POJ 3687 Labeling Balls()

Labeling Balls

Time Limit: 1000MS
Memory Limit: 65536K

Total Submissions: 9641
Accepted: 2636

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

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::做这道题，原先的思路就是从前面往后扫，入度为0的就赋予未被使用的最小值，可是这样无法得到字典序最小。

看了题解才知道，这道题要逆向建图，且从后往前扫，入度为0的赋予当前未使用的最大值。

 

   1: #include <iostream>

   2: #include <cstdio>

   3: #include <cstring>

   4: #include <algorithm>

   5: using namespace std;

   6: typedef long long ll;

   7: const int maxn=40010;

   8: int head[maxn],in[220],ans[220],vis[220];

   9: int cnt,n,m;

  10:  

  11: struct node

  12: {

  13:     int u,v,p;

  14: }e[maxn];

  15:  

  16: int topo()

  17: {

  18:     memset(vis,0,sizeof(vis));//标记数组

  19:     int k=n;

  20:     int i,j,u;

  21:     for(i=1; i<=n; i++)

  22:     {

  23:         for(u=n; u>0; u--)

  24:         {

  25:             if(!vis[u]&&in[u]==0)

  26:                break;

  27:         }

  28:         if(u<=0) return 0;

  29:         vis[u]=1;

  30:         ans[u]=k--;

  31:         for(j=head[u]; j!=-1; j=e[j].p)

  32:         {

  33:             in[e[j].v]--;

  34:         }

  35:     }

  36:     return 1;

  37: }

  38:  

  39: int main()

  40: {

  41:     int T;

  42:     scanf("%d",&T);

  43:     while(T--)

  44:     {

  45:         int ok=1,a,b;

  46:         cnt=0;

  47:         scanf("%d%d",&n,&m);

  48:         memset(head,-1,sizeof(head));

  49:         memset(in,0,sizeof(in));

  50:         while(m--)

  51:         {

  52:             scanf("%d%d",&a,&b);

  53:             e[cnt].u=b;

  54:             e[cnt].v=a;

  55:             e[cnt].p=head[b];

  56:             head[b]=cnt++;

  57:             in[a]++;

  58:             if(a==b) ok=0;

  59:         }

  60:         if(ok==0||topo()==0)

  61:            printf("-1\n");

  62:         else

  63:         {

  64:             for(int i=1; i<=n; i++)

  65:                 printf("%d ",ans[i]);

  66:             printf("\n");

  67:         }

  68:         //printf("\n");

  69:     }

  70:     return 0;

  71: }