1146 Topological Order (25分)

最新推荐文章于 2021-03-13 09:46:30 发布

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本文解析了一道2018年研究生入学考试中的编程题目，该题要求编写程序判断给定的顶点序列是否为有向图的有效拓扑排序。输入包括图的顶点数、边数、边的起点和终点，以及一系列顶点排列，输出不符合拓扑排序的查询索引。

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

#include<iostream>
#include<vector>

using namespace std;

int main(){

    int n, m;

    cin >> n >> m;
    // vector<int> in_graph[1001];
    vector<int> out_graph[1001];
    vector<int> pre_vec(n + 1);
    for(int i = 0; i < m; i++){

        int start, end;
        scanf("%d %d", &start, &end);
        out_graph[start].push_back(end);
        // in_graph[end].push_back(start);
        pre_vec[end]++;


    }

    int k;
    cin >> k;
    
    vector<int> out_result;
    for(int i = 0; i < k; i++){
        vector<int> temp_pre_vec(pre_vec.begin(), pre_vec.end());
        bool flag = true;
        vector<int> in_result(n);
        for(int z = 0; z < n; z++)
            scanf("%d", &in_result[z]);
            

        for(int z = 0; z < n; z++){

            int temp = in_result[z];

            if(temp_pre_vec[temp] > 0){
                flag = false;
                break;
            }
            for(int j = 0; j < out_graph[temp].size(); j++){

                temp_pre_vec[out_graph[temp][j]]--;

            }
            
            

        }
        if(!flag){
            out_result.push_back(i);
        }


        
    }
    printf("%d", out_result[0]);
    for(int i = 1; i < out_result.size(); i++)
        printf(" %d", out_result[i]);

    return 0;
}