1146 Topological Order (25 分)
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
解析
#include<iostream>
#include<string>
#include<vector>
#include<cstdio>
#include<algorithm>
#include<set>
using namespace std;
vector<vector<int>> G;
vector<int> InDegree;
bool Topological(const vector<int>& top) {
vector<int> In(InDegree.cbegin(), InDegree.cend());
set<int> Q;
for (int i = 1; i < In.size(); i++) {
if (In[i] == 0)
Q.insert(i);
}
auto it = top.begin();
while (!Q.empty()) {
if (Q.find(*it) != Q.cend()) {
Q.erase(*it);
for (auto x : G[*it]) {
In[x]--;
if (In[x] == 0)
Q.insert(x);
}
}
else
return false;
it++;
}
return it == top.cend();
}
int main()
{
int N, M;
scanf("%d %d", &N, &M);
G.resize(N+1);
InDegree.resize(N+1, 0);
int v1, v2;
for (int i = 0; i < M; i++) {
scanf("%d %d", &v1, &v2);
G[v1].push_back(v2);
InDegree[v2]++;
}
int K;
scanf("%d", &K);
vector<int> top(N);
vector<int> NOT;
for (int i = 0; i < K; i++) {
for (int i = 0; i < N; i++)
scanf("%d", &top[i]);
if (Topological(top) == false)
NOT.push_back(i);
}
for (int i = 0; i < NOT.size(); i++) {
printf("%d%c", NOT[i], i + 1 == NOT.size() ? '\n': ' ');
}
}