1118 Birds in Forest

Some scientists took pictures of thousands of birds in a forest. Assume that all the birds appear in the same picture belong to the same tree. You are supposed to help the scientists to count the maximum number of trees in the forest, and for any pair of birds, tell if they are on the same tree.

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive number N (≤) which is the number of pictures. Then N lines follow, each describes a picture in the format:

K B​1​​ B​2​​ ... B​K​​

where K is the number of birds in this picture, and B​i​​'s are the indices of birds. It is guaranteed that the birds in all the pictures are numbered continuously from 1 to some number that is no more than 1.

After the pictures there is a positive number Q (≤) which is the number of queries. Then Q lines follow, each contains the indices of two birds.

Output Specification:

For each test case, first output in a line the maximum possible number of trees and the number of birds. Then for each query, print in a line Yes if the two birds belong to the same tree, or No if not.

Sample Input:

4
3 10 1 2
2 3 4
4 1 5 7 8
3 9 6 4
2
10 5
3 7

 

Sample Output:

2 10
Yes
No

题意：

　　给出一组图画，问画中有几棵树，几只鸟，同一组画中的鸟在同一棵树上。

思路：

　　这道题用到了并查集的知识，并查集中有两个重要的函数

void findFather(int x) {
    int a = x;
    while (x != fa[x]) {
        x = fa[x];
    }
    while (a != fa[a]) {       // 路径压缩
        int z = a;
        a = fa[a];
        fa[z] = x;
    }
    return x;
}

　　这个函数是用来寻找某个节点的祖先结点的

void Union(int x, int y) {
    int faA = findFather(x);
    int faB = findFather(y);
    if (faA != faB) fa[faA] = faB;
}    
        

　　用来合并两个不同的子集。

　　

---

　　首先，我们要初始化fa数组，使每个结点的祖先结点是其本身。　

　　然后要做的就是将同一副画中的每一个结点通过Union函数合并到一个集合中。

　　用exist数组来存储结点是否存在。cnt数组来存储以某一个节点为祖先结点的集合中节点的个数。

Code：

#include <bits/stdc++.h>

using namespace std;

const int maxn = 10005;
vector<int> fa(10005);

int findFather(int x) {
    int a = x;
    while (x != fa[x]) {
        x = fa[x];
    }
    while (a != fa[a]) {
        int z = a;
        a = fa[a];
        fa[z] = x;
    }
    return x;
}

void Union(int x, int y) {
    int faX = findFather(x);
    int faY = findFather(y);
    if (faX != faY) fa[faY] = faX;
}

int main() {
    int n;
    cin >> n;

    for (int i = 0; i < maxn; ++i) {
        fa[i] = i;
    }

    int k, id, m;
    vector<bool> exist(10005, false);
    vector<int> cnt(10005, 0);
    int numTrees = 0;
    int numBirds = 0;
    for (int i = 0; i < n; ++i) {
        cin >> k >> id;
        exist[id] = true;
        for (int j = 1; j < k; ++j) {
            cin >> m;
            exist[m] = true;
            Union(id, m);
        }
    }
    for (int i = 1; i < maxn; ++i) {
        if (exist[i]) {
            int root = findFather(i);
            cnt[root]++;
        }
    }
    for (int i = 1; i < maxn; ++i) {
        if (exist[i] && cnt[i] > 0) {
            numTrees++;
            numBirds += cnt[i];
        }
    }
    cout << numTrees << " " << numBirds << endl;
    int querys;
    cin >> querys;
    for (int i = 0; i < querys; ++i) {
        int bird1, bird2;
        cin >> bird1 >> bird2;
        if (findFather(bird1) == findFather(bird2))
            cout << "Yes" << endl;
        else
            cout << "No" << endl;
    }
    return 0;
}

参考：https://www.liuchuo.net/archives/2482