1118. Birds in Forest (25)-PAT甲级真题(并查集)

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 (<= 104) which is the number of pictures. Then N lines follow, each describes a picture in the format:
K B1 B2 … BK
where K is the number of birds in this picture, and Bi’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 104.

After the pictures there is a positive number Q (<= 104) 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


题目理解

给出n幅画,每幅画中有k只小鸟且在同一个树上,现在给出每幅画中小鸟的个数k和小鸟的id,求树和小鸟的总数量。

分析

  • 讲每幅画的小鸟进行Union操作,使其归属于同一个根
  • 设置findFather函数用于查找小鸟所在集合的根,并把回溯过程中的节点的根都设置为对应的根节点
  • 用exist数组记录遍历过的小鸟id,exist[id]=true
  • cnt数组用于存储隶属同一个根节点的小鸟个数
  • 遍历cnt数组并累加cnt数组不为0的个数,即为树的个数
  • 判断两鸟是否在同一颗树上,只需要知道这两只鸟的根节点是否相等即可
#include 
using namespace std;

int n, m, k;
const int maxn = 10010;
int fa[maxn] = {0}, cnt[maxn] = {0};

int findFather(int x) {
    int a = x;
    while (x != fa[x])
        x = fa[x];   //一层层回溯
    while (a != fa[a]) {
        int z = a;
        a = fa[a];
        fa[z] = x;  //更新fa数组
    }
    return x;
}

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

bool exist[maxn];

int main() {
    cin >> n;
    for (int i = 1; i <= maxn; i++)
        fa[i] = i;
    int id, tmp;
    for (int i = 0; i < n; i++) {
        cin >> k >> id;
        exist[id]=true;
        for (int j = 0; j < k - 1; j++) {
            cin >> tmp;
            Union(id, tmp);
            exist[tmp] = true;
        }
    }

    for (int i = 1; i <= maxn; i++) {
        if (exist[i]) {
            int root = findFather(i);
            cnt[root]++;
        }
    }

    int numTrees = 0, numBirds = 0;
    for (int i = 1; i <= maxn; i++) {
        if (exist[i] && cnt[i] != 0) {
            numTrees++;
            numBirds += cnt[i];
        }
    }
    cout << numTrees << " " << numBirds << endl;
    cin >> m;
    int ida, idb;
    for (int i = 0; i < m; i++) {
        cin >> ida >> idb;
        if (findFather(ida) == findFather(idb))
            cout << "Yes" << endl;
        else
            cout << "No" << endl;
    }
    return 0;
}

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