CodeForces 621B Wet Shark and Bishops

最新推荐文章于 2019-04-04 01:08:47 发布

原创最新推荐文章于 2019-04-04 01:08:47 发布 · 384 阅读

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本文介绍了一个编程问题的解决方案，该问题要求计算在一个1000x1000的网格中放置的主教（类似于国际象棋中的主教）之间互相攻击的对数。通过确定主教位于哪些对角线并使用二分查找来计算攻击对的数量。

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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

题目大意：给出一个1000 * 1000的网格，现在向格子里放鲨鱼，已知处于同一对角线上的鲨鱼会互相攻击，如果它们之间还有第三条鲨鱼的话，那么这三条鲨鱼都会互相攻击（都在对角线上），给出有鲨鱼的格子，问有几对鲨鱼互相攻击。

解题思路：lrj白书193页指出，对于一个棋盘，格子(x,y)的y - x标识了主对角线，y + x标识了副对角线。根据这一性质，利用二分函数求出每个对角线上有ai鲨鱼，每条对角线上互相攻击的鲨鱼对数等于ai(ai - 1) / 2，最后主对角线副对角线上鲨鱼对数相加即为结果。（理解题意是解题的关键！two bishops may attack each other (according to Wet Shark) even if there is another bishop located between them这句话卡了很长时间。。）

代码如下：

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>

using namespace std;

int ans1[200005],ans2[200005];

int fun(int n)
{
    return n * (n - 1) / 2;
}

int main()
{
    int n;
    int x,y;
    scanf("%d",&n);
    memset(ans1,0xff,sizeof(ans1));
    memset(ans2,0xff,sizeof(ans2));
    for(int i = 0;i < n;i++){
        scanf("%d %d",&x,&y);
        ans1[i] = y - x;
        ans2[i] = x + y;
    }
    int res1 = 0,res2 = 0;
    sort(ans1,ans1 + n);
    sort(ans2,ans2 + n);
    for(int i = -999;i < 1000;i++){
        int temp = upper_bound(ans1,ans1 + n,i) - lower_bound(ans1,ans1 + n,i);
        if(temp > 1)
            res1 += fun(temp);
    }
    for(int i = 0;i < 2001;i++){
        int temp = upper_bound(ans2,ans2 + n,i) - lower_bound(ans2,ans2 + n,i);
        if(temp > 1)
            res2 += fun(temp);
    }
    printf("%d\n",res1 + res2);
    return 0;
}