1118 Birds in Forest (25 分)

2019-04-27

1118 Birds in Forest (25 分)

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 … 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 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:

1
2
3

2 10
Yes
No

作者: CHEN, Yue

单位: 浙江大学

时间限制: 150 ms

内存限制: 64 MB

代码长度限制: 16 KB

题目大意

给出森林中鸟的照片，出现在同一张照片的鸟是同一颗树上的。求有几棵树，以及鸟的总数。然后给出任两只鸟，要求判断是否同一颗树。

分析

并查集，每次都与照片中出现的第一只鸟合并。注意必须降低树的深度，否则测试点4会超时。

代码

#include <iostream>
#include <vector>
#include <stack>
#include <algorithm>
#include <deque>
#include <queue>
#include <map>
#include <set>
#include <cstring>
#include <cmath>
#include <sstream>

using namespace std;
int n;
int par[10010],showup[10010],tshowup[10010];
int findpar(int x){
    int t=x;
    while(x!=par[x])
        x=par[x];
    while(par[t]!=x){
        int temp=par[t];
        par[t]=x;
        t=temp;
    }
    return x;
}
void makeunion(int a,int b){
    int pa=findpar(a);
    int pb=findpar(b);
    par[pa]=pb;
}
int main() {
    int k,t,q;
    cin>>n;
    for(int i=0;i<10010;i++){
        par[i]=i;
    }
    for(int i=0;i<n;i++){
        int x;
        cin>>k;
        for(int j=0;j<k;j++){
            cin>>t;
            showup[t]=1;
            if(j==0)
                x=t;
            else
                makeunion(x, t);
        }
    }
    int sum=0,tsum=0;
    for(int i=0;i<10010;i++){
        if(showup[i]){
            sum++;
            tshowup[findpar(i)]=1;
        }
    }
    for(int i=0;i<10010;i++)
        tsum+=tshowup[i];
    cout<<tsum<<" "<<sum<<endl;
    
    cin>>q;
    int b1,b2;
    for(int i=0;i<q;i++){
        cin>>b1>>b2;
        if(findpar(b1)==findpar(b2))
            cout<<"Yes"<<endl;
        else
            cout<<"No"<<endl;
    }
    return 0;
}