一.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 (≤ 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
二.题意
首先输入m（表示有m个数）和n（有八条有向边），接下来n行为有向图的边，接下来输入k（表示有k行），每一行有m个数，然后判断是否是一个拓扑序列，将不是拓扑序列的部分输出。
三.代码
#include
using namespace std;
int main(){
int n,m,a,b,k,flag=0,in[1010]={0};
vector vec[1010];//类似于二维的数组（kv型）
cin>>n>>m;
for(int i=0;i>a>>b;
vec[a].push_back(b);//将kv键值对push进去
in[b]++;//in数组表示入度
}
cin>>k;
for(int i=0;i tin(in,in+n+1);//每次循环将in部分复制到tin中，in+n+1
for(int j=0;j>a;
if(tin[a]!=0)
judge=0;//用于判断是不是拓扑排序
for(int it:vec[a])
tin[it]--;//将对应元素的入度减一
}
if(judge==1)
continue;//continue关键字可以进行下一次的循环
printf("%s%d",flag==1?" ":"",i);//输出
flag=1;//用于判断是不是第一个
}
return 0;
}
四.类似题目
1.Leetcode 207
Course Schedule
There are a total of n courses you have to take, labeled from 0 to n-1.
Some courses may have prerequisites, for example to take course 0 you have to first take course 1, which is expressed as a pair: [0,1]
Given the total number of courses and a list of prerequisite pairs, is it possible for you to finish all courses?
Example 1:
Input: 2, [[1,0]]
Output: true
Explanation: There are a total of 2 courses to take.
To take course 1 you should have finished course 0. So it is possible.
Example 2:
Input: 2, [[1,0],[0,1]]
Output: false
Explanation: There are a total of 2 courses to take.
To take course 1 you should have finished course 0, and to take course 0 you should
also have finished course 1. So it is impossible.
Note:
- The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about how a graph is represented.
- You may assume that there are no duplicate edges in the input prerequisites.
class Solution {
public:
bool canFinish(int n, vector>& pre) {
vector> adj(n, vector());
vector degree(n, 0);
for (auto &p: pre) {
adj[p[1]].push_back(p[0]);
degree[p[0]]++;
}
queue q;
for (int i = 0; i < n; i++)
if (degree[i] == 0) q.push(i);
while (!q.empty()) {
int curr = q.front(); q.pop(); n--;
for (auto next: adj[curr])
if (--degree[next] == 0) q.push(next);
}
return n == 0;
}
};