zoukankan      html  css  js  c++  java
  • Legal or Not

    Legal or Not

    Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)
    Total Submission(s) : 43   Accepted Submission(s) : 26
    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
    代码:两种方法:
      1 /*#include<stdio.h>
      2 #include<string.h>
      3 const int MAXN=110;
      4 int map[MAXN][MAXN];
      5 int que[MAXN];
      6 int N,M,flot;
      7 void topsort(){int k;
      8     for(int i=0;i<N;i++){
      9             int temp=-1;
     10         for(int j=0;j<N;j++){
     11             if(!que[j]){
     12             k=j;
     13             flot++;
     14         que[k]=-1;
     15         temp=0;
     16             break;
     17         }
     18         }
     19         if(temp==-1)break;
     20         for(int j=0;j<N;j++){
     21             if(map[k][j])que[j]--;
     22         }
     23     }
     24     if(flot==N)puts("YES");
     25     else puts("NO");
     26 }
     27 void initial(){
     28     memset(map,0,sizeof(map));
     29     memset(que,0,sizeof(que));
     30     flot=0;
     31 }
     32 int main(){
     33     int a,b;
     34         while(~scanf("%d%d",&N,&M),N){
     35                 initial();
     36             while(M--){
     37                 scanf("%d%d",&a,&b);
     38                 if(!map[a][b]){
     39                     map[a][b]=1;
     40                     que[b]++;
     41                 }
     42             }
     43             topsort();
     44         }
     45     return 0;
     46     }
     47 */
     48 #include<stdio.h>
     49 #include<string.h>
     50 #include<queue>
     51 using namespace std;
     52 const int MAXN=110;
     53 struct Node{
     54     int to,next;
     55 };
     56 Node edg[MAXN];
     57 int head[MAXN];
     58 int que[MAXN],N,flot;
     59 priority_queue<int,vector<int>,greater<int> >dl;
     60 void topsort(){
     61    // puts("fasdf/--");
     62     for(int i=0;i<N;i++){
     63             //printf("%d
    ",i);
     64         if(!que[i])dl.push(i);
     65     }
     66     //puts("fasdf/..");
     67     while(!dl.empty()){
     68         int k=dl.top();
     69         dl.pop();
     70         flot++;
     71         for(int j=head[k];j!=-1;j=edg[j].next){
     72             que[edg[j].to]--;
     73             if(!que[edg[j].to])dl.push(edg[j].to);
     74         }
     75     }
     76    // puts("fasdf/++");
     77     if(flot==N)puts("YES");
     78     else puts("NO");
     79 }
     80 void initial(){
     81     memset(head,-1,sizeof(head));
     82     memset(que,0,sizeof(que));
     83     while(!dl.empty())dl.pop();
     84     flot=0;
     85 }
     86 int main(){
     87     int M,a,b;
     88     while(~scanf("%d%d",&N,&M),N){
     89             initial();
     90         for(int i=0;i<M;i++){
     91                 scanf("%d%d",&a,&b);
     92             edg[i].to=b;
     93             edg[i].next=head[a];
     94             head[a]=i;
     95             que[b]++;
     96         }
     97         //puts("fasdf/**");
     98         topsort();
     99     }
    100     return 0;
    101 }
  • 相关阅读:
    数字化工厂仿真系统-易景空间数字孪生工厂
    会议小程序-智能会议助手在会务系统中的优势
    商场室内地图制作-商场导航-智慧商业综合体
    室内定位室内导航到底能带来什么?
    医院导航系统-智慧医院室内导航-院内导航系统
    室内地图制作-首款实时室内绘制室内地图-3D室内地图
    城市综合三维管网管理-城市三维GIS管线系统-易景地图三维管线地图制作平台
    如何制作好看的室内地图-室内电子地图-在线制作室内地图
    jQuery ui中sortable draggable droppable的使用
    综合CSS3 transition、transform、animation写的一个动画导航
  • 原文地址:https://www.cnblogs.com/handsomecui/p/4732553.html
Copyright © 2011-2022 走看看