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
题意:根据给定的有向图判断给定的拓扑排序序列是否正确
解题思想:以邻接表的形式存储有向图,用数组存储每一个点的入度,遍历给点拓扑排序序列中的结点,如果其入度为0,则是拓扑排序,否则不是,遍历的结点选定后要将其所指向的结点的入度减一。在遍历的过程中若出现入度不为零的点则可判定此非拓扑排序,可输出其序列编号;同时输出之前还要判断之前是否有过输出,若有则要加以空格相隔。
#include <iostream>
#include <vector>
#include <stdio.h>
using namespace std;
int main()
{
int n, m, a, b, k, flag = 0, in[1010];
vector<int> v[1010];
scanf("%d %d", &n, &m);
for (int i = 0; i < m; i++)
{
scanf("%d %d", &a, &b);
v[a].push_back(b);
in[b]++;
}
scanf("%d", &k);
for (int i = 0; i < k; i++)
{
int judge = 1;
vector<int> tin(in, in+n+1);
for (int j = 0; j < n; j++)
{
scanf("%d", &a);
if (tin[a] != 0) judge = 0;
for (int it : v[a]) tin[it]--;
}