Uva 10305 - Ordering Tasks

最新推荐文章于 2021-04-27 23:32:31 发布

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本文解析了一道关于任务依赖关系的拓扑排序问题，通过记录每个任务的入度数，采用迭代方式寻找并移除入度为0的任务，直至所有任务完成排序。此方法适用于解决任务执行顺序的问题。

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

拓扑排序的模板题。

解题思路：

读入图结构，记录每个节点的入度数（也就是有多少条边的终点为该节点）。
每次寻找入度为0的点，push_back到ans，然后寻找到的节点的入度减1（入度为-1则后续不会再被考虑），然后将该点连接的点的入度减1。这一步的目的可以理解为将入度为0的点从图中移除，将问题变为求剩下的结点的拓扑排序。
循环第二步直到所有结点均已进入ans

代码：

#include <bits/stdc++.h>
using namespace std;

const int maxn = 505;
int G[maxn][maxn];
int indegree[maxn];
int n,m,u,v;

int main(int argc, char const *argv[])
{
	while(cin >> n >> m && n != 0)
	{
		memset(indegree,0,sizeof(indegree));
		memset(G,0,sizeof(G));
		
		for(int i = 0;i < m; i++)
		{
			cin >> u >> v;
			G[u][v] = 1;
			indegree[v]++;
		}
		
		vector<int> ans;
		while(ans.size() != n)
		{
			for(int i = 1;i <= n; i++)
			{
				if(indegree[i] == 0)           
				{
					indegree[i]--;              // 将入度变为-1就不再有机会被加入ans 
					ans.push_back(i);
					for(int j = 1;j <= n; j++)   //将i连接的点的入度都减1 
						if(G[i][j])
							indegree[j]--;
				}
			}	
		}
		for(int i = 0;i < ans.size(); i++)
			cout << ans[i] << " ";
		cout << endl;
	}
	return 0;
}