题目
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 $N(\le1,000)$, the number of vertices in the graph, and $M(\le10,000)$, 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 $K(\le100)$. 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
题目大意
题目给出一个拓扑图,然后给出若干个拓扑序列,要求判断哪些拓扑序列是不合法的;
思路
线将有向图存起来,只需要存储某个节点指向的下一节点即可,同时记录每个节点的入度,在判断是否合法时将入度信息复制一个备份,没遍历一个节点,将该节点的下一了节点的入度减一,当当前节点的入度值不为0时,就表示还有前驱节点,那么这个拓扑序列就是不合法的;
代码
#include <cstdio>
#include <iostream>
#include <cmath>
#include <vector>
using namespace std;
int main(){
int n, m;
vector<vector<int> > mp;
vector<int> indegree;
scanf("%d%d", &n, &m);
mp.resize(n+1), indegree.resize(n+1);
fill(indegree.begin(), indegree.end(), 0);
for(int i=0, v1, v2; i<m; i++){
scanf("%d%d", &v1, &v2);
mp[v1].push_back(v2);
indegree[v2]++;
}
int k;
vector<int> ans, order;
order.resize(n);
scanf("%d", &k);
for(int i=0; i<k; i++){
vector<int> temp = indegree;
for(int j=0; j<n; j++){
scanf("%d", &order[j]);
}
bool flag = true;
for(int j=0; j<n; j++){
if(temp[order[j]] != 0){
flag = false;
break;
}
for(int t=0; t<mp[order[j]].size(); t++)
temp[mp[order[j]][t]]--;
}
if(!flag)
ans.push_back(i);
}
for(int i=0; i<ans.size(); i++){
if(i)
printf(" ");
printf("%d", ans[i]);
}
printf("\n");
return 0;
}