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

题目大意:给出一个有向图,并且给定K个序列,判断这个序列是否是拓扑序列。

//我一看见我想的就是,得用邻接表存储图,然后对每一个输入的序列,都进行判断,基本上复杂度是非常高的,就是对每一个序列中的数,判断其之前出现的每一个数是否是它的next,这样来判断,然后写的不对。

#include <iostream>
#include<vector>
#include<map>
#include<algorithm>
using namespace std; vector<vector<int>> graph;
int main(){
int n,m;
cin>>n>>m;
graph.resize(n+);
int f,t;
for(int i=;i<m;i++){
cin>>f>>t;
graph[f].push_back(t);//因为是单向图
}
int u;
cin>>u;
vector<int> vt(n);
vector<int> ans;
for(int i=;i<u;i++){//复杂度是O(n^2),稍微有点高啊。
for(int j=;j<n;j++)
cin>>vt[j];
//检查在其之前出现的是否是在这个图的next里。
bool flag=true;
for(int j=;j<n;j++){
for(int k=;k<j;k++){//在这还得遍历vt[j]
for(int v=;v<graph[j].size();v++){
if(vt[k]==graph[j][v]){
cout<<vt[k]<<" "<<graph[j][v]<<"\n";
ans.push_back(i);
flag=false;
break;
}
}
if(!flag)break;
}
if(!flag)break;
}
}
for(int i=;i<ans.size();i++){
cout<<ans[i];
if(i!=ans.size()-)cout<<' ';
} return ;
}

结果:

//真的很奔溃啊,怎么每个都是不对的,那个2 2 到底是什么意思?我明天再看看吧。

//柳神的代码:

//根据入度出度来判断,非常可以了。。

学习了,要多复习。

PAT 1146 Topological Order[难]的更多相关文章

  1. [PAT] 1146 Topological Order(25 分)

    This is a problem given in the Graduate Entrance Exam in 2018: Which of the following is NOT a topol ...

  2. PAT 1146 Topological Order

    This is a problem given in the Graduate Entrance Exam in 2018: Which of the following is NOT a topol ...

  3. PAT 甲级 1146 Topological Order (25 分)(拓扑较简单,保存入度数和出度的节点即可)

    1146 Topological Order (25 分)   This is a problem given in the Graduate Entrance Exam in 2018: Which ...

  4. PAT 甲级 1146 Topological Order

    https://pintia.cn/problem-sets/994805342720868352/problems/994805343043829760 This is a problem give ...

  5. PAT甲级——1146 Topological Order (25分)

    This is a problem given in the Graduate Entrance Exam in 2018: Which of the following is NOT a topol ...

  6. PAT A1146 Topological Order (25 分)——拓扑排序,入度

    This is a problem given in the Graduate Entrance Exam in 2018: Which of the following is NOT a topol ...

  7. 1146. Topological Order (25)

    This is a problem given in the Graduate Entrance Exam in 2018: Which of the following is NOT a topol ...

  8. 1146 Topological Order

    题意:判断序列是否为拓扑序列. 思路:理解什么是拓扑排序就好了,简单题.需要注意的地方就是,因为这里要判断多个,每次判断都会改变入度indegree[],因此记得要把indegree[]留个备份.ps ...

  9. PAT_A1146#Topological Order

    Source: PAT A1146 Topological Order (25 分) Description: This is a problem given in the Graduate Entr ...

随机推荐

  1. 并查集 - UVALive 6889 City Park

    City Park Problem's Link: http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=129725 Mean: 在 ...

  2. 【BZOJ】1046: [HAOI2007]上升序列(dp)

    http://www.lydsy.com/JudgeOnline/problem.php?id=1046 一直看错题....................... 这是要求位置的字典序啊QQQAAAQ ...

  3. (转)Spring AOP编程原理、Demo

    转自: http://pandonix.iteye.com/blog/336873/ AOP原理: http://blog.csdn.net/moreevan/article/details/1197 ...

  4. RabbitMQ之Queues-5

    工作队列的主要任务是:避免立刻执行资源密集型任务,然后必须等待其完成.相反地,我们进行任务调度:我们把任务封装为消息发送给队列.工作进行在后台运行并不断的从队列中取出任务然后执行.当你运行了多个工作进 ...

  5. Struts2_day02--Struts2封装获取表单数据方式

    Struts2封装获取表单数据方式 原始方式获取表单封装到实体类对象 属性封装(会用) 1 直接把表单提交属性封装到action的属性里面 2 实现步骤 (1)在action成员变量位置定义变量 - ...

  6. iOS开发之--iOS APP打包的时候出现的四个选项

  7. shell脚本学习总结01--文件描述符和重定向

    文件描述符是与文件输入和输出的相关联的整数,它们用来追踪已打开的文件,文件描述符0,1,2是系统预留的. 0 --> stdin (标准输入) 1 --> stdout (标准输出) 2 ...

  8. 《C++ Primer Plus》第6章 学习笔记

    使用引导程序选择不同操作的语句后,程序和编程将更有趣.C++提供了if 语句 .if else 语 句 和 switch 语句来管理选项.if 语句使程序有条件地执行语句或语句块,也就是说,如果满足特 ...

  9. JZOJ.5288【NOIP2017模拟8.17】球场大佬

    Description       每天下午,古猴都会去打羽毛球.但是古猴实在是太强了,他必须要到一些比较强的场去打.但是每个羽毛球场都有许多的人排着队,每次都只能上四个人,每个人都有自己的能力值,然 ...

  10. eclipse控制台不限制显示的行数

    在Preferences中搜索Console,设置Limit console output没有限制即可.