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 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
題目:給出一個(gè)矩陣,矩陣上會(huì)有傘兵濺落,每次可殺死某一行的所有兵或者殺死某一列的所有兵。每個(gè)行、每個(gè)列都有一個(gè)花費(fèi)。
求:殺掉所有兵的最少投資花費(fèi)(投資花費(fèi)為被選上的花費(fèi)(可以是行,可以是列)的乘積)。
分析:殺死一個(gè)兵,可通過(guò)選行或者選列,最小點(diǎn)權(quán)覆蓋模型,每個(gè)兵拆成兩個(gè)點(diǎn)(u,v),s->u為選擇行的花費(fèi)的對(duì)數(shù),v->t為選擇列的花費(fèi)的對(duì)數(shù),u->v為無(wú)窮。最大流后結(jié)果用exp(maxflow)就是最小花費(fèi)。
代碼:
#include <stdio.h>
#include 
<string.h>
#include 
<queue>
#include 
<cmath>
#include 
<algorithm>
#include 
<iostream>
#define Min(a, b) (a) < (b) ? a : b
#define Max(a, b) (a) > (b) ? a : b
using namespace std;
const int MAXN = 2005;
const int MAXM = 700000;
const double INF = 1100000000;//double 
struct Edge
{
    
int st, ed, next;
    
double flow, cap;//double 
}edge[MAXM];
int head[MAXN], level[MAXN], que[MAXN], stk[MAXN], pre[MAXN], del[MAXN], E;
void add(int u, int v, double w)
{
    edge[E].flow 
= 0, edge[E].cap = w, edge[E].st = u, edge[E].ed = v;
    edge[E].next 
= head[u], head[u] = E++;
    edge[E].flow 
= 0, edge[E].cap = 0, edge[E].st = v, edge[E].ed = u;
    edge[E].next 
= head[v], head[v] = E++;
}
int dinic_bfs(int src, int dest, int ver)
{
    
int i, j;
    
for (i = 0; i <= ver; i++)
    {
        level[i] 
= -1;
    }
    
int rear = 1;
    que[
0= src, level[src] = 0;
    
for (i = 0; i < rear; i++)
    {
        
for (j = head[que[i]]; j != -1; j = edge[j].next)
        {
            
if (level[edge[j].ed] == -1 && edge[j].cap > edge[j].flow)
            {
                level[edge[j].ed] 
= level[que[i]] + 1;
                que[rear
++= edge[j].ed;
            }
        }
    }
    
return level[dest] >= 0;
}
double dinic_dfs(int src, int dest, int ver)//double返回值 
{
    
int top = 0, cur, ptr, i;
    
double ret = 0, minf;//double
    for (i = 0; i <= ver; i++)
    {
        del[i] 
= 0;
    }
    stk[top
++= src;
    pre[src] 
= src;
    cur 
= src;
    
while (top)
    {
        
while (cur != dest && top)
        {
            
for (i = head[cur]; i != -1; i = edge[i].next)
            {
                
if (level[edge[i].ed] == level[cur] + 1 && edge[i].cap > edge[i].flow && !del[edge[i].ed])
                {
                    stk[top
++= edge[i].ed;
                    cur 
= edge[i].ed;
                    pre[edge[i].ed] 
= i;
                    
break;
                }
            }
            
if (i == -1)
            {
                del[cur] 
= 1;
                top
--;
                
if (top) cur = stk[top-1];
            }
        }
        
if (cur == dest)
        {
            minf 
= INF;
            
while (cur != src)
            {
                cur 
= pre[cur];
                minf 
= Min(minf, edge[cur].cap - edge[cur].flow);
                cur 
= edge[cur].st;
            }
            cur 
= dest;
            
while (cur != src)
            {
                cur 
= pre[cur];
                edge[cur].flow 
+= minf;
                edge[cur
^1].flow -= minf;
                
if (edge[cur].cap - edge[cur].flow == 0)
                {
                    ptr 
= edge[cur].st;
                }
                cur 
= edge[cur].st;
            }
            
while (top > 0 && stk[top-1!= ptr) top--;
            
if (top) cur = stk[top-1];
            ret 
+= minf;
        }
    }
    
return ret;
}
double Dinic(int src, int dest, int ver)//double返回值 
{
    
double ret = 0, t;//double 
    while (dinic_bfs(src, dest, ver))//bfs 
    {
        t 
= dinic_dfs(src, dest, ver);//dfs 
        if (t) ret += t;
        
else break;
    }
    
return ret;
}
int main()
{
    
int ca, n, m, k, i, j;
    
double w;
    scanf(
"%d"&ca);
    
while (ca--)
    {
        scanf(
"%d%d%d"&n, &m, &k);
        
int s = 0, t = n + m + 1, ver = t + 1;
        E 
= 0;
        
for (i = 0; i <= ver; i++)
        {
            head[i] 
= -1;
        }
        
for (i = 1; i <= n; i++)
        {
            scanf(
"%lf"&w);
            add(s, i, log(w));
        }
        
for (i = 1; i <= m; i++)
        {
            scanf(
"%lf"&w);
            add(i 
+ n, t, log(w));
        }
        
while (k--)
        {
            scanf(
"%d%d"&i, &j);
            add(i, j 
+ n, INF);
        }
        
        
double ans = Dinic(s, t, ver);
        printf(
"%.4f\n", exp(ans));
    }
    
return 0;
}
/*
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
*/