【PAT甲级A1146】Topological Order (25分)（c++）

1146 Topological Order (25分)

作者：CHEN, Yue
单位：浙江大学
代码长度限制：16 KB
时间限制：400 ms
内存限制：64 MB

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：

根据所给有向无环图，判断是否是正确的拓扑顺序输出。

思路：

回顾有向无环图的拓扑顺序输出，是记录每个点的入度，将入度为零的结点放入队列，然后从队列中输出时，将其子结点的入度-1，如有入度为零的结点那么继续放入队列，直到队列为空。那么这题一样，根据所给图，记录每个点的入度，顺序读入所给数列，判断其入度是否为零，并将其子结点入度-1，若不为零，那么这个顺序是错误的。

参考代码：

#include <cstdio>
#include <algorithm>
#include <vector>

using namespace std;
const int maxn = 1010;
vector<int> G[maxn], indegree(maxn), v;

int main() {
    int n, m, k, a, b;
    scanf("%d%d", &n, &m);
    for (int i = 0; i < m; i++) {
        scanf("%d%d", &a, &b);
        G[a].push_back(b);
        indegree[b]++;
    }
    scanf("%d", &k);
    for (int i = 0; i < k; i++) {
        bool flag = 0;
        vector<int> ind = indegree;
        for (int j = 0; j < n; j++) {
            scanf("%d", &a);
            if (ind[a] != 0)flag = 1;
            for (int t = 0; t < G[a].size(); t++)
                ind[G[a][t]]--;
        }
        if (flag)v.push_back(i);
    }
    for (int i = 0; i < v.size(); i++) {
        if (i != 0)printf(" ");
        printf("%d", v[i]);
    }
    return 0;
}

如有错误，欢迎指正