POJ 3308 Paratroopers,pojparatroopers

POJ 3308 Paratroopers,pojparatroopers

Paratroopers

Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 8940   Accepted: 2696

网赌平台哪个信誉好,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 × ngrid 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 ri (resp. ci ) 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 ri and then, a line with n positive real numbers
greater or equal to 1.0 comes where the ith number is ci.
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

Source

Amirkabir University of Technology Local Contest 2006      
二分图最小点权覆盖集 对于一个敌人拆成两个点$x,y$
从$S$向$x$连给定权值的边, 从$y$向$T$连给定权值的边,
从$x$向$y$连$INF$的边 二分图最小点权覆盖集=最小割=最大流
乘法取log变加法 mmp精度坑死人  

// luogu-judger-enable-o2
#include<cstdio>
#include<cstring>
#include<queue>
#include<cmath>
#define AddEdge(x,y,z) add_edge(x,y,z),add_edge(y,x,0);
using namespace std;
const int MAXN=100001;
double INF=2000000000;
const double eps=1e-9;
int N,M,P,S,T;
struct node
{
    int u,v,nxt;
    double flow;
}edge[MAXN*5];
int head[MAXN],cur[MAXN],num=0;
inline void add_edge(int x,int y,double z)
{
    edge[num].u=x;
    edge[num].v=y;
    edge[num].flow=z;
    edge[num].nxt=head[x];
    head[x]=num++;
}
int deep[MAXN];
inline bool BFS()
{
    memset(deep,0,sizeof(deep));
    deep[S]=1;
    queue<int>q;
    q.push(S);
    while(q.size()!=0)
    {
        int p=q.front();
        q.pop();
        for(int i=head[p];i!=-1;i=edge[i].nxt)
            if(!deep[edge[i].v]&&edge[i].flow>eps)
            {
                deep[edge[i].v]=deep[p]+1;q.push(edge[i].v);
                if(edge[i].v==T) return 1;
            }
    }
    return deep[T];
}
double DFS(int now,double nowflow)
{
    if(now==T||nowflow<eps)    return nowflow;
    double totflow=0;
    for(int &i=cur[now];i!=-1;i=edge[i].nxt) 
    {
        if(deep[edge[i].v]==deep[now]+1&&edge[i].flow>eps)
        {
            double canflow=DFS(edge[i].v,min(nowflow,edge[i].flow));
            if(canflow>eps)
            {
                   edge[i].flow-=canflow;
                edge[i^1].flow+=canflow;
                   totflow+=canflow;
                   nowflow-=canflow;    
            }
            if(nowflow<eps) break;
        }
    }
    return totflow;
}
double Dinic()
{
    double ans=0;
    while(BFS())
    {
        memcpy(cur,head,sizeof(head)); 
        ans+=DFS(S,INF);
    }
    return ans;    
}
double valr,valc;
int main()
{
    #ifdef WIN32
    freopen("a.in","r",stdin);
    #else
    #endif
    int test;
    scanf("%d",&test);
    while(test--)
    {
        memset(head,-1,sizeof(head));
        scanf("%d%d%d",&N,&M,&P);S=0;T=N+M+P;
        for(int i=1;i<=N;i++) scanf("%lf",&valr),AddEdge(S,i,log(valr) );
        for(int i=1;i<=M;i++) scanf("%lf",&valc),AddEdge(i+N,T,log(valc) );
        for(int i=1;i<=P;i++) 
        {
            int x,y;
            scanf("%d%d",&x,&y);
            AddEdge(x,y+N,INF);
        }
        printf("%.4lf\n",exp(Dinic()));
    }

    return  0;
}

 

http://www.bkjia.com/cjjc/1301454.htmlwww.bkjia.comtruehttp://www.bkjia.com/cjjc/1301454.htmlTechArticlePOJ 3308 Paratroopers,pojparatroopers
Paratroopers Time Limit: 1000MS Memory Limit: 65536K Total Submissions:
8940 Accepted: 2696 Description It is year 2500 A.D. and there is
a…

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