1146 Topological Order（25 分）

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

代码:

#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<algorithm>
using namespace std;
vector<int>vec[1005];
vector<int>v;
int num[1005];
int main()
{int n,m,a,b;scanf("%d%d",&n,&m);for(int i = 1;i<=m;i++){scanf("%d%d",&a,&b);vec[a].push_back(b);num[b]++;//表明完成这个任务需要先完成哪些 }int k;scanf("%d",&k);int dd[1005],gg[1005];for(int p = 0; p < k ; p++){for(int i = 1;i <= n ;i++){dd[i] = num[i];scanf("%d",&gg[i]); }for(int i = 1;i <= n; i++){if(dd[gg[i]] != 0){v.push_back(p); break;}else  //表示他等于0 了 那么把他vector里面的东西各自减一 {int size = vec[gg[i]].size();for(int j = 0;j < size; j++)dd[ vec[gg[i]][j] ]--;}}   }int o = v.size();for(int i = 0;i<o;i++)printf("%d%c",v[i]," \n"[i==o-1]);return 0;
}