POJ 3687 Labeling Balls-拓扑排序

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题目：

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

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

方法：拓扑排序

代码：

#include<iostream>using namespace std;int adjMat[201][201] = { 0 };//0表示两者没有关系,下标从1开始，也是球的标号int finishOrder[201] = { 0 };//节点完成搜索的顺序int beginOrder[201] = { 0 }; //节点开始搜索的顺序int num, N, M, t1, t2;               //N(1 ≤ N ≤ 200):N个球(N个节点) and M(0 ≤ M ≤ 40000)(M个限制关系)int dfs(int label_n) //标号n{if (beginOrder[label_n] > 0) //如果该节点已经被开始搜索，则说明出现了环路return -1;beginOrder[label_n] = ++t1;int j = 0;for (j = 1; j <= N; ++j){if (adjMat[label_n][j] == 1 && finishOrder[j] == 0){if (dfs(j) == -1)return -1;}}if (j - 1 == N)finishOrder[label_n] = ++t2;return 0;}int topoSort(){for (int i = 1; i <= N; ++i)if (finishOrder[i] == 0)if (dfs(i) == -1)return -1;return 0;}void main(){int i, j;cin >> num;while (num--){t1 = 0;t2 = 0;cin >> N >> M;for (int m = 0; m < M; m++){cin >> i >> j;    //i标号的重量小于j标号的重量，通常i>jadjMat[j][i] = 1; //反过来就是，j标号的重量大于i标号的重量}if (topoSort() == -1)cout << "出现了环路，无解！";elsefor (int k = 1; k <= N; ++k)cout << finishOrder[k] << ' ';cout << endl;memset(adjMat, 0, sizeof(adjMat));memset(finishOrder, 0, 4 * (N + 1));memset(beginOrder, 0, 4 * (N + 1));}}

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