)
{
    int i,u,v;
    queue<int>q;
    for(int i=1; i<=n; i++) vis[i]=0,dis[i]=inf,pre[i]=-1;
    vis[s]=1;
    dis[s]=0;
    q.push(s);
    while(!q.empty())
    {
        u=q.front();
        q.pop();
        vis[u]=0;
        for(int i=first[u]; i!=-1; i=e[i].nex)
        {
            v=e[i].v;
            if(e[i].cap>e[i].flow&&dis[v]>dis[u]+e[i].cost+eps)
            {
                dis[v]=dis[u]+e[i].cost;
                pre[v]=i;
                if(!vis[v])
                {
                    q.push(v);
                    vis[v]=1;
                }
            }
        }
    }
    if(dis[t]==inf)
        return 0;
    return 1;
}
double MCMF(int s,int t,int n)
{
    int d;
    double mincost=0;
    while(spfa(s,t,n))
    {
        d=inf;
        for(int i=pre[t]; i!=-1; i=pre[e[i^1].v])
            d=min(d,e[i].cap-e[i].flow);
        maxflow+=d;
        for(int i=pre[t]; i!=-1; i=pre[e[i^1].v])
        {
            e[i].flow+=d;
            e[i^1].flow-=d;
        }
        mincost+=dis[t]*d;
    }
    return mincost;
}

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        init();
        int n,m,x,y,w;
        double z;
        scanf("%d%d",&n,&m);
        int st=n+1,ed=n+2;
        for(int i=1;i<=n;i++)
        {
            scanf("%d%d",&x,&y);
            add(st,i,x,0);
            add(i,ed,y,0);
        }
        for(int i=0;i<m;i++)
        {
            scanf("%d%d%d%lf",&x,&y,&w,&z);
            z=-log2(1-z); 
            if(w) add(x,y,1,0);
            if(w>1) add(x,y,w-1,z);
        }
        double mincost=MCMF(st,ed,n+2);
        mincost=pow(2,-mincost);
        printf("%.2f\n",1-mincost);
    }
    return 0;
}