PAT甲级A1146 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

题意：给定一个图，然后给出n个序列，让你判断给定的这些序列是否为该图的一个拓扑排序，如果不是则输出其序号。

思路1：（最后一个Case超内存）但写法是标准的递归版拓扑排序。定义一个indegree[]存储每个节点的入度，然后根据结点入度使用拓扑排序函数求所有的拓扑序列，由于要把所有的拓扑序列，求出来并保存然后在与给定的序列逐个比较，所以需要在求出每个拓扑序列后，恢复indegree[]和vis[]数组，然后接着求下一个数组。求完后与给定的每个序列逐个比较对应输出即可。

参考代码：

#include<cstdio>
#include<queue>
#include<vector>
using namespace std;
const int maxn=1010;
int n,m,k;
int indegree[maxn],temp_in[maxn];
bool vis[maxn]={false};
vector<int> adj[maxn];
vector<vector<int>> ans;
void TopSort(int u,int d,bool vis[],int indegree[],vector<int> temp){
	temp.push_back(u);
	vis[u]=true;
	if(d==n){
		ans.push_back(temp);
	}
	for(int i=0;i<adj[u].size();i++){
		int v=adj[u][i];
		indegree[v]--;
	}
	for(int i=1;i<=n;i++){
		if(indegree[i]==0&&!vis[i])
			TopSort(i,d+1,vis,indegree,temp);
	}
	for(int i=0;i<adj[u].size();i++){
		int v=adj[u][i];
		indegree[v]++;
	}
	temp.pop_back();
	vis[u]=false;
}
int main()
{
	int u,v;
	scanf("%d%d",&n,&m);
	for(int i=0;i<m;i++){
		scanf("%d%d",&u,&v);
		adj[u].push_back(v);
		indegree[v]++;
	}
	for(int i=1;i<=n;i++){
		if(indegree[i]==0){
			vector<int> temp;
			TopSort(i,1,vis,indegree,temp);
		}
	}
	scanf("%d",&k);
	bool flag=false;//flag=false表示为第一次输出
	for(int i=0;i<k;i++){
		vector<int> temp;
		for(int j=1;j<=n;j++){
			scanf("%d",&u);
			temp.push_back(u);
		}
		int l=0;
		while(l<ans.size()&&ans[l]!=temp) 
			l++;
		if(l==ans.size()&&!flag){
			printf("%d",i);
			flag=true;
		}else if(l==ans.size())
			printf(" %d",i);
	}
	return 0;
}

 

思路2：不求拓扑序列，直接按拓序列性质，对所给序列按从前往后顺序依次判断该点的入度是否为零，若为零则将图中对应的后继定点indegree[]减一，继续求下一顶点，直到所有顶点均符合要求或存在某个当前点的入度不为零，输出。

参考代码：

#include<cstdio>
#include<vector>
using namespace std;
const int maxn=1010;
int n,m,k,u,v,flag=0,indegree[maxn],in[maxn];
vector<int> adj[maxn];
bool isTopSeq(vector<int> t,int in[]){
	for(int i=0;i<t.size();i++){
		int u=t[i];
		if(in[u])
			return false;
		for(int j=0;j<adj[u].size();j++) in[adj[u][j]]--;
	}
	return true;
}
int main()
{
	scanf("%d%d",&n,&m);
	for(int i=0;i<m;i++){
		scanf("%d%d",&u,&v);
		adj[u].push_back(v);
		indegree[v]++;
	}
	scanf("%d",&k);
	for(int i=0;i<k;i++){
		vector<int> temp(n);
		for(int j=0;j<n;j++){
			scanf("%d",&temp[j]);
		}
		for(int j=1;j<=n;j++)	in[j]=indegree[j];
		if(!isTopSeq(temp,in)){
			printf("%s%d",flag?" ":"",i);
			flag=true;
		}
	}
	return 0;
}