PAT-A 真题 – 1146 Topological Order

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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.

gre.jpg

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

题目大意:拓扑排序,给定一个有向图,判断哪个序列是拓扑序列,把给定的序列从0 - n-1进行编号,输出不是拓扑排序的编号。

这道题用邻接表来存储有向图会比较方便,因为涉及到入度的计算。建立两个vector,一个存放邻接表,另一个存放入度。之后给定的序列中,每读入一个数字,判断其入度是否为0,若不是,则不是拓扑排序,如果是,将其出边的所有点的入度-1,不断循环即可。

代码:

#include <iostream>
#include <bits/stdc++.h>
using namespace std;

int main(){
  int pcnt, vcnt;
  cin >> pcnt >> vcnt;
  vector<int> inv(pcnt + 1, 0);
  vector<vector<int> > G(pcnt + 1);
  for(int i = 0; i < vcnt; i++){
    int l, r;
    cin >> l >> r;
    inv[r]++;
    G[l].push_back(r);
  }
  int qcnt;
  cin >> qcnt;
  vector<int> ans;
  for(int i = 0; i < qcnt; i++){
    vector<int> tmpinv = inv;
    int flag = false;
    for(int i = 0; i < pcnt; i++){
      int n;
      cin >> n;
      if(tmpinv[n] != 0) flag = true;
      if(flag == false)
        for(int it : G[n]) tmpinv[it]--;
    }
    if(flag) ans.push_back(i);
  }
  for(int i = 0; i < ans.size(); i++){
    if(i) cout << ' ';
    cout << ans[i];
  }
  cout << endl;
  return 0;
}

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