Codeforces Round #341 (Div. 2)B

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.
제목: 바둑판에서 코끼리는 서로 공격할 수 있다(코끼리는 사율 1-1에서만 공격할 수 있다)
n개 코끼리를 입력하여 몇 가지 공격 배합이 있는지 판단합니다
문제풀이: 경사율이 1인 x-y의 값이 같다
경사율이 -1인 x+y의 값이 같다(모두 같은 대각선에 있다)
배열 태그 또는 맵 태그
각 라인의 짝짓기 수량fun() 구현

#include<bits/stdc++.h>
using namespace std;
#define ll __int64
int flag[4005];
int flag1[4005];
ll fun(ll n,ll m)
{
    ll sum=1;
    for(ll i=1,j=n;i<=m;i++,j--)
        sum=sum*j/i;
    return sum;
}
int main()
{
    ll x,y,z,i;
    scanf("%I64d",&x);
    for(i=0;i<x;i++)
    {
        scanf("%I64d%I64d",&y,&z);
        flag[y+z]++;
        flag1[y-z+1000]++;
    }
    ll ans=0;
    for(i=0;i<=3000;i++)
    {
        if(flag[i])
        ans+=fun(flag[i],2);
        if(flag1[i])
        ans+=fun(flag1[i],2);
    }
    printf("%I64d
",ans);
    return 0;
}