1146 Topological Order (25 point(s))

1146 Topological Order (25 point(s))

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

拓扑排序。

思路：存储有向图，记录结点的入度。然后依次读取给出的排列，每读取一个结点a，其入度应该为0，否则不是拓扑排序。然后去掉这个结点a，需要去掉从这个结点a出发的边，并修改到达结点的入度。

注意点：

1. 由于有多个查询，不能在结点图上直接修改，需要拷贝到临时空间上；

2. 修改入度很容易忘。

 重要总结：https://blog.youkuaiyun.com/coderwait/article/details/89185481

#include<iostream>
#include<cstring>
using namespace std;
const int MAX = 1007;
int G[MAX][MAX]={0};
int graph[MAX][MAX];
int I[MAX]={0};
int inDegree[MAX];
int order[MAX];
int N,M,K;
bool topo(){
	for(int i=1;i<=N;i++){
		int cur = order[i];
		if(inDegree[cur]==0){
			for(int j=1;j<=N;j++){
				if(graph[cur][j]==1){
					graph[cur][j]=0;
					inDegree[j]--;
				}
			}
		}
		else return false;
	}
	return true;
}
int main(void){
	cin>>N>>M;
	int a,b;
	while(M--){
		cin>>a>>b;
		G[a][b]=1;
		I[b]++;
	}
	cin>>K;bool first = true;
	for(int i=0;i<K;i++){
		for(int i=1;i<=N;i++) cin>>order[i];
		memcpy(graph,G,sizeof(G));
		memcpy(inDegree,I,sizeof(I));
		if(!topo()){
			if(first) cout<<i,first=false;
			else cout<<" "<<i;
		}
	}
}