A1146. Topological Order
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
#include<iostream>
#include<cstdio>
using namespace std;
int G[][] = {}, dele[] = {};
int N, M, K;
int main(){
scanf("%d%d", &N, &M);
for(int i = ; i < M; i++){
int v1, v2;
scanf("%d%d", &v1, &v2);
G[v1][v2] = ;
}
scanf("%d", &K);
int ans[], pt = ;
for(int i = ; i < K; i++){
fill(dele, dele + , );
int isTopl = ;
for(int j = ; j <= N; j++){
int v;
int tag = ;
scanf("%d", &v);
for(int k = ; k <= N; k++){
if(dele[k] == && G[k][v] != ){
tag = ;
break;
}
}
if(tag == ){
isTopl = ;
}else{
dele[v] = ;
}
}
if(isTopl == ){
ans[pt++] = i;
}
}
for(int i = ; i < pt; i++){
if(i == pt - )
printf("%d", ans[i]);
else printf("%d ", ans[i]);
}
cin >> N;
}
总结:
1、题意:给出一个有向图,检验给出的序列是否是拓扑排序。
2、拓扑排序要求每次删除一个入度为0的节点。
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