A1146 Topological Order (25 point(s))

拓扑排序

1. 原文

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

2. 解析

拓扑排序：

输出入度为0的结点，删除以该结点为起点的边，输出现在入度为0的结点，continue…

建立vector记录结点边信息，记录终点indegree[]++入度数

遍历记录查询的点 当indegree[i]>0 说明不能组成一组拓扑排序

3. AC代码

#include<cstdio>
#include<vector>
using namespace std;
const int maxn=1010;
vector<int> v[maxn];
int indegree[maxn]={};
int temp[maxn]={};
vector<int> ans;
int main()
{
	int n,m;
	scanf("%d%d",&n,&m);
	int a,b;
	for (int i = 0; i < m; ++i)
	{
		scanf("%d%d",&a,&b);
		v[a].push_back(b);
		indegree[b]++;
		temp[b]++;
	}
	int k; int key=0;
	scanf("%d",&k);
	for(int t=0;t<k;t++){
		int flag=0;
		for (int i = 0; i < n; ++i)
		{
			scanf("%d",&a);
			if (indegree[a]!=0)
			{
				flag=1;
				continue;
			}else{
				for (int j = 0; j < v[a].size(); ++j)
				{
					indegree[v[a][j]]--;
				}
			}
		}
		if (flag==1)
		{
			if (key==0)
			{
				printf("%d",t);
				key++;
			}else{
				printf(" %d",t);
			}
		}
		for (int i = 1; i <= n; ++i)
		{
			indegree[i]=temp[i];
		}
	}
	printf("\n");
	return 0;
}

indegree数组需要在本次查询后重新赋值为原始状态