PAT (Advanced Level) Practice 1146 Topological Order (25 分) 拓撲排序
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
程式碼如下:
#include <cstdio> #include <cstring> #include <algorithm> #include <iostream> #include <vector> using namespace std; const int maxn=1005; int n,m,q; int in[maxn],out[maxn]; int tin[maxn],re[maxn]; vector<int> ve[maxn],no; void init() { memset (in,0,sizeof(in)); } void tinit() { for (int i=1;i<=n;i++) { tin[i]=in[i]; } } int main() { init(); scanf("%d%d",&n,&m); while (m--) { int x,y; scanf("%d%d",&x,&y); ve[x].push_back(y); in[y]++; } scanf("%d",&q); for (int k=0;k<q;k++) { tinit(); for (int i=0;i<n;i++) scanf("%d",&re[i]); for (int i=0;i<n;i++) { if(tin[re[i]]) { no.push_back(k); break; } else { for (int j=0;j<ve[re[i]].size();j++) { tin[ve[re[i]][j]]--; } } } } int Size=no.size(); for (int i=0;i<Size;i++) printf("%d%c",no[i],i==Size-1?'\n':' '); return 0; }