标签：body mes ++i nta one col 输入 ati fir
hihocoder-1497-Queen Attack
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?
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
One integer, the number of pairs may attack each other.
5 1 1 2 2 3 3 1 3 3 1
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\n", ans );
}
标签：body mes ++i nta one col 输入 ati fir
原文地址：https://www.cnblogs.com/zhang-yd/p/11022811.html
