POJ 3308 Paratroopers （最小割）

POJ 3308 Paratroopers （最小割）

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

14 4 52.0 7.0 5.0 2.01.5 2.0 2.0 8.01 12 23 34 41 4

Sample Output

16.0000

题意

n∗m 的地图，给出 L

思路

因为总代价是所有激光枪的代价之积，并不是之和，所以在这里我们要用点小技巧： logn×logm=log(n+m)

找一个源点，它到所有 ​​x​​ 坐标的容量是相应行的代价。

找一个汇点，所有 ​​y​​ 坐标到它的容量是相应列的代价。

然后输入的点的坐标也就是 ​​x​​​ 到 ​​y​​​ 的一条边，容量为 ​​INF​​ ，此时这一条边就代表了这一个点。

想要消灭所有的点，也就是找一个最小的割边集合让源汇分离，根据最大流最小割定理，求出图的最大流即可。

AC 代码

#include#include#include#include#include#includeusing namespace std;#include#include

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