hihocoder-1497-Queen Attack

hihocoder-1497-Queen Attack

#1497 : Queen Attack

时间限制:10000ms

单点时限:1000ms

内存限制:256MB

描述

There are N queens in an infinite chessboard. We say two queens may attack each other if they are in the same vertical line, horizontal line or diagonal line even if there are other queens sitting between them.

Now given the positions of the queens, find out how many pairs may attack each other?

输入

The first line contains an integer N.

Then N lines follow. Each line contains 2 integers Ri and Ci indicating there is a queen in the Ri-th row and Ci-th column.  

No two queens share the same position.  

For 80% of the data, 1 <= N <= 1000

For 100% of the data, 1 <= N <= 100000, 0 <= Ri, Ci <= 1000000000

输出

One integer, the number of pairs may attack each other.

样例输入

5  
1 1  
2 2  
3 3   
1 3
3 1

样例输出

10

题解：

　　使用unordered_map, 记录之前添加过的position，因为不可能同时两个position重叠，两两attack的position必定只存在着一种相交方式。

#include <cstdio> 
#include <cstdlib> 

#include <iostream>
#include <unordered_map> 
using namespace std;  

int main()
{
    int n, x, y;
    long long ans = 0; 
    scanf("%d", &n); 
    unordered_map<int, int> hor; 
    unordered_map<int, int> vet; 
    unordered_map<int, int> dx; 
    unordered_map<int, int> vdx; 
    for(int i=0; i<n; ++i)
    {
        scanf("%d %d", &x, &y); 
        if(hor.find(x) != hor.end())
        {
            ans += hor[x]; 
            hor[x] += 1; 
        }else{
            hor[x] = 1; 
        }

        if(vet.find(y) != vet.end())
        {
            ans += vet[y]; 
            vet[y] += 1; 
        }else{
            vet[y] = 1; 
        }

        if(dx.find(x-y) != dx.end())
        {
            ans += dx[x-y]; 
            dx[x-y] += 1; 
        }else{
            dx[x-y] = 1; 
        }

        if(vdx.find(x+y) != vdx.end())
        {
            ans += vdx[x+y]; 
            vdx[x+y] += 1; 
        }else{
            vdx[x+y] = 1; 
        }

    }
    printf("%lld
", ans );
}