uva 10305 Ordering Tasks （简单拓扑）

最新推荐文章于 2019-12-06 04:49:34 发布

SPZn_up

最新推荐文章于 2019-12-06 04:49:34 发布

阅读量789

点赞数

CC 4.0 BY-SA版权

分类专栏： ====ACM===== 图文章标签： uva C语言

本文链接：https://blog.youkuaiyun.com/llx523113241/article/details/43370691

====ACM===== 同时被 2 个专栏收录

312 篇文章

订阅专栏

图

13 篇文章

订阅专栏

本文解决UVA10305问题——任务排序，通过输入任务数量及任务间的依赖关系，使用拓扑排序算法求解可能的任务执行顺序。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

uva 10305 Ordering Tasks

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：拓扑排序。

解题思路：简单拓扑。

#include<stdio.h>
#include<string.h>
int gra[105][105], degree[105];
int n, m;
int main() {
	while (scanf("%d %d", &n, &m) == 2, (n || m)) {
		memset(gra, 0, sizeof(gra));
		memset(degree, 0, sizeof(degree));
		int a, b;
		for (int i = 0; i < m; i++) {
			scanf("%d %d", &a, &b);
			if (!gra[a][b]) {
				gra[a][b] = 1;
				degree[b]++;
			}
		}
		for (int i = 1; i <= n; i++) {
			for (int j = 1; j <= n; j++) {
				if (degree[j] == 0) {
					degree[j]--;
					if (i != n) printf("%d ", j);
					else printf("%d", j);
					for (int k = 1; k <= n; k++) {
						if (gra[j][k]) {
							degree[k]--;
						}
					}
					break;
				}
			}
		}
		printf("\n");
	}
}