PAT 天梯赛 L2-024. 部落 【并查集】

题目链接

https://www.patest.cn/contests/gplt/L2-024

题意

给出 几个不同的圈子，然后 判断 有哪些人 是属于同一个部落的，或者理解为 ，有哪些人 是有关系的， 朋友的朋友 也属于同一个部落

思路

用并查集 并，然后最后查一下 有几个连通块，就可以输出有几个互不相交的部落，然后最后判断两个人是否是同一部落的，只要查找一下两个人是否属于同一祖宗就可以

AC代码

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <deque>
#include <vector>
#include <queue>
#include <string>
#include <cstring>
#include <map>
#include <stack>
#include <set>
#include <cstdlib>
#include <ctype.h>
#include <numeric>
#include <sstream>
using namespace std;

typedef long long LL;
const double PI  = 3.14159265358979323846264338327;
const double E   = 2.718281828459;
const double eps = 1e-6;
const int MAXN   = 0x3f3f3f3f;
const int MINN   = 0xc0c0c0c0;
const int maxn   = 1e4 + 5;
const int MOD    = 1e9 + 7;
int pre[maxn], arr[maxn];
int find(int x)
{
    int r = x;
    while (pre[r] != r)
        r = pre[r];
    pre[x] = r;
    return r;
}
void join(int x, int y)
{
    int fx = find(x), fy = find(y);
    if (x != fy)
        pre[fx] = fy;
}   

int main()
{
    int n, mi;
    int num, vis;
    cin >> n;
    int i, j, k = 0;
    map <int, int> q;
    map <int, int> m;
    map <int, int> flag;
    flag.clear();
    m.clear();
    q.clear();
    for (i = 0; i < maxn; i++)
        pre[i] = i;
    for (i = 0; i < n; i++)
    {
        cin >> mi;
        cin >> num;
        flag[num] = 1;
        for (j = 1; j < mi; j++)
        {
            scanf("%d", &vis);
            join(num, vis);
            flag[vis] = 1;
        }
    }
    map <int, int>::iterator it;
    for (it = flag.begin(); it != flag.end(); it++)
        arr[k++] = it -> first;
    int len = k;
    for (i = 0; i < k; i++)
        q[find(arr[i])] = 1;
    cout << len << " " << q.size() << endl;
    scanf("%d", &mi);
    for (i = 0; i < mi; i++)
    {
        int a, b;
        scanf("%d%d", &a, &b);
        if (find(a) == find(b))
            printf("Y\n");
        else
            printf("N\n");

    }
}