PAT 1146 Topological Order（25 分）

1146 Topological Order（25 分）

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

Approach

题目大意：给出一个图，再给几组数据，让你判断这几组数据是否符合拓扑排序
思路方法：首先要熟悉拓扑排序的过程，就是每次将入度为零的点放入队列中，由此枚举数据判断此时的点是否入度已为零。

Code

void PrintAnswer(vector<int> &err) {
    for (int i = 0; i < err.size(); i++) {
        if (i == 0)
            printf("%d", err[i]);
        else
            printf(" %d", err[i]);
    }
}
void deindegree(int node,vector<int> &indegree,vector<vector<int>> &graph) {
    for (auto &v : graph[node]) {
        indegree[v]--;
    }
}
int main() {
    int N, M;
    scanf("%d%d", &N, &M);
    vector<vector<int>>graph(N + 1, vector<int>());
    vector<int>indegree(N + 1, 0);
    for (int i = 0; i < M; i++) {
        int a, b;
        scanf("%d%d", &a, &b);
        indegree[b]++;
        graph[a].push_back(b);
    }
    queue<int>q;
    for (int i = 1; i <= N; i++) {
        if (indegree[i] == 0)
            q.push(i);
    }
    int K;
    scanf("%d", &K);
    vector<int>err;
    for (int i = 0; i < K; i++) {
        vector<int> vindegree(indegree.begin(),indegree.end());
        vector<int> nums;
        for (int j = 0; j < N; j++) {
            int a;
            scanf("%d", &a);
            nums.push_back(a);
        }
        for (int j = 0; j < N; j++) {
            if (vindegree[nums[j]] == 0) {
                deindegree(nums[j], vindegree,graph);
            }
            else {
                err.push_back(i);
                break;
            }
        }
    }
    PrintAnswer(err);
    return 0;
}