思路：维护主对角线和副对角线上的点的数量，然后加起来就好了
#include <cstdio>
#include <queue>
#include <cstring>
#include <iostream>
#include <cstdlib>
#include <algorithm>
#include <vector>
#include <map>
#include <string>
#include <set>
#include <ctime>
#include <cmath>
#include <cctype>
using namespace std;
#define maxn 1010
#define LL long long
int cas=1,T;
LL dpl[maxn][maxn];
LL dpr[maxn][maxn];
int main()
{
int n;
scanf("%d",&n);
for (int i = 1;i<=n;i++)
{
int x,y;
scanf("%d%d",&x,&y);
dpl[x][y]=1;
dpr[x][y]=1;
}
LL ans = 0;
for (int i = 1;i<=1000;i++)
for (int j = 1;j<=1000;j++)
{
if (dpl[i][j])
{
ans+=dpl[i-1][j-1];
ans+=dpr[i-1][j+1];
}
dpl[i][j]+=dpl[i-1][j-1];
dpr[i][j]+=dpr[i-1][j+1];
}
printf("%lld\n",ans);
//freopen("in","r",stdin);
//scanf("%d",&T);
//printf("time=%.3lf",(double)clock()/CLOCKS_PER_SEC);
return 0;
}
Description
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 Input
Input
5 1 1 1 5 3 3 5 1 5 5
Output
6
Input
3 1 1 2 3 3 5
Output
0
Hint
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)
