poj3308 Paratroopers

 

Description

It is year 2500 A.D. and there is a terrible war between the forces of the Earth and the Mars. Recently, the commanders of the Earth are informed by their spies that the invaders of Mars want to land some paratroopers in the m × n grid yard of one their main weapon factories in order to destroy it. In addition, the spies informed them the row and column of the places in the yard in which each paratrooper will land. Since the paratroopers are very strong and well-organized, even one of them, if survived, can complete the mission and destroy the whole factory. As a result, the defense force of the Earth must kill all of them simultaneously after their landing.

In order to accomplish this task, the defense force wants to utilize some of their most hi-tech laser guns. They can install a gun on a row (resp. column) and by firing this gun all paratroopers landed in this row (resp. column) will die. The cost of installing a gun in the ith row (resp. column) of the grid yard is r_i (resp. c_i ) and the total cost of constructing a system firing all guns simultaneously is equal to the product of their costs. Now, your team as a high rank defense group must select the guns that can kill all paratroopers and yield minimum total cost of constructing the firing system.

Input

Input begins with a number T showing the number of test cases and then, T test cases follow. Each test case begins with a line containing three integers 1 ≤ m ≤ 50 , 1 ≤ n ≤ 50 and 1 ≤ l ≤ 500 showing the number of rows and columns of the yard and the number of paratroopers respectively. After that, a line with m positive real numbers greater or equal to 1.0 comes where the ith number is r_i and then, a line with n positive real numbers greater or equal to 1.0 comes where the ith number is c_i. Finally, l lines come each containing the row and column of a paratrooper.

Output

For each test case, your program must output the minimum total cost of constructing the firing system rounded to four digits after the fraction point.

Sample Input

1
4 4 5
2.0 7.0 5.0 2.0
1.5 2.0 2.0 8.0
1 1
2 2
3 3
4 4
1 4

Sample Output

16.0000

 

題解：

最小點權覆蓋，最小割模型，能夠用最大流去作，而後就是板子題，重點是轉換過程，

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <queue>
#define inf 0x3f3f
#define ll long long
#define MAXN 30000
using namespace std;
int n,m;//點數、邊數
int X[MAXN],y[MAXN];
int sp,tp;//原點、匯點
struct node
{
    int v,next;
    double cap;
}mp[MAXN*10];
int pre[MAXN],dis[MAXN],cur[MAXN];//cur爲當前弧優化，dis存儲分層圖中每一個點的層數（即到原點的最短距離），pre建鄰接表
int cnt=0;
void init()//不要忘記初始化
{
    cnt=0;
    memset(pre,-1,sizeof(pre));
}
void add(int u,int v,double w)//加邊
{
    mp[cnt].v=v;
    mp[cnt].cap=w;
    mp[cnt].next=pre[u];
    pre[u]=cnt++;
    mp[cnt].v=u;
    mp[cnt].cap=0;
    mp[cnt].next=pre[v];
    pre[v]=cnt++;
}


bool bfs()//建分層圖
{
    memset(dis,-1,sizeof(dis));
    queue<int>q;
    while(!q.empty())
        q.pop();
    q.push(sp);
    dis[sp]=0;
    int u,v;
    while(!q.empty())
    {
        u=q.front();
        q.pop();
        for(int i=pre[u];i!=-1;i=mp[i].next)
        {
            v=mp[i].v;
            if(dis[v]==-1&&mp[i].cap>0)
            {
                dis[v]=dis[u]+1;
                q.push(v);
                if(v==tp)
                    break;
            }
        }
    }
    return dis[tp]!=-1;
}


double dfs(int u,double cap)//尋找增廣路
{
    if(u==tp||cap==0)
        return cap;
    double res=0,f;
    for(int i=cur[u];i!=-1;i=mp[i].next)
    {
        int v=mp[i].v;
        if(dis[v]==dis[u]+1&&(f=dfs(v,min(cap-res,mp[i].cap)))>0)
        {
            mp[i].cap-=f;
            mp[i^1].cap+=f;
            res+=f;
            if(res==cap)
                return cap;
        }
    }
    if(!res)
        dis[u]=-1;
    return res;
}



double dinic()
{
    double ans=0;
    while(bfs())
    {
        for(int i=0;i<=tp;i++)
            cur[i]=pre[i];
        ans+=dfs(sp,inf);
    }
    return ans;
}
int main()
{

    int _;
    scanf("%d",&_);
    int l;
    while(_--) {
        init();
        int len=0;
        scanf("%d%d%d",&n,&m,&len);
        sp=0;tp=n+m+1;
        memset(pre,-1,sizeof pre);
        cnt=0;
        double w;
        for (int i = 1; i <=n ; ++i) {
            scanf("%lf",&w);
            add(sp,i,log(w));
        }
        for (int i = 1; i <=m ; ++i) {
            scanf("%lf",&w);
            add(n+i,tp ,log(w));
        }
        int a,b;
        for (int i = 0; i <len ; ++i) {
            scanf("%d%d",&a,&b);
            add(a,b+n,inf);
        }

        double kk=dinic();
        printf("%.4f\n",exp(kk));
    }
    return 0;
}