2015
04-16

# Cover The Enemy

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.

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

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

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

18
30

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define N 100010

struct enemy
{
int x, y, da, db;
}p[N], c1, c2;

bool cmp(enemy a, enemy b)
{
return a.da < b.da;
}

int dist(enemy a, enemy b)
{
return ((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
}

int main()
{
int ncase;
int num;
int ans, res, onlyb;
scanf("%d", &ncase);
while(ncase--)
{
scanf("%d%d%d%d", &c1.x, &c1.y, &c2.x, &c2.y);
scanf("%d", &num);
for(int i = 0; i < num; ++i)
{
scanf("%d%d", &p[i].x, &p[i].y);
p[i].da = dist(c1, p[i]);
p[i].db = dist(c2, p[i]);
}
sort(p, p + num, cmp);
res = p[num - 1].da; //b不用
ans = 0;
for(int i = num - 2; i >= 0; --i)
{
ans = max(ans, p[i + 1].db);
res = min(res, ans + p[i].da);
}
ans = max(ans, p[0].db);
res = min(res, ans); //a不用
printf("%d\n", res);
}
return 0;
}

1. bottes vernies blanches

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2. 第一句可以忽略不计了吧。从第二句开始分析，说明这个花色下的所有牌都会在其它里面出现，那么还剩下♠️和♦️。第三句，可以排除2和7，因为在两种花色里有。现在是第四句，因为♠️还剩下多个，只有是♦️B才能知道答案。