codeforces-650A Watchmen(思维)

题意:给n个坐标,问有多少个坐标对的x值相等或y值相等
题解:分别排序,分别求出x相等和y相等的组合,最后减去x和y均相等的
#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>

using namespace std;

int n;
struct node{
    int x, y;
}p[200010];

bool cmp1(node a, node b){
    return a.x<b.x;
}
bool cmp2(node a, node b){
    return a.y<b.y;
}
bool cmp3(node a, node b){
    if(a.x == b.x) return a.y<b.y;
    return a.x<b.x;
}

int main(){
    scanf("%d", &n);
    for(int i = 1; i<=n; i++){
        scanf("%d%d", &p[i].x, &p[i].y);
    }
    sort(p+1, p+1+n, cmp1);

    long long ans = 0;
    long long cnt = 1;
    for(int i = 2; i<=n; i++){
        if(p[i].x == p[i-1].x){
            cnt++;
        }
        else{
            ans+=cnt*(cnt-1)/2;
            //cout<<" "<<ans<<endl;
            cnt = 1;
        }
    }
    ans+=cnt*(cnt-1)/2;
    //cout<<ans<<endl;

    sort(p+1, p+1+n, cmp2);
    cnt = 1;
    for(int i = 2; i<=n; i++){
        if(p[i].y == p[i-1].y){
            cnt++;
        }
        else{
            ans+=cnt*(cnt-1)/2;
            cnt = 1;
        }
    }
    ans+=cnt*(cnt-1)/2;
    //cout<<ans<<endl;

    sort(p+1, p+1+n, cmp3);
    cnt = 1;
    int sum = 0;
    for(int i = 2; i<=n; i++){
        if(p[i].x == p[i-1].x && p[i].y == p[i-1].y){
            cnt++;
        }
        else{
            ans-=cnt*(cnt-1)/2;
            cnt = 1;
        }
    }
    ans -= cnt*(cnt-1)/2;
    //cout<<sum<<endl;
    printf("%I64d\n", ans);
    return 0;
}

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转载自blog.csdn.net/grimcake/article/details/80364430