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[1001][1001] = {0}, dele[1001] = {0}; int N, M, K; int main(){ scanf("%d%d", &N, &M); for(int i = 0; i < M; i++){ int v1, v2; scanf("%d%d", &v1, &v2); G[v1][v2] = 1; } scanf("%d", &K); int ans[10000], pt = 0; for(int i = 0; i < K; i++){ fill(dele, dele + 1001, 0); int isTopl = 1; for(int j = 1; j <= N; j++){ int v; int tag = 1; scanf("%d", &v); for(int k = 1; k <= N; k++){ if(dele[k] == 0 && G[k][v] != 0){ tag = 0; break; } } if(tag == 0){ isTopl = 0; }else{ dele[v] = 1; } } if(isTopl == 0){ ans[pt++] = i; } } for(int i = 0; i < pt; i++){ if(i == pt - 1) 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。