CodeForces - 621B （思维题）

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.

题意：意思就是有一个表格，每个格子为1或0（代表是否存在鲨鱼） 求出每条与对角线平行的斜线上1的对数

只要发现左上到由下的斜线上每个单元的横纵坐标的差事不变的，右上到左下的斜线上横纵坐标的和事不变的这点，再用map记录一下每条斜线上点的个数，就能A了

#include <vector>
#include <stdio.h>
#include <map>
#include <stdlib.h>
#include<ctype.h>
#include <iostream>
#include <queue>
#include <algorithm>
#define LL long long
using namespace std;

const int MAX = 1e3 + 50;

map<int, int> mp1;
map<int, int> mp2;
int main(int argc, char const *argv[]){
	int n;
	scanf("%d", &n);

	for(int i = 0; i < n; i++){
		int x, y;
		scanf("%d%d", &x, &y);
		mp1[x - y]++;
		mp2[x + y]++;
	}

	LL ans = 0;
	
	map<int, int>::iterator it;
	for(it = mp1.begin(); it != mp1.end(); it++){
		
		int x = it -> second;
		
		if(x != 0){
			ans += x * (x - 1) / 2;
		}
	}
	

	for(it = mp2.begin(); it != mp2.end(); it++){
		int x = it -> second;
		if(x != 0){
			ans += x * (x - 1) / 2;
		}
	}
	printf("%lld\n", ans);
    return 0;
}