PAT 甲级 1146 Topological Order（25 分）

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This is a problem given in the Graduate Entrance Exam in 2018: Which of the following is NOT a topological order obtained from the given directed graph? Now you are supposed to write a program to test each of the options.

Input Specification:

Each input file contains one test case. For each case, the first line gives two positive integers N (≤ 1,000), the number of vertices in the graph, and M (≤ 10,000), the number of directed edges. Then M lines follow, each gives the start and the end vertices of an edge. The vertices are numbered from 1 to N. After the graph, there is another positive integer K (≤ 100). Then K lines of query follow, each gives a permutation of all the vertices. All the numbers in a line are separated by a space.

Output Specification:

Print in a line all the indices of queries which correspond to “NOT a topological order”. The indices start from zero. All the numbers are separated by a space, and there must no extra space at the beginning or the end of the line. It is graranteed that there is at least one answer.

Sample Input:

6 8
1 2
1 3
5 2
5 4
2 3
2 6
3 4
6 4
5
1 5 2 3 6 4
5 1 2 6 3 4
5 1 2 3 6 4
5 2 1 6 3 4
1 2 3 4 5 6

Sample Output:

3 4

题意：给出一个有向图，再给出一个所有结点排列的序列，判断这个序列是否是这个有向图的拓扑排序序列。 解题思路：拓扑排序是输出有向图中入度的结点，并将这个结点指向的结点入度减1，循环输出，直到所有结点入度为0

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#include <cstdio>
#include <set>
#include <vector>

using namespace std;

vector<set<int>> vs(1010);

bool is_topological(int *query, int n) {
    int *in = new int[n + 1];
    for (int i = 1; i <= n; i++) {
        in[i] = vs[i].size();
    }

    for (int i = 0; i < n; i++) {
        if (in[query[i]] != 0) return false;
        for (int j = 1; j <= n; j++) {
            if (vs[j].find(query[i]) != vs[j].end()) {
                in[j]--;
            }
        }
    }
    return true;
}

int main() {
    int n = 0, m = 0, start = 0, end = 0, k = 0;
    scanf("%d %d", &n, &m);
    for (int i = 0; i < m; i++) {
        scanf("%d %d", &start, &end);
        vs[end].insert(start);
    }
    scanf("%d", &k);
    vector<int> ans;
    for (int i = 0; i < k; i++) {
        int *query = new int[n + 1];
        for (int j = 0; j < n; j++) {
            scanf("%d", &query[j]);
        }
        if (!is_topological(query, n)) {
            ans.push_back(i);
        }
        delete[] query;
    }
    printf("%d", ans[0]);
    for (int i = 1; i < ans.size(); i++) {
        printf(" %d", ans[i]);
    }
    return 0;
}