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:
给出一个有向图,并给出k个查询,输出所有不是拓扑排序的查查询。
思路:
1.记录每个点的next,入度。
2.vector拷贝的相关操作:C++ vector拷贝使用总结
题解:
1 #include<cstdlib> 2 #include<cstdio> 3 #include<vector> 4 using namespace std; 5 //定义每个节点的入度和对应出去的节点 6 struct node 7 { 8 int in = 0; 9 vector<int> out; 10 }; 11 int main() { 12 int n, m; 13 scanf("%d %d", &n, &m); 14 vector<node> nodes(n + 1); 15 int a, b; 16 for (int i = 0; i < m; i++) { 17 scanf("%d %d", &a, &b); 18 nodes[a].out.push_back(b); 19 nodes[b].in++; 20 } 21 int query; 22 scanf("%d", &query); 23 int cnt = 0; 24 for (int i = 0; i < query; i++) { 25 bool flag = true; 26 //每次查询对nodes的副本进行修改。 27 vector<node> tNodes(nodes); 28 for (int j = 0; j < n; j++) { 29 int t; 30 scanf("%d", &t); 31 //即使已经判断出来不是拓扑排序了,仍然需要接受剩下的输入。 32 if (!flag) continue; 33 if (tNodes[t].in == 0) { 34 for (int k = 0; k < tNodes[t].out.size(); k++) { 35 tNodes[tNodes[t].out[k]].in--; 36 } 37 } 38 else { 39 if (cnt != 0) printf(" "); 40 printf("%d", i); 41 cnt++; 42 flag = false; 43 } 44 } 45 } 46 return 0; 47 }