一、题目描述
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
二、解题思路
一个有向无环图的拓扑序列，即对于这个序列中的每个结点，它的前驱结点一定在它之前。由于数据量不大，我们这道题可以使用邻接矩阵存储，G[tmp1][tmp2]=1表示从结点tmp1指向tmp2，即前一个结点是后一个结点的前驱结点，同时将G[tmp2][tmp1]设置为-1。那么对于输入的数据，我们只需要用一个简单的二重循环，看一看有没有两个点的邻接矩阵对应值为-1的，如果有，那么就不是一个拓扑序列，将标号存入ans中，最后打印输出即可，详情可见代码注释。
三、AC代码
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<vector>
using namespace std;
const int maxn = 1010;
int G[maxn][maxn] = {
0}; //邻接矩阵
int main()
{
int N, M, K, tmp1, tmp2;
scanf("%d%d", &N, &M);
for(int i=0; i<M; i++)
{
scanf("%d%d", &tmp1, &tmp2);
G[tmp1][tmp2] = 1; //tmp1是tmp2的前驱
G[tmp2][tmp1] = -1;
}
vector<int> ans;
scanf("%d", &K);
for(int i=0; i<K; i++)
{
vector<int> test;
bool flag = true;
for(int j=0; j<N; j++)
{
scanf("%d", &tmp1);
test.push_back(tmp1);
}
for(int j=0; j<N; j++)
{
for(int k=j+1; k<N; k++)
{
if(G[test[j]][test[k]] == -1) //结点test[k]是G[test[j]的前驱结点，但却排在它的后面
{
flag = false;
break;
}
}
if(!flag) //flag为false，存入ans中
{
ans.push_back(i);
break;
}
}
}
for(int i=0; i<ans.size(); i++)
{
if(i==0) printf("%d", ans[i]);
else printf(" %d", ans[i]);
}
return 0;
}