TOJ Right Triangles II

最新推荐文章于 2021-01-14 11:24:30 发布

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最新推荐文章于 2021-01-14 11:24:30 发布

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本文介绍了一种算法，用于计算平面上N个点中能构成直角三角形的组合数量，这些三角形的两条直角边需平行于坐标轴。文章提供了完整的C++代码实现，并解释了输入输出格式。

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Right Triangles II

Time Limit(Common/Java):1000MS/3000MS Memory Limit:65536KByte
Total Submit: 53 Accepted: 14

Description

N points are placed in the coordinate plane.
Write a program that calculates how many ways we can choose three points so that they form a right triangle with legs parallel to the coordinate axes.
A right triangle has one 90-degree internal angle. The legs of a right triangle are its two shorter sides.

Input

The first line of input contains the integer N (3 ≤ N ≤ 100 000), the number of points.
Each of the following N lines contains two integers X and Y (1 ≤ X, Y ≤ 100 000), the coordinates of one point.
No pair of points will share the same pair of coordinates.

Output

Output the number of triangles.

Sample Input

Sample Output

Uploader

crq

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;

const int maxn = 100008;

vector<__int64> x[maxn];
__int64 y[maxn];

int n;

void work() {
    __int64 i, j;
    __int64 cnt = 0;
    for(i = 1; i < maxn; i++) {//maxn --> n ！！！
        __int64 Size = x[i].size();
        if(Size <= 1) continue;
        for(j = 0; j < Size; j++) {
            __int64 id = x[i][j];
            if(y[id] > 0)
                cnt += (Size - 1) * (y[id] - 1);
        }
    }
    printf("%I64d\n", cnt);
}

void init() {
    int i;
    for(i = 0; i < maxn; i++) {
        x[i].clear();
        y[i] = 0;
    }
}

int main()
{
    int i;
    int dx, dy;
    while(scanf("%d", &n) != EOF) {
        init();
        for(i = 0; i < n; i++) {
            scanf("%d%d", &dx, &dy);
            x[dx].push_back(dy);
            y[dy]++;
        }
        work();
    }
    return 0;
}