zoukankan      html  css  js  c++  java
  • HDU 1937 J

    J - Justice League
    Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

    Description

    Thirty five years ago, a group of super heroes was chosen to form the Justice League, whose purpose was to protect the planet Earth from the villains. After all those years helping mankind, its members are retiring and now it is time to choose the new members of the Justice League. In order to keep their secret identity, let’s say, secret, super heroes usually use an integer number to identify themselves. There are H super heroes on Earth, identified with the integers from 1 to H. With a brief look at the newspapers anyone can find out if two super heroes have already worked together in a mission. If this happened, we say that the two heroes have a relationship. 

    There must be only one Justice League in the world, which could be formed by any number of super heroes (even only one). Moreover, for any two heroes in the new league, they must have a relationship. 

    Besides, consider the set of the heroes not chosen to take part in the Justice League. For any two heroes on that set, they must not have a relationship. This prevents the formation of unofficial justice leagues. 

    You work for an agency in charge of creating the new Justice League. The agency doesn’t know if it is possible to create the League with the restrictions given, and asked for your programming skills. Given a set of super heroes and their relationships, determine if it is possible to select any subset to form the Justice League, according to the given restrictions. 
     

    Input

    The input is composed of several test cases. The first line of each test case contains two integers separated by a single space, H (2 <= H <= 5×10 4) and R (1 <= R <= 10 5), indicating, respectively, the number of heroes and the number of relationships. Each of the following R lines contains two integers separated by a single space, A and B (1 <= A < B <= H), indicating that super hero A has a relationship with super hero B. Note that if A has a relationship with B, B also has a relationship with A. A relationship is never informed twice on a test case. 
    The end of input is indicated by H = R = 0. 
     

    Output

    For each test case in the input print a single line, containing the uppercase letter “Y” if it is possible to select a subset of heroes to form the Justice League according to the given restrictions, or the uppercase letter “N” otherwise.
     

    Sample Input

    5 5
    1 2
    2 3
    1 3
    1 4
    3 5
    9 8
    1 2
    2 3
    3 4
    4 5
    5 6
    6 7
    7 8
    8 9
    4 3
    1 2
    2 3
    3 4
    0 0
     

    Sample Output

    Y N Y
     
     
     
     
    #include<stdio.h>
    #include<string.h>
    #include<iostream>
    #include<vector>
    #include<algorithm>
    using namespace std;
    int ind[60000];
    int vis[60000];
    int t[60000];
    bool cmp(int a,int b){
       return ind[a]<ind[b];
    }
    int main(){
       int n,m;
       while(scanf("%d%d",&n,&m)!=EOF){
            if(n==0&&m==0)
            break;
           memset(ind,0,sizeof(ind));
           memset(t,0,sizeof(t));
           memset(vis,0,sizeof(vis));
           for(int i=0;i<=n;i++){
            t[i]=i;
           }
           int u,v;
           vector<int>q[60000];
           for(int i=1;i<=m;i++){
              scanf("%d%d",&u,&v);
              ind[u]++;
              ind[v]++;
              q[u].push_back(v);
              q[v].push_back(u);
           }
          sort(t+1,t+n+1,cmp);
    
          for(int i=1;i<=n;i++){
                int temp=t[i];
            if(vis[temp]==0){
                    for(int j=0;j<q[temp].size();j++){
                        vis[q[temp][j]]=1;
                        ind[q[temp][j]]--;
                    }
            }
          }
         int ans=0;
         int tmin=60000;
         for(int i=1;i<=n;i++){
            if(vis[i]){
                ans++;
                tmin=min(tmin,ind[i]);
            }
         }
    
         if(ans==tmin+1)
            printf("Y
    ");
         else
            printf("N
    ");
    
       }
       return 0;
    }
     
     
     
     
     
     
  • 相关阅读:
    【Luogu】P3369 【模板】普通平衡树(树状数组)
    文艺平衡树 lg3391(splay维护区间入门)
    普通平衡树 lg3369
    noip2018游记
    webview与壳交互的几种方式
    iOS、Android 之类的如何调试 Web APP
    box-sizing属性
    Hybridapp /webapp调试工具
    DOS 批处理高级教程精选合编
    瀑布流Masonry学习
  • 原文地址:https://www.cnblogs.com/13224ACMer/p/4678482.html
Copyright © 2011-2022 走看看