pat 甲级 1118 森林中的鸟

最新推荐文章于 2021-12-06 10:17:01 发布

原创最新推荐文章于 2021-12-06 10:17:01 发布 · 283 阅读

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本文介绍了一个利用并查集算法解决鸟类图片计数的问题，通过处理多个图片中鸟的位置信息来确定森林中树木的最大数量及鸟是否位于同一棵树上。文章详细展示了输入输出规范，并给出了一段实现该算法的C++代码。

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题目描述：

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.

之前都没有写过并查集，这一次经过学习后，第一次使用并查集，之前做这道题没有用并查集，只得到了17分，后来考察了一些数据果然还是出了问题，并查集主要两个基本操作，find和union操作，即发现根和合并操作，经后还要多多训练。

#include<iostream>
#include<algorithm>
#include<memory.h>
using namespace std;
int fa[10010]={0};
int cnt[10010]={0};
int findfather(int x){
    if(fa[x]==x)
        return x;
    else{
        return fa[x]=findfather(fa[x]);
    }
}
void Union(int x,int y){
    x=findfather(x);
    y=findfather(y);
    if(x!=y){
        fa[x]=y;
    }
}
bool exist[10010];
int main(){
    int n;
    cin>>n;
    for(int i=0;i<10010;i++){
        fa[i]=i;
    }
    memset(exist,false,sizeof(exist));
    for(int i=0;i<n;i++){
        int id,k;
        scanf("%d%d",&k,&id);
        exist[id]=true;
        for(int j=0;j<k-1;j++){
            int temp;
            cin>>temp;
            Union(id,temp);
            exist[temp]=true;
        }
    }
    int treecount=0,birdcount=0;
    for(int i=1;i<10010;i++){
        if(exist[i]==true){
            int root=findfather(i);
            cnt[root]++;
        }
        else{
            break;
        }
    }
    for(int i=1;i<10010;i++){
        if(cnt[i]!=0){
            treecount++;
            birdcount+=cnt[i];
        }
    }
    cout<<treecount<<" "<<birdcount<<endl;
    int q;
    cin>>q;
    for(int i=0;i<q;i++){
        int left,right;
        cin>>left>>right;
        if(findfather(left)==findfather(right))
            cout<<"Yes"<<endl;
        else
            cout<<"No"<<endl;
    }
    return 0;
}