POJ 3687 Labeling Balls

Labeling Balls

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 14617		Accepted: 4278

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:

No two balls share the same label.
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

Sample Output

1 2 3 4
-1
-1
2 1 3 4
1 3 2 4
-------------------------------------------------------------------------------------------------------------------手动分割线-----------------------------------------------------------------------------------------------------------------
题目大意：
　　给你1~N个小球，要求从轻到重排序。
　　第一行输入T（T表示测试数据个数）
　　第二行输入N,M （N是小球数量，M是想描述的个数）
　　下面接着M行，每一行有两个数a,b  表示a比b轻。
　　
　　最后从轻到重输出小球编号
　　无法排序则输出-1
大致思路：
　　逆序的拓扑排序。
代码送上：

 1 #include<stdio.h>
 2 #include<algorithm>
 3 #include<string.h>
 4 #include<queue>
 5 using namespace std;
 6 #define N 205
 7 int e[N][N];
 8 int in[N];
 9 int ans[N];
10 int sum;
11 bool tuopu(int n)
12 {
13     priority_queue<int> q;
14     for(int i=1;i<=n;i++)
15         if(in[i]==0)q.push(i);
16     for(int i=1;i<=n;i++)
17     {
18         if(q.empty())return false;
19         else
20         {
21             int x=q.top();q.pop();
22             ans[x]=sum--;
23             for(int j=1;j<=n;j++)
24                 if(e[x][j]&&(--in[j])==0)q.push(j);
25         }
26     }
27     return true;
28 }
29 int main()
30 {
31     int t;
32     int n,m;
33     int a,b;
34     scanf("%d",&t);
35     while(t--)
36     {
37         scanf("%d%d",&n,&m);
38         sum=n;
39         memset(in,0,sizeof(in));
40         memset(e,0,sizeof(e));
41         for(int i=0;i<m;i++)
42         {
43 
44             scanf("%d%d",&a,&b);
45             if(e[b][a]==0)in[a]++;
46             e[b][a]=1;
47         }
48        if(tuopu(n))
49         {
50             for(int i=1;i<=n;i++)
51             {
52                 printf("%d",ans[i]);
53                 if(i!=n)printf(" ");
54             }
55             printf("
");
56         }
57         else printf("-1
");
58     }
59         return 0;
60 }