Codeforces 621B Wet Shark and Bishops(判定对角线点+组合数统计)

B. Wet Shark and Bishops

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Today, Wet Shark is given n bishops on a 1000 by 1000 grid. Both rows and columns of the grid are numbered from 1 to 1000. Rows are numbered from top to bottom, while columns are numbered from left to right.

Wet Shark thinks that two bishops attack each other if they share the same diagonal. Note, that this is the only criteria, so two bishops may attack each other (according to Wet Shark) even if there is another bishop located between them. Now Wet Shark wants to count the number of pairs of bishops that attack each other.

Input

The first line of the input contains n (1 ≤ n ≤ 200 000) — the number of bishops.

Each of next n lines contains two space separated integers xi and yi (1 ≤ xi, yi ≤ 1000) — the number of row and the number of column where i-th bishop is positioned. It's guaranteed that no two bishops share the same position.

Output

Output one integer — the number of pairs of bishops which attack each other.

Sample test(s)

input

5
1 1
1 5
3 3
5 1
5 5

output

6

input

3
1 1
2 3
3 5

output

0

Note

In the first sample following pairs of bishops attack each other: (1, 3), (1, 5), (2, 3), (2, 4), (3, 4) and (3, 5). Pairs (1, 2), (1, 4),(2, 5) and (4, 5) do not attack each other because they do not share the same diagonal.

题意：有1000*1000的棋盘上有n个点，给出n个点的坐标。判断在同一对角线上的点有多少对，两点在同一对角线上，中间隔有其他点也算一对。

题解：开始是O(n^2)算法暴力枚举每个点的关系。直接TLe。  这题要O(n)算法做。我们可以统计每条对角线上有多少点。然后求取2的组合数，相加。PS：（思路对了，但是cf的时候各种傻逼错误，最后五分钟才对。。卒）

代码如下：

#include<cstdio>
#include<cstring>
#define LL __int64
int a[2010],b[2010]; 

__int64 C(int n)//组合数函数，求C(n,2)的值 
{
	if(n<2)
		return 0;
	else
	{
		__int64 x=1;
		int i;
		for(i=1;i<=2;++i)
		{
			x=x*(n-i+1)/i;
		}
		return x;
	}
}

int main()
{
	int n,i,j,x,y;
	LL ans;
	while(scanf("%d",&n)!=EOF)
	{
		memset(a,0,sizeof(a));
		memset(b,0,sizeof(b));
		for(i=0;i<n;++i)
		{
			scanf("%d%d",&x,&y);
			a[x+y]++;//统计每条主对角线线上的点 
			b[(x-y)+1000]++;//统计每条副对角线上面的点 
		}
		ans=0;
		for(i=0;i<2010;++i)
		{
			ans+=C(a[i]);
			ans+=C(b[i]);
		}
		printf("%I64d\n",ans);
	}
	return 0;
}