Paratroopers

Time Limit: 1000MS Memory Limit: 65536K

Total Submissions: 7881 Accepted: 2373

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 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

好恶心的题啊,一直超时,后来也没有怎么改经过一大波的TLE后就过了,好奇怪,难道精度没有控制好?

#include <map>
#include <cmath>
#include <queue>
#include <stack>
#include <string>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std; const double INF = 10000.0; const double eps = 1e-8; const int Max = 3000; struct Edge
{
int v;
int next;
double cap;
}E[Max]; int Head[120];
int Du[120];
int top;
int n,m,L;
int s,t;
void AddEdge(int u,int v,double w)
{
E[top].cap=w;E[top].v=v;
E[top].next=Head[u];Head[u]=top++;
E[top].cap=0;E[top].v=u;
E[top].next=Head[v];Head[v]=top++;
}
double Eps(double s)
{
return fabs(s)<eps?0:s;
}
double min(double a,double b)
{
return a<b?a:b;
}
int bfs()
{
memset(Du,0,sizeof(Du));
queue<int>Q;
Du[s]=1;
Q.push(s);
while(!Q.empty())
{
int a=Q.front();
Q.pop();
for(int i=Head[a];i!=-1;i=E[i].next)
{
if(Du[E[i].v]==0&&Eps(E[i].cap)>0)
{
Du[E[i].v]=Du[a]+1;
Q.push(E[i].v);
}
}
}
return Du[t];
} double dfs(int star,double num)
{ if(star==t)
{
return num;
}
double S=0;
double ant;
for(int i=Head[star];i!=-1;i=E[i].next)
{
if(Du[star]+1==Du[E[i].v]&&Eps(E[i].cap)>0)
{
ant=dfs(E[i].v,min(E[i].cap,num));
E[i].cap-=ant;
E[i^1].cap+=ant;
num-=ant;
S+=ant;
if(Eps(num)==0)
{
break;
}
}
}
return S;
}
double Dinic()
{
double ant=0;
while(bfs())
{
ant+=dfs(0,INF);
}
return ant;
} int main()
{
int T;
double w;
int u,v;
scanf("%d",&T);
while(T--)
{
scanf("%d %d %d",&n,&m,&L);
s=0;
t=n+m+1;
top=0;
memset(Head,-1,sizeof(Head));
for(int i=1;i<=n;i++)
{
scanf("%lf",&w);
AddEdge(s,i,log(w));
}
for(int i=1;i<=m;i++)
{
scanf("%lf",&w);
AddEdge(n+i,t,log(w));
}
for(int i=1;i<=L;i++)
{
scanf("%d %d",&u,&v);
AddEdge(u,v+n,INF);
}
printf("%.4f\n",exp(Dinic()));
}
return 0;
}

Paratroopers的更多相关文章

  1. POJ3308 Paratroopers(网络流)(最小割)

                                                     Paratroopers Time Limit: 1000MS   Memory Limit: 655 ...

  2. POJ 3308 Paratroopers(最小割EK(邻接表&矩阵))

    Description It is year 2500 A.D. and there is a terrible war between the forces of the Earth and the ...

  3. 伞兵(Paratroopers)

    伞兵(Paratroopers) 时间限制: 1 Sec  内存限制: 128 MB 题目描述 公元 2500 年,地球和火星之间爆发了一场战争.最近,地球军队指挥官获悉火星入侵者将派一些伞兵来摧毁地 ...

  4. POJ 3308 Paratroopers 最大流,乘积化和 难度:2

    Paratroopers Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 7267   Accepted: 2194 Desc ...

  5. POJ 3308 Paratroopers(最小点权覆盖)(对数乘转加)

    http://poj.org/problem?id=3308 r*c的地图 每一个大炮可以消灭一行一列的敌人 安装消灭第i行的大炮花费是ri 安装消灭第j行的大炮花费是ci 已知敌人坐标,同时消灭所有 ...

  6. POJ 3308 Paratroopers(最大流最小割の最小点权覆盖)

    Description It is year 2500 A.D. and there is a terrible war between the forces of the Earth and the ...

  7. poj3308 Paratroopers

    Description It is year 2500 A.D. and there is a terrible war between the forces of the Earth and the ...

  8. poj 3308 Paratroopers(二分图最小点权覆盖)

    Paratroopers Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 8954   Accepted: 2702 Desc ...

  9. POJ - 3308 Paratroopers(最大流)

    1.这道题学了个单词,product 还有 乘积 的意思.. 题意就是在一个 m*n的矩阵中,放入L个敌军的伞兵,而我军要在伞兵落地的瞬间将其消灭.现在我军用一种激光枪组建一个防御系统,这种枪可以安装 ...

随机推荐

  1. MAX(A,B)

    MAX(A,B)  可以把x,或者y的变量扔进去比较,会自动放出比较结果,这样就避免的三目运算.

  2. Nginx简介

    序言Nginx 是 lgor Sysoev 为俄罗斯访问量第二的 rambler.ru 站点设计开发的.从 2004 年发布至今,凭借开源的力量,已经接近成熟与完善.Nginx 功能丰富,可作为 HT ...

  3. Codeforces Round #312 (Div. 2) E. A Simple Task

    题目大意就是给一个字符串,然后多个操作,每次操作可以把每一段区间的字符进行升序或者降序排序,问最终的字符串是多少. 一开始只考虑字符串中字符'a'的情况,假设操作区间[L,R]中有x个'a',那么一次 ...

  4. java 中多线程和锁的使用

    关键词: implements  实现  Runnable 类 run()  方法 注意点 : 创建类的实例 InterfaceController inter=new InterfaceContro ...

  5. Oracle角色

    一 .3种标准角色 Qracle为了兼容以前的版本,提供了三种标准的角色(role):CONNECT.RESOURCE和DBA. 1. CONNECT Role(连接角色) 临时用户,特别是那些不需要 ...

  6. 夺命雷公狗jquery---6属性选择器

    <!DOCTYPE html> <html> <head> <meta charset="utf-8"> <title> ...

  7. SqlServer中使用Select语句给变量赋值的时候需要注意的一个问题

    我们知道在SqlServer中可以用Select语句给变量赋值,比如如下语句就为int类型的变量@id赋值 ; select @id=id from ( as id union all as id u ...

  8. Reporting Service报表项默认可见+号和-号的显示问题

    在Reporting Service里面可以设置报表项(组.tablix行.tablix列.文本框等所有SSRS报表项)的可见性,并且可以设置某个报表项的可见性由点击另外一个报表项来控制,比如报表项A ...

  9. OpenStack collectd的从零安装客户端

    1.查看是否需要增加yum 源 1 2 3 4 5 6 7 8 9 10 11 12 13 14 [root@node-12 ~]# yum search collectd Loaded plugin ...

  10. android 应用架构随笔六(Loading加载页面)

    import java.util.concurrent.ExecutorService; import java.util.concurrent.Executors; import com.heima ...