題目鏈接：https://cn.vjudge.net/problem/UVA-10305（忍不住uva連接滿）

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

?

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

題意：給你n個點（從1開始），還有要求一些點i必須在j前面。輸出任意一種排序。

題解：裸的拓撲排序，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

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轉載于:https://www.cnblogs.com/denghaiquan/p/6668417.html

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