源哥每日一题第十八弹 poj 1182 并查集

Posted 2020-11-18 fengyuzhicheng

题目链接：http://poj.org/problem?id=1182

题意：看不懂？退群吧

比平常的并查集加了一个判断集合间关系的操作；

开一个数组记录当前点所在集合的次序（第几个集合）用012表示

比较简单的思路，不过体现了并查集的精妙

#include <iostream>
#include <cstring>
#include <cstdio>

using namespace std;
int dad[500005];
int st[50005];
int find(int x) {
    if(dad[x] != x) {
        int t = dad[x];
        dad[x] = find(t);
        st[x] += st[t];
        st[x] %= 3;
    }
    return dad[x];
}

int main() {
#ifndef ONLINE_JUDGE
    freopen("D:\fengyu\Jiang_C\.vscode\in.txt","r",stdin);
	freopen("D:\fengyu\Jiang_C\.vscode\out.txt","w",stdout);
#endif
    int n, k;
    cin >> n >> k;
    for (int i = 0; i <= n; i++) {
        dad[i] = i;
        st[i] = 0;
    }
    int d, x, y;
    int ans = 0;
    while (k--) {
        scanf("%d%d%d", &d, &x, &y);
        if (x>n|| y>n || d==2 && x==y) {
            ans++;
            continue;
        }
        int fx = find(x);
        int fy = find(y);
        if(fx==fy) {
            if(d==1 && st[x]!=st[y] || d==2 &&st[x]!=(st[y]+2)%3) {
                ans++;
            }
        } else {
            dad[fy] = fx;
            st[fy] = (st[x]+d-1+3-st[y])%3;
        }
    }
    cout << ans << endl;

    return 0;
}