PAT(A) - 1118. Birds in Forest (25)

最新推荐文章于 2024-11-08 01:00:00 发布

这个好玩吗

最新推荐文章于 2024-11-08 01:00:00 发布

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分类专栏： ----并查集 PAT

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本文介绍了一个使用并查集算法解决鸟类计数问题的实例。该问题涉及根据图片中出现的鸟类来计算森林中树木的最大数量，并判断任意两只鸟是否位于同一棵树上。文章提供了一个详细的C++实现方案，包括初始化、查找父亲节点和合并等核心步骤。

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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:

Sample Output:

2 10
Yes
No

我刚开始想到的方法是散列，用两个数组进行存储，只拿了17分，后来想到并查集，AC了。看来甲级每次都会有并查集的题，这个我没怎么练过，多训练这方面吧！

#include <cstdio>

#define MAX 10000

int father[MAX];
bool isRoot[MAX];

int findFather( int x ) {
    int a = x;
    while( x != father[x] ) {
        x = father[x];
    }

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

void Union( int a, int b ) {
    int faA = findFather( a );
    int faB = findFather( b );
    //printf( "%d %d\n", faA, faB );
    if( faA != faB ) {
        father[faA] = faB;
    }
}

void init( int n ) {
    for( int i = 1; i <= n; i++ ) {
        father[i] = i;
        isRoot[i] = false;
    }
}

int main() {

    int n;
    scanf( "%d", &n );
    init( MAX );
    int max = -1;
    for( int i = 1; i <= n;i ++ ) {
        int m;
        scanf( "%d", &m );
        for( int j = 1; j <= m; j++ ) {
            int index;
            int temp;
            scanf( "%d", &index );
            if( j == 1 ) {
                temp = index;
            }
            if( max < index ) max = index;
            Union( temp, index );
        }
    }



    for( int i = 1; i <= max; i++ ) {
        //printf( "根节点: %d\n", findFather( i ) );
        isRoot[findFather( i )] = true;
    }

    int ans = 0;
    for( int i = 1; i <= max; i++ ) {
        if( isRoot[i] ) {
            ans++;
        }
    }

    printf( "%d %d\n", ans, max );

    int q;
    scanf( "%d", &q );

    for( int i = 0; i < q; i++ ) {
        int a, b;
        scanf( "%d%d", &a, &b );
        int faA = findFather( a );
        int faB = findFather( b );

        if( faA == faB ) {
            printf( "Yes\n" );
        }
        else {
            printf( "No\n" );
        }
    }
    return 0;
}