hdu 4145 Cover The Enemy

Cover The Enemy

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 1165    Accepted Submission(s): 293

Problem Description

 

  Now, the war is coming, no one want to fight with others, but you have to when you are facing.
  As the king of a country, you should defend your country. The condition is that “There are N enemy armies around you, you have two towers and you want to build cannon on each tower to cover all the enemy armies”. The cannon has an “attack distance” R with a cost of R^2.
  You are facing a problem, to cover all the armies and make the cost R1^2 + R2^2 minimal, where R1 and R2 are the attack distance of the two towers.

 

 

Input

 

The first line contains an integer t, number of test cases, following with t groups of test cases described as following.
For each case, the first line contains four integers: x1, y1, x2, y2, the Coordinates of the two towers.
The next line is an integer n, the number of enemy armies, following with n lines of coordinates: x, y, which is the coordinate of a group of enemy army, where:
  t <= 100
1 <= n <= 100000,  
-1000 <= x, y <= 1000  
-1000 <= x1, y1, x2, y2 <= 1000

 

 

Output

 

The minimal cost of the two tower to cover all the enemy armies.

 

 

Sample Input

 

2 0 0 10 0 2 -3 3 10 0 0 0 6 0 5 -4 -2 -2 3 4 0 6 -2 9 1

 

 

Sample Output

 

18 30

 

计算出每个点到两个塔x1,x2的距离，然后，把按照x1的距离从大到小排序。r1先取距x1最远的点的平方，点被完全覆盖。

然后，将r1依次递减，把r1不能覆盖的点让r2覆盖。这样，这些点始终都被完全覆盖着，ans取r1+r2的最小值。

 

#include
#include
#include
#include"stdlib.h"
#define N 100005
struct node
{
	int x,y;
	int d1,d2;       //记录平方和
}f[N];
int dis(int x1,int y1,int x2,int y2)
{
	int x=x1-x2,y=y1-y2;
	return x*x+y*y;
}
int cmp(const void*a,const void*b)
{
	return (*(struct node*)a).d1>(*(struct node*)b).d1?-1:1;
}
int min(int a,int b)
{
	return a