PAT 1146 Topological Order[难]
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
题目大意:给出一个有向图,并且给定K个序列,判断这个序列是否是拓扑序列。
//我一看见我想的就是,得用邻接表存储图,然后对每一个输入的序列,都进行判断,基本上复杂度是非常高的,就是对每一个序列中的数,判断其之前出现的每一个数是否是它的next,这样来判断,然后写的不对。
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include <iostream> #include<vector> #include<map> #include<algorithm> using namespace std; vector<vector<int>> graph; int main(){ int n,m; cin>>n>>m; graph.resize(n+1); int f,t; for(int i=0;i<m;i++){ cin>>f>>t; graph[f].push_back(t);//因为是单向图 } int u; cin>>u; vector<int> vt(n); vector<int> ans; for(int i=0;i<u;i++){//复杂度是O(n^2),稍微有点高啊。 for(int j=0;j<n;j++) cin>>vt[j]; //检查在其之前出现的是否是在这个图的next里。 bool flag=true; for(int j=1;j<n;j++){ for(int k=0;k<j;k++){//在这还得遍历vt[j] for(int v=0;v<graph[j].size();v++){ if(vt[k]==graph[j][v]){ cout<<vt[k]<<" "<<graph[j][v]<<"\n"; ans.push_back(i); flag=false; break; } } if(!flag)break; } if(!flag)break; } } for(int i=0;i<ans.size();i++){ cout<<ans[i]; if(i!=ans.size()-1)cout<<' '; } return 0; }
结果:
//真的很奔溃啊,怎么每个都是不对的,那个2 2 到底是什么意思?我明天再看看吧。
//柳神的代码:
//根据入度出度来判断,非常可以了。。
学习了,要多复习。