UVA10305 Ordering Tasks

题目链接：https://cn.vjudge.net/problem/UVA-10305（忍不住uva连接满）

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

***************************************************************************************

题意：给你n个点（从1开始），还有要求一些点i必须在j前面。输出任意一种排序。

题解：裸的拓扑排序，直接套模板。

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cstdlib>
 5 #include <string>
 6 #include <vector>
 7 #include <map>
 8 #include <set>
 9 #include <queue>
10 #include <sstream>
11 #include <algorithm>
12 using namespace std;
13 #define pb push_back
14 #define mp make_pair
15 #define ms(a, b)  memset((a), (b), sizeof(a))
16 //#define LOCAL
17 typedef long long LL;
18 const int inf = 0x3f3f3f3f;
19 const int maxn = 100+10;
20 const int mod = 1e9+7;
21 int gap[maxn][maxn];
22 int topo[maxn], c[maxn], t;
23 int n, m;
24 bool dfs(int u){
25     c[u] = -1;
26     for(int v = 1; v<=n; v++)   if(gap[u][v]){
27         if(c[v]< 0) return false;
28         else if(!c[v] && !dfs(v) )  return false;
29     }
30     c[u] = 1;
31     topo[--t] = u;
32     return true;
33 }
34 bool toposort(){
35     t = n+1;
36     ms(c, 0);
37     for(int u = 1; u<=n ;u++)   if(!c[u])
38         if(!dfs(u)) return false;
39     return true;
40 }
41 int main()
42 {
43     #ifdef LOCAL
44         freopen("input.txt" , "r", stdin);
45     #endif // LOCAL
46     while(~scanf("%d%d", &n, &m)){
47         if(n==0 && m==0)    break;
48         ms(gap, 0);
49         for(int i = 0;i<m;i++){
50             int a, b;
51             scanf("%d%d", &a, &b);
52             gap[a][b] = 1;
53         }
54         if(toposort()){
55             for(int i =1 ; i<=n ;i++)
56                 if(i==1)    printf("%d", topo[i]);
57                 else    printf(" %d", topo[i]);
58             printf("\n");
59         }
60     }
61     return 0;
62 }

View Code

转载于:https://www.cnblogs.com/denghaiquan/p/6668417.html