Codeforces Round #341 Div.2 B. Wet Shark and Bishops

题意:处在同一对角线上的主教(是这么翻译没错吧= =)会相互攻击 求互相攻击对数

由于有正负对角线 因此用两个数组分别保存每个主教写的 x-y 和 x+y

然后每个数组中扫描重复数字k ans加上kC2就行了

wa了两发的原因是没考虑到如果整个数组都是重复的 那要最后额外加一次

#include <cstdio>
#include <cmath>
#include <cstring>
#include <queue>
#include <vector>
#include <algorithm>
#define INF 0x3f3f3f3f
#define mem(str,x) memset(str,(x),sizeof(str))
#define STOP puts("Pause");
using namespace std;
typedef long long LL;

int n, node[], node2[];
LL ans;

int main()
{
    ans = ;
    scanf("%d", &n);
    for(int i = ; i <= n; i++){
        int x, y;
        scanf("%d%d", &x, &y);
        node[i] = x - y;
        node2[i] = x + y;
    }
    sort(node + , node +  + n);
    sort(node2 + , node2 +  + n);
    LL count = ;
    for(int i = ; i <= n; i++){
        if(node[i] == node[i-]) count++;
        else{
            ans += (count * (count - ) / );
            count = ;
        }
    }
    ans += (count * (count - ) / );
    count = ;
    for(int i = ; i <= n; i++){
        if(node2[i] == node2[i-]) count++;
        else{
            ans += (count * (count - ) / );
            count = ;
        }
    }
    ans += (count * (count - ) / );
    printf("%I64d\n", ans);
    return ;
}