uva 10305 Ordering Tasks(拓扑排序）

最新推荐文章于 2019-07-25 14:28:15 发布

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最新推荐文章于 2019-07-25 14:28:15 发布

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分类专栏：图论-拓扑排序 UVA

本文链接：https://blog.youkuaiyun.com/keshuai19940722/article/details/9672157

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该博客介绍了如何解决UVA在线判题系统中的10305题目，即任务排序问题。通过使用拓扑排序的方法，处理任务之间的依赖关系，确保每个任务在所有前置任务完成后才能执行。输入包含多个实例，每个实例包含任务数量n和直接依赖关系m，输出应为一个可能的任务执行顺序。博客提供了解题思路和样例输入输出。

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Ordering Tasks

Input: standard input

Output: standard output

Time Limit: 1 second

Memory Limit: 32 MB

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：给出n组数据，前面的的序号表示这个任务先于后面序号的任务，要求将任务先后排序（不一定有一种）

解题思路：拓扑排序，先将数据存成有向图，遍历，输出入读为0的点，并且将该点所有出入抹掉。再遍历，知道所有点都已经输出。

#include<stdio.h>
#include<string.h>
#define N 105
int m, n, cnt;
int map[N][N], bo[N];

int judge(int k){
	for (int i = 1; i <= n; i++)
		if (map[k][i])
			return 0;
	return 1;
}

int clear(int k){
	for (int i = 1; i <= n; i++)
		map[i][k] = 0;
}

int main(){
	while (scanf("%d%d", &n, &m), n || m){
		// Init.
		memset(map, 0, sizeof(map));
		memset(bo, 0, sizeof(bo));
		cnt = 0;

		// Read.
		for (int i = 0; i < m; i++){
			int a, b; 
			scanf("%d%d", &a, &b);
			map[b][a] = 1;
		}

		// Handle.
		while (cnt < n){
			for (int i = 1; i <= n; i++){
				if (bo[i])	continue;
				if (judge(i)){
					if (cnt)
						printf(" ");
					printf("%d", i);
					bo[i] = 1;
					cnt++;
					clear(i);
					break;
				}
			}
		}
		printf("\n");
	}
	return 0;}