Ordering Tasks(拓扑排序)

最新推荐文章于 2019-10-19 20:01:14 发布

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最新推荐文章于 2019-10-19 20:01:14 发布

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分类专栏：拓扑排序

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本文介绍了一个典型的任务排序问题，通过图论中的拓扑排序算法来解决任务间的依赖关系。使用C++实现，详细展示了如何读取任务之间的依赖关系并进行排序。

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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,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 n is h 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

题意

john 有n个任务要完成，但是每个任务完成有先后顺序，即完成一个任务必须先完成他的准备任务；

第一行 输入任务数量 n ,任务先后关系 m

代码如下

#include<stdio.h>
#include<string.h>
#include<queue>
using namespace std;
int n,m;
int map[150][150];     //存放关系矩阵
int book[150][150];    //记录之前是否存过，防止重复存关系，造成入度增加
int r[150];            //记录点的入度
int main(){
    while(scanf("%d%d",&n,&m)!=EOF){
        if(n==0&&m==0) break;
        memset(map,0,sizeof(map));     //初始化为0
        memset(book,0,sizeof(book));
        memset(r,0,sizeof(r));
        int i;
        for(i=0;i<m;i++){              //读入关系
            int a,b;
            scanf("%d%d",&a,&b);
            if(book[a][b]==0){
                map[a][b]=1;
                r[b]++;
            }
        }
        queue<int>p;
        for(i=1;i<=n;i++){        //此题是 先入度为0 的点优先 后小的点优先 因此要用到队列
            if(r[i]==0)
                p.push(i);
        }
        int cnt=0;
        int t[150];
        while(!p.empty()){
            int temp=p.front();
            p.pop();
            t[cnt++]=temp;
            int j;
            for(j=1;j<=n;j++){
                if(map[temp][j]==1){
                    r[j]--;
                    if(r[j]==0)
                        p.push(j);
                }
            }
        }
        for(i=0;i<cnt-1;i++)
            printf("%d ",t[i]);
        printf("%d\n",t[i]);
    }
return 0;
}