poj 36876 Labeling Balls

Labeling Balls

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5819		Accepted: 1455

Description

Windy has N balls of distinct weights from 1 unit to N units. Now he tries to label them with 1 to N in such a way that:

1. No two balls share the same label.
2. The labeling satisfies several constrains like "The ball labeled with a is lighter than the one labeled with b".

Can you help windy to find a solution?

Input

The first line of input is the number of test case. The first line of each test case contains two integers, N (1 ≤ N ≤ 200) and M (0 ≤ M ≤ 40,000). The next M line each contain two integers a and b indicating the ball labeled with a must be lighter than the one labeled with b. (1 ≤ a, b ≤ N) There is a blank line before each test case.

Output

For each test case output on a single line the balls' weights from label 1 to label N. If several solutions exist, you should output the one with the smallest weight for label 1, then with the smallest weight for label 2, then with the smallest weight for label 3 and so on... If no solution exists, output -1 instead.

Sample Input

5

4 0

4 1
1 1

4 2
1 2
2 1

4 1
2 1

4 1
3 2

Sample Output

1 2 3 4
-1
-1
2 1 3 4
1 3 2 4

Source

POJ Founder Monthly Contest – 2008.08.31, windy7926778

 

/* 诡异的拓扑排序，要求编号最小的数字尽可能早被输出 诡异啊诡异 遵循一个原则 小的头不一定在前面，但是大的尾巴一定在后面 反向建图，找编号大的 赞，http://imlazy.ycool.com/post.2144071.html */ #include<cstdio> #include<cstring> int map[201][201]; int in[201],cou[201]; int n,m,num; int main() { int t; scanf("%d",&t); while(t--) { int a,b; scanf("%d%d",&n,&m); memset(in,0,sizeof(in)); memset(cou,0,sizeof(cou)); num=0; for(int i=1;i<=n;i++) map[i][0]=0; for(int i=0;i<m;i++) { scanf("%d%d",&a,&b); map[b][0]++; map[b][map[b][0]]=a; in[a]++; } while(1) { bool flag=0; for(int i=n;i>=1;i--) if(in[i]==0) { in[i]=-1; cou[i]=++num; flag=1; if(map[i][0]>0) { for(int j=1;j<=map[i][0];j++) in[map[i][j]]--; } break; } if(!flag) break; } if(num<n) printf("-1/n"); else { for(int i=1;i<=n-1;i++) printf("%d ",n-cou[i]+1); printf("%d/n",n-cou[n]+1); } } return 0; }