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 6Sample Output:
3 4
Straight to the point.
Code:
#include <iostream> #include <cstring> #include <cstdio> #include <map> #define Max 1005 using namespace std; int n,m,k,limit[Max],exa[Max],a,b,mp [Max][Max],ans[Max],ant; /// There are several points ahead of the limit stored in a certain point int check() { int p[Max]; for(int i = 1;i <= n;i ++)///limit转到p { p[i] = limit[i]; } for(int i = 0;i < n;i ++) { if (p[exa[i]]) /// p is positive means the order is invalid { return 0 ; } for ( int j = 1 ;j <= n;j ++) /// If it is legal to reduce the points p restricted by it by 1 { if (mp[exa[i]][j])p[j] - - ; } } return 1; } intmain () { scanf("%d%d",&n,&m); for(int i = 0;i < m;i ++) { scanf("%d%d",&a,&b); mp [a] [b] = 1 ; limit[b] ++; } scanf("%d",&k); for(int i = 0;i < k;i ++) { for(int j = 0;j < n;j ++) { scanf("%d",&exa[j]); } if(!check()) { ans[ant ++] = i; } } for(int i = 0;i < ant;i ++) { if(i)printf(" %d",ans[i]); else printf("%d",ans[i]); } }