PAT日志 1146「建议收藏」

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顽强的小白

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.

在这里插入图片描述

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

题目解析

判断所给的序列是不是拓扑排序 我的思路很暴力，拓扑排序本来就是一个一个拿掉图上的顶点，这些被拿掉的顶点只有一个要求，就是不能有其他顶点指向它，因此我的思路就是，判断当点要拿掉的结点有没有被谁指着，如果有就不是拓扑排序。

代码实现

#include <cstdio> 
#include <vector> 
#include <algorithm> 

using namespace std; 

const int maxn=1005;

bool vis[maxn];
int G[maxn][maxn];
vector<int> ans;int main(){ 
   
 fill(G[0],G[0]+maxn*maxn,0);
 int n,e,u,v;
 scanf("%d%d",&n,&e);
 for(int i=0;i<e;++i){ 
   
  scanf("%d%d",&u,&v);
  G[u][v]=1;
 }
 int m;
 scanf("%d",&m);
 for(int i=0;i<m;++i){ 
   
  fill(vis,vis+n+1,true); //初始状态 灯全亮 
  int flag=0;
  for(int j=0;j<n;++j){ 
   
   scanf("%d",&u);
   for(int k=1;k<=n;++k){ 
   
    if(vis[k]==true&&G[k][u]==1){ 
   
     flag=1;
     break; 
    }
   }
   vis[u]=false;   //把这个顶点灭灯 
  }
  if(flag==1) ans.push_back(i);
 }
 for(int i=0;i<ans.size();++i) { 
   
  printf("%d",ans[i]);
  if(i<ans.size()-1) printf(" ");
  else printf("\n");
 }
} 

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