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Ordering Tasks From:UVA, 10305(拓扑排序)

2024-08-03 19:16:47   220人阅读



 Ordering Tasks

From:UVA, 10305

Submit

Time Limit: 3000 MS      Special Judge

 

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

题目大意：

用N份工作，M组关系，例如关系A,B表示要先完成B，才能完成A，让你输出一组工作完成的顺序。

解题思路：

  拓扑排序。


代码：

#include<iostream>
#include<cstdio>
#include<queue>
#include<string>
using namespace std;

const int maxn=110,maxm=100000;

struct edge{
    int u,v,next;
    edge(int u0=0,int v0=0,int next0=0){u=u0,v=v0,next=next0;}
}e[maxm];

int n,m,inDegree[maxn],head[maxn],cnt;
queue <int> path;

void initial(){
    cnt=0;
    for(int i=0;i<=n;i++){
        inDegree[i]=0;
        head[i]=-1;
    }
}

void adde(int u,int v){
    e[cnt].u=u,e[cnt].v=v,e[cnt].next=head[u],head[u]=cnt++;
}

void input(){
    int a,b;
    for(int i=0;i<m;i++){
        scanf("%d%d",&a,&b);
        adde(a,b);
        inDegree[b]++;
    }
}

void tuopu(){
    queue <int> q;
    for(int i=1;i<=n;i++)if(inDegree[i]==0) q.push(i);
    while(!q.empty()){
        int s=q.front();
        path.push(s);
        q.pop();
        for(int i=head[s];i!=-1;i=e[i].next){
            inDegree[e[i].v]--;
            if(inDegree[e[i].v]==0) q.push(e[i].v);
        }
    }
}

void outResult(){
    while(!path.empty()){
        printf("%d",path.front());
        path.pop();
        if(!path.empty()) printf(" ");
    }
    printf("\n");
}

int main(){
    while(cin>>n>>m&&(n||m)){
        initial();
        input();
        tuopu();
        outResult();
    }
    return 0;
}


Ordering Tasks From:UVA, 10305(拓扑排序)


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