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.

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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，并给出k个查询，输出所有不是拓扑排序的查查询。

思路：
1.记录每个点的next，入度。
2.vector拷贝的相关操作：C++ vector拷贝使用总结

题解：

#include
#include
#include
using namespace std;
//定义每个节点的入度和对应出去的节点
struct node
{
    int in = 0;
    vector out;
};
int main() {
    int n, m;
    scanf("%d %d", &n, &m);
    vector nodes(n + 1);
    int a, b;
    for (int i = 0; i < m; i++) {
        scanf("%d %d", &a, &b);
        nodes[a].out.push_back(b);
        nodes[b].in++;
    }
    int query;
    scanf("%d", &query);
    int cnt = 0;
    for (int i = 0; i < query; i++) {
        bool flag = true;
        //每次查询对nodes的副本进行修改。
        vector tNodes(nodes);
        for (int j = 0; j < n; j++) {
            int t;
            scanf("%d", &t);
            //即使已经判断出来不是拓扑排序了，仍然需要接受剩下的输入。
            if (!flag) continue;
            if (tNodes[t].in == 0) {
                for (int k = 0; k < tNodes[t].out.size(); k++) {
                    tNodes[tNodes[t].out[k]].in--;
                }
            }
            else {
                if (cnt != 0) printf(" ");
                printf("%d", i);
                cnt++;
                flag = false;
            }
        }
    }
    return 0;
}