TOJ Right Triangles II

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

34 22 11 3

Sample Output

0

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;}