1146 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.
gre.jpg
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
分析:
判断顶点序列是否为拓扑序列。依次判断顶点序列中的顶点,若顶点入度为0,则顶点符合输出要求;不为1,则说明不符合要求,则不是拓扑序列。如果所有顶点都是在满足入度为0的情况下输出,那么顶点序列为拓扑序列。
参考代码:
#include<vector>
#include<iostream>
using namespace std;
vector<vector<int>>graph;
vector<int>in, intemp, v;
int n, m;
int main() {
int a, b, query, vex, flag;
scanf_s("%d%d", &n, &m);
graph.resize(n + 2);
in.resize(n + 2, 0);
intemp.resize(n + 2, 0);
for (int i = 1; i <= m; i++) {
scanf_s("%d%d", &a, &b);
graph[a].push_back(b);
intemp[b]++;
}
scanf_s("%d", &query);
for (int i = 0; i < query; i++) {
in = intemp; flag = 1;
for (int j = 1; j <= n; j++) {
scanf_s("%d", &vex);
if (flag) {
if (in[vex] == 0)
for (int w = 0; w < graph[vex].size(); w++) {
in[graph[vex][w]]--;
}
else flag = 0;
}
}
if (!flag)v.push_back(i);
}
flag = 0;
for (auto val : v) {
if (flag)cout << " ";
cout << val; flag = 1;
}
return 0;
}