Codeforces 621B - Wet Shark and Bishops(思維)

B. Wet Shark and Bishops
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard 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.

Examples
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.

題意:
給出n個點,有多少對在同一對角線上.

解題思路:
因爲對角線的畫斜率只有1或-1,所以記錄截距出現的次數就可以.

AC代碼:

#include <bits/stdc++.h>
using namespace std;
const int maxn = 3e3+5;
int a[maxn];
int b[maxn];
int main()
{
    ios::sync_with_stdio(false);
    int n;
    cin>>n;
    long long cnt = 0;
    while(n--)
    {
        int x,y;
        cin>>x>>y;
        a[x+y]++;
        b[y-x+1000]++;
    }
    for(int i = 1;i <= 3000;i++)    cnt += a[i]*(a[i]-1)/2+b[i]*(b[i]-1)/2;
    cout<<cnt;
    return 0;
}