POJ 3687-Labeling bils (reverse topological sorting)

POJ 3687-Labeling bils (reverse topological sorting)

Labeling bils

Time Limit:1000 MS		Memory Limit:65536 K
Total Submissions:11256		Accepted:3230

Description

Windy hasNBils of distinct weights from 1 unitNUnits. Now he tries to label them with 1NIn such a way that:

No two bils share the same label. The labeling satisfies several constrains like "The ball labeled AIs lighter than the one labeled 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) andM(0 ≤M≤ 40,000). The nextMLine each contain two integersAAndBIndicating the ball labeledAMust be lighter than the one labeledB. (1 ≤A, B≤N) There is a blank line before each test case.

Output

For each test case output on a single line the Ball' weights from label 1 to labelN. If several solutions exist, you shoshould 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

54 04 11 14 21 22 14 12 14 13 2

Sample Output

1 2 3 4-1-12 1 3 41 3 2 4

There are T groups of test data. The first row of each group of test data has two numbers n and m. It indicates that there are n balls. In the next m rows, each row has two numbers a and B, indicating that the ball is lighter than the B ball, so that you can output the weight of the ball from small to large. If it is the same weight, output in Lexicographic Order.

Thought: This is slightly different from the previous topological sorting, not considering the inbound degree of 0. Because assume n is 4, 4 <1. then, the output columns are, instead of 4, 1, 2, and 3. therefore, we should first put the heaviest one behind it, and then output it in the lexicographically ascending order.

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        Using namespace std; int map [210] [210]; int out [210]; void topo (int n) {int I; int cnt = n; int weight [210]; priority_queue
       
         Q; // priority queue, sorted in ascending order. For (I = 1; I <= n; I ++) {if (out [I] = 0) q. push (I) ;}while (! Q. empty () {int k = q. top (); q. pop (); weight [k] = cnt --; for (I = 1; I <= n; I ++) {if (map [I] [k]) {out [I] --; if (out [I] = 0) q. push (I) ;}}if (cnt> 0) printf ("-1 \ n"); else {for (I = 1; I