zoukankan      html  css  js  c++  java
  • hdu 3342

    Legal or Not

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 4322    Accepted Submission(s): 1926


    Problem Description
    ACM-DIY is a large QQ group where many excellent acmers get together. It is so harmonious that just like a big family. Every day,many "holy cows" like HH, hh, AC, ZT, lcc, BF, Qinz and so on chat on-line to exchange their ideas. When someone has questions, many warm-hearted cows like Lost will come to help. Then the one being helped will call Lost "master", and Lost will have a nice "prentice". By and by, there are many pairs of "master and prentice". But then problem occurs: there are too many masters and too many prentices, how can we know whether it is legal or not?

    We all know a master can have many prentices and a prentice may have a lot of masters too, it's legal. Nevertheless,some cows are not so honest, they hold illegal relationship. Take HH and 3xian for instant, HH is 3xian's master and, at the same time, 3xian is HH's master,which is quite illegal! To avoid this,please help us to judge whether their relationship is legal or not. 

    Please note that the "master and prentice" relation is transitive. It means that if A is B's master ans B is C's master, then A is C's master.
     
    Input
    The input consists of several test cases. For each case, the first line contains two integers, N (members to be tested) and M (relationships to be tested)(2 <= N, M <= 100). Then M lines follow, each contains a pair of (x, y) which means x is y's master and y is x's prentice. The input is terminated by N = 0.
    TO MAKE IT SIMPLE, we give every one a number (0, 1, 2,..., N-1). We use their numbers instead of their names.
     
    Output
    For each test case, print in one line the judgement of the messy relationship.
    If it is legal, output "YES", otherwise "NO".
     
    Sample Input
    3 2
    0 1
    1 2
    2 2
    0 1
    1 0
    0 0
     
    Sample Output
    YES
    NO
    拓扑模板题
     1 #include<string>
     2 #include<cstdio>
     3 #include<iostream>
     4 #include<vector>
     5 #include<queue>
     6 #include<stack>
     7 #include<algorithm>
     8 #include<cstring>
     9 #include<stdlib.h>
    10 #include<string>
    11 #include<cmath>
    12 using namespace std;
    13 #define pb push_back
    14 vector<int >p[10010];
    15 int in[10010],n,m,deep[10010],num[10010];
    16 void init(){
    17     memset(in,0,sizeof(in));
    18     memset(deep,0,sizeof(deep));
    19     memset(num,0,sizeof(num));
    20     for(int i=0;i<=n;i++) p[i].clear();
    21 }
    22 void tuopu(){
    23     queue<int >ak_47;
    24     int cnt=0;
    25     for(int i=0;i<n;i++)
    26     if(in[i]==0)
    27         ak_47.push(i),deep[i]=1,num[1]++;
    28     while(!ak_47.empty()){
    29         int pos=ak_47.front();
    30         cnt++;
    31         for(int i=0;i<p[pos].size();i++){
    32             int to=p[pos][i];
    33             if(--in[to]==0) ak_47.push(to),deep[to]=deep[pos]+1,num[deep[to] ]++;
    34         }
    35         ak_47.pop();
    36     }
    37     if(cnt<n) cout<<"NO"<<endl;
    38     else cout<<"YES"<<endl;
    39 
    40 }
    41 int main(){
    42     while(cin>>n>>m,n){
    43         init();
    44         for(int i=1;i<=m;i++){
    45             int a,b;
    46             scanf("%d%d",&a,&b);
    47             in[b]++;
    48             p[a].pb(b);
    49         }
    50         tuopu();
    51     }
    52 }
  • 相关阅读:
    forEach 不能跳出循环;用some 或者every 代替
    echarts图表不重新渲染
    vue 的el-tree获取选中节点的集合执行多次问题
    vue 2.6版本 手动配置json文件显示隐藏
    echart category series 数据多个 长度不对应 对应的数据一定要用字符串 不要用数字
    nginx前端配置后端
    UCOS多任务下有效的喂狗的方式
    判断数据类型
    PDFJS插件带添加header以及携带授权
    vue中控制浏览器前进和后退
  • 原文地址:https://www.cnblogs.com/ainixu1314/p/3896055.html
Copyright © 2011-2022 走看看