nyoj 349&Poj 1094 Sorting It All Out——————【拓扑应用】
Sorting It All Out
- 描述
-
An ascending sorted sequence of distinct values is one in which some form of a less-than operator is used to order the elements from smallest to largest. For example, the sorted sequence A, B, C, D implies that A < B, B < C and C < D. in this problem, we will give you a set of relations of the form A < B and ask you to determine whether a sorted order has been specified or not.
- 输入
- Input consists of multiple problem instances. Each instance starts with a line containing two positive integers n and m. the first value indicated the number of objects to sort, where 2 <= n <= 26. The objects to be sorted will be the first n characters of the uppercase alphabet. The second value m indicates the number of relations of the form A < B which will be given in this problem instance. Next will be m lines, each containing one such relation consisting of three characters: an uppercase letter, the character "<" and a second uppercase letter. No letter will be outside the range of the first n letters of the alphabet. Values of n = m = 0 indicate end of input.
- 输出
- For each problem instance, output consists of one line. This line should be one of the following three:
Sorted sequence determined after xxx relations: yyy...y.
Sorted sequence cannot be determined.
Inconsistency found after xxx relations.where xxx is the number of relations processed at the time either a sorted sequence is determined or an inconsistency is found, whichever comes first, and yyy...y is the sorted, ascending sequence.
- 样例输入
-
- 4 6
- A<B
- A<C
- B<C
- C<D
- B<D
- A<B
- 3 2
- A<B
- B<A
- 26 1
- A<Z
- 0 0
- 4 6
- 样例输出
-
- Sorted sequence determined after 4 relations: ABCD.
- Inconsistency found after 2 relations.
- Sorted sequence cannot be determined.
- 题目大意:给你n个点,给你m条边代表大小关系。问你在第几条边加入后有矛盾(有环)或能确定关系,或者不能确定关系。
解题思路:首先每次加入一条边,就用floyd传递闭包,之后再判断是否形成环。如果没有环,就判断是否能确定唯一大小关系,这里有一个重要的判断条件即如果所有的结点的度等于n-1,则拓扑排序记录路径。
- #include<bits/stdc++.h>
- using namespace std;
- int Map[50][50],indegree[50],outdegree[50];
- char S_ord[50];
- bool floyd(int n){
- for(int k=0;k<n;k++){ //传递闭包
- for(int i=0;i<n;i++){
- for(int j=0;j<n;j++){
- if(Map[i][k]&&Map[k][j])
- Map[i][j]=1;
- }
- }
- }
- for(int i=0;i<n;i++) //判断是否形成环
- if(Map[i][i])
- return 1;
- return 0;
- }
- bool calcu_is_ord(int n){ //计算目前是否有序
- memset(indegree,0,sizeof(indegree));
- memset(outdegree,0,sizeof(outdegree));
- for(int i=0;i<n;i++){
- for(int j=0;j<n;j++){
- if(Map[i][j]){
- indegree[j]++;
- outdegree[i]++;
- }
- }
- }
- for(int i=0;i<n;i++){
- if(indegree[i]+outdegree[i]!=n-1){
- /*如果所有结点都满足入度加出度等于结点总数减一,说明已经有序。因为如果有序,必然
- 会有入度为0~n-1,相应的出度为n-1~0。所以只要所有的结点度都为n-1,则说明已经有序。
- */
- return 0;
- }
- }
- return 1;
- }
- void topo_sort(int n){ //拓扑排序求大小顺序
- int que_[50],vis[50],top=0,cnt=0,u;
- for(int i=0;i<n;i++){
- if(indegree[i]==0){
- que_[++top]=i;
- }
- }
- memset(vis,0,sizeof(vis));
- while(top){
- u=que_[top--];
- vis[u]=1;
- S_ord[cnt++]=u+'A';
- for(int i=0;i<n;i++){
- if(!vis[i]&&Map[u][i]){
- indegree[i]--;
- }
- if(!vis[i]&&indegree[i]==0){
- que_[++top]=i;
- }
- }
- }
- S_ord[cnt++]='\0';
- }
- int main(){
- int n,m;
- char str[10];
- while(scanf("%d%d",&n,&m)!=EOF&&(n+m)){
- memset(Map,0,sizeof(Map));
- int flag_cir=0,flag_ord=0; //记录在第几组关系输入时形成环或有序
- for(int i=1;i<=m;i++){
- scanf("%s",str);
- Map[str[0]-'A'][str[2]-'A']=1;
- if(flag_cir||flag_ord)
- continue;
- if(floyd(n)){ flag_cir=i;continue;}
- else if(calcu_is_ord(n)){topo_sort(n);flag_ord=i;continue;}
- }
- if(flag_cir)
- printf("Inconsistency found after %d relations.\n",flag_cir);
- else if(flag_ord){
- printf("Sorted sequence determined after %d relations: %s.\n",flag_ord,S_ord);
- }else{
- printf("Sorted sequence cannot be determined.\n");
- }
- }
- return 0;
- }
- Sorted sequence determined after 4 relations: ABCD.
nyoj 349&Poj 1094 Sorting It All Out——————【拓扑应用】的更多相关文章
- ACM: poj 1094 Sorting It All Out - 拓扑排序
poj 1094 Sorting It All Out Time Limit:1000MS Memory Limit:10000KB 64bit IO Format:%lld & ...
- poj 1094 Sorting It All Out (拓扑排序)
http://poj.org/problem?id=1094 Sorting It All Out Time Limit: 1000MS Memory Limit: 10000K Total Su ...
- nyoj 349 (poj 1094) (拓扑排序)
Sorting It All Out 时间限制:3000 ms | 内存限制:65535 KB 难度:3 描述 An ascending sorted sequence of distinct ...
- POJ 1094 Sorting It All Out 拓扑排序 难度:0
http://poj.org/problem?id=1094 #include <cstdio> #include <cstring> #include <vector& ...
- [ACM_模拟] POJ 1094 Sorting It All Out (拓扑排序+Floyd算法 判断关系是否矛盾或统一)
Description An ascending sorted sequence of distinct values is one in which some form of a less-than ...
- POJ 1094 Sorting It All Out (拓扑排序) - from lanshui_Yang
Description An ascending sorted sequence of distinct values is one in which some form of a less-than ...
- poj 1094 Sorting It All Out_拓扑排序
题意:是否唯一确定顺序,根据情况输出 #include <iostream> #include<cstdio> #include<cstring> #include ...
- poj 1094 Sorting It All Out(nyoj 349)
点击打开链接 Sorting It All Out Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 24544 Accep ...
- poj 1094 Sorting It All Out(图论)
http://poj.org/problem?id=1094 这一题,看了个大牛的解题报告,思路变得非常的清晰: 1,先利用floyd_warshall算法求出图的传递闭包 2,再判断是不是存在唯一的 ...
随机推荐
- ASP.NET Session原理及处理方法
session是怎么存储,提取的 1.在服务器端有一个session池,用来存储每个用户提交session中的数据,Session对于每一个客户端(或者说浏览器实例)是“人手一份”,用户首次与Web服 ...
- WM_QUERYENDSESSION与WM_ENDSESSION
此文已由作者王荣涛授权网易云社区发布. 欢迎访问网易云社区,了解更多网易技术产品运营经验. 首先XP系统和Vista以后的系统,这两个消息的处理方式是不同的. XP系统 系统发送WM_QUERYEND ...
- C# 小球100米自由落下
//一球从N 米高自由落下,每次落地后反跳回原高度的一般:再录下,求它在第十次落地时,共经过多少米?第10次反弹多高 static string ballDsitance(float height1, ...
- javascript点击变绿色再点击变红色
<!DOCTYPE html> <html lang="en"> <head> <meta charset="UTF-8&quo ...
- 基于openstack搭建百万级并发负载均衡器的解决方案
最近,喜欢研究一些国外技术大咖们的文章,而这篇文章是基于openstack负载均衡器的解决方案,做的一些总结~希望能够给小伙伴带来一些灵感或者帮助. openstack现有的负载均衡解决方案,无论是l ...
- JDBC_时间处理_Date_Time_Timestamp区别_随机日期生成
import java.sql.Connection; import java.sql.DriverManager; import java.sql.PreparedStatement;import ...
- 当AVPlayer在被释放之后,Player一直监听的时间没有被移除,提示错误的解决办法
Xcode Consolu打印出来的提示: An instance 0x156608c0 of class AVPlayer was deallocated while key value obser ...
- Centos7服务器启动jar包项目最佳方式
jar后台运行:nohup java -jar xx.jar >/dev/null & 此处的“>/dev/null”作用是将终端输出信息输出到空洞中,即不保存输出信息,若要查看输 ...
- Qt 学习之路 2(68):访问网络(4)
Home / Qt 学习之路 2 / Qt 学习之路 2(68):访问网络(4) Qt 学习之路 2(68):访问网络(4) 豆子 2013年11月7日 Qt 学习之路 2 19条评论 前面几章我们了 ...
- class __init__()
python 先定义函数才能调用 类是客观对象在人脑中的主观映射,生产对象的模板,类相当于盖房的图纸,生产工具的模具 模板 类:属性.方法 __init__() 这个方法一般用于初始化一个类但是 当实 ...