zoukankan      html  css  js  c++  java
  • 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:

    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
    
     
    #include <bits/stdc++.h>
    using namespace std;
    int p[10010];
    int visit[10010];
    int a[10010];
    int n,m,k;
    int found(int a)
    {
        if(a==p[a]){
            return a;
        }
        return p[a]=found(p[a]);
    }
    void unite(int a,int b)
    {
        int x = found(a);
        int y = found(b);
        if(x!=y){
            p[x] = y;
        }
        return ;
    }
    bool istrue(int a,int b){
        return found(a) == found(b);
    }
    int main()
    {
        int x,y;
        set<int> s;
    
        for(int i=1;i<=10010;i++){
            p[i] = i;
        }
        scanf("%d",&n);
        for(int i=1;i<=n;i++){
            scanf("%d",&m);
            for(int j=1;j<=m;j++)
            {
                scanf("%d",&a[j]);
                s.insert(a[j]);
            }
            for(int j=2;j<=m;j++)
            {
                unite(a[j-1],a[j]);
            }
        }
        int sum = s.size();
        set<int> ss;
        for(int i=1;i<=sum;i++)
        {
            ss.insert(found(i));
        }
        cout<<ss.size()<<" "<<sum<<endl;
        scanf("%d",&k);
        while(k--)
        {
            scanf("%d%d",&x,&y);
            if(istrue(x,y)){
                cout<<"Yes"<<endl;
            }else{
                cout<<"No"<<endl;
            }
        }
        return 0;
    }
  • 相关阅读:
    web网页端上传用户头像,后端获取后,返回路径给前端做展示
    获取时间戳后按要求转换为要使用的条件
    本地测试环境获取微信授权的,不用在手动跳过
    php批量压缩指定目录的图片,优点-比工具快好多陪。
    git 生成本地密钥
    商品活动抽奖概率算法
    thinkadmin-controller下面的api接口访问形式
    SpringMVC的请求和响应
    SpringMVC注解解析和项目配置
    SpringMVC 概述
  • 原文地址:https://www.cnblogs.com/tonyyy/p/10545377.html
Copyright © 2011-2022 走看看