1146 Topological Order (25 分)

2019-05-19

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

作者: CHEN, Yue

单位: 浙江大学

时间限制: 200 ms

内存限制: 64 MB

代码长度限制: 16 KB

题目大意

给出一个有向图，要求判断所给序列是否为拓扑排序

分析

用in,out记录对应出入关系。对于给出的序列，判断其入度是否为0，然后删除该点的出边。

代码

#include <iostream>
#include <algorithm>
#include <vector>
#include <set>
#include <cmath>

using namespace std;
int n,m,k,t,first;
vector<set<int>> out;
vector<set<int>> in;
vector<int> temp;

int main() {
    cin>>n>>m;
    out.resize(n+1);
    in.resize(n+1);
    int a,b;
    for(int i=0;i<m;i++){
        cin>>a>>b;
        out[a].insert(b);
        in[b].insert(a);
    }
    
    cin>>k;
    for(int j=0;j<k;j++){
        vector<set<int>> tout=out;
        vector<set<int>> tin=in;
        temp.clear();
        for(int j=0;j<n;j++){
            cin>>t;
            temp.push_back(t);
        }
        for(int i=0;i<n;i++){
            if(tin[temp[i]].size()!=0){
                first++==0?cout<<"":cout<<" ";
                cout<<j;
                break;
            }
        for(int k:tout[temp[i]])
            tin[k].erase(temp[i]);
        }
    }
    return 0;
}