zoukankan      html  css  js  c++  java
  • uva 10305


    /*************************************************************************
         File Name: 10305four.cpp
         Author: yubo
         Mail: yuzibode@126.com 
         Created Time: 2014年06月02日 星期一 00时23分06秒
         学习重点:
    reference
    John has n tasks to do. Unfortunately, the tasks are not independent and the execution of one task is only possible if other tasks have already been executed.
    
    Input
    
    The input will consist of several instances of the problem. Each instance begins with a line containing two integers, 1 <= n <= 100 and m. n is the number of tasks (numbered from 1 to n) and m is the number of direct precedence relations between tasks. After this, there will be m lines with two integers i and j, representing the fact that task i must be executed before task j. An instance with n = m = 0 will finish the input.
    
    Output
    For each instance, print a line with n integers representing the tasks in a possible order of execution.
    Sample Input
    
    5 4
    
    1 2
    
    2 3
    
    1 3
    
    1 5
    
    0 0
    
    Sample Output
    
    1 4 2 5 3:
    http://sbp810050504.blog.51cto.com/2799422/901022
     ************************************************************************/
    
    #include<cstring>
    #include<cstdio>
    #include<iostream>
    #include<stack>
    using namespace std;
    #define MAXN 105
    //have time according to text to write
    
    struct ArcNode{
    	int to;
    	struct ArcNode* next;
    };
    ArcNode* List[MAXN];
    int n,m;
    int counts[MAXN];
    int output[MAXN];
    void toposort()
    {
    stack<int> s;
    	memset(output,0,sizeof(output));
    	for(int i=0;i<n;i++){
    		if(counts[i]==0){
    
    		
    			s.push(i);
    		}
    	}
    	bool bcycle=false;
    	ArcNode* temp=new ArcNode;
    	int k=0;
    	for(int i=0;i<n;i++){
    		if(s.empty()){
    			bcycle=true;
    			break;
    
    		}
    		else{
    			int j=s.top();s.pop();
    			output[k++]=j+1;
    			temp=List[j];
    			while(temp!=NULL){
    				int f=temp->to;
    				counts[f]--;
    				if(counts[f]==0){
    					s.push(f);
    				}
    				temp=temp->next;
    			}
    		}
    	}if(bcycle)
    	printf("The Network has a cycle
    ");
    	else
    	{
    		for(int l=0;l<n-1;l++)
    			printf("%d ",output[l]);
    			printf("%d
    ",output[n-1]);
    	}
    }
    void del()
    {
    	ArcNode* temp=new ArcNode;
    	for(int i=0;i<n;i++)
    	{
    		temp=List[i];
    		while(temp!=NULL){
    			List[i]=temp->next;
    			delete temp;
    			temp=List[i];
    		}
    
    	}
    
    }
    int main()
    {
    	//freopen("in.txt","r",stdin);
    	int u,v;
    	while(scanf("%d%d",&n,&m)){
    		if(!n&&!m)
    			break;
    		memset(List,0,sizeof(List));
    		memset(counts,0,sizeof(counts));
    		for(int i=0;i<m;i++){
    			scanf("%d%d",&u,&v);
    			u--;v--;
    			counts[v]++;
    			ArcNode* temp=new ArcNode;
    			temp->to=v;
    			temp->next=NULL;
    			if(temp==NULL)
    				List[u]=temp;
    			else{
    				temp->next=List[u];
    				List[u]=temp;
    			}
    		}
    
    		ArcNode* temp=new ArcNode;
    		toposort();
    		del();
    	}
    }
    


    版权声明:本文博客原创文章,博客,未经同意,不得转载。

  • 相关阅读:
    springnodejs
    js CacheQueue
    权重练习
    架构师速成8.3-可用性之分库分表
    架构师速成8.3-可用性之分布式
    架构师速成8.3-可用性
    架构师速成8.3-架构师必须要了解的规则(转)
    架构师速成6.15-开发框架-单点登录
    架构师速成6.14-开发框架-异常处理
    架构师速成6.13-开发框架-前后结合
  • 原文地址:https://www.cnblogs.com/bhlsheji/p/4728572.html
Copyright © 2011-2022 走看看