UVA140 ——bandwidth(搜索)
Given a graph (V,E) where V is a set of nodes and E is a set of arcs in VxV, and an ordering on the elements in V, then the bandwidth of a node v is defined as the maximum distance in the ordering between v and any node to which it is connected in the graph. The bandwidth of the ordering is then defined as the maximum of the individual bandwidths. For example, consider the following graph:
This can be ordered in many ways, two of which are illustrated below:
For these orderings, the bandwidths of the nodes (in order) are 6, 6, 1, 4, 1, 1, 6, 6 giving an ordering bandwidth of 6, and 5, 3, 1, 4, 3, 5, 1, 4 giving an ordering bandwidth of 5.
Write a program that will find the ordering of a graph that minimises the bandwidth.
Input
Input will consist of a series of graphs. Each graph will appear on a line by itself. The entire file will be terminated by a line consisting of a single #. For each graph, the input will consist of a series of records separated by `;'. Each record will consist of a node name (a single upper case character in the the range `A' to `Z'), followed by a `:' and at least one of its neighbours. The graph will contain no more than 8 nodes.
Output
Output will consist of one line for each graph, listing the ordering of the nodes followed by an arrow (->) and the bandwidth for that ordering. All items must be separated from their neighbours by exactly one space. If more than one ordering produces the same bandwidth, then choose the smallest in lexicographic ordering, that is the one that would appear first in an alphabetic listing.
Sample input
A:FB;B:GC;D:GC;F:AGH;E:HD
#
Sample output
A B C F G D H E -> 3
求出排列好后,相连的两个值之间存在的最大值,然后找出最大值最小的那一组
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std; int maps[30][30];
int hav[30];
int p[10],a[10];
int ans[10],n,pmax,sum;
int work() //如果相连,求它们之间距离的最大值
{
int tmax = 0;
for(int i=1; i<n; i++)
{
for(int j=i+1; j<n; j++)
{
if(maps[a[i]][a[j]])
{
if(j - i>tmax)
tmax=j-i;
}
}
}
return tmax;
}
void dfs(int cur)
{
int flag;
if(cur==n)
{
sum=work();
if(pmax>sum) //找出最大距离最小的那一组
{
pmax=sum;
memcpy(ans,a,sizeof(a));
}
return ;
}
else
{
for(int i=1; i<n; i++)
{
flag=1;
a[cur]=p[i];
for(int j=1; j<cur; j++)
{
if(a[j]==a[cur])
{
flag=0;
break;
}
}
if(flag)
dfs(cur+1);
}
}
}
int main()
{
char str[100];
char c;
int len,i,pre,now;
while(gets(str)&&strcmp(str,"#"))
{
n=1,pmax = 0x3f3f3f3f;
len=strlen(str);
memset(maps,0,sizeof(maps));
memset(hav,0,sizeof(hav));
memset(p,0,sizeof(p));
for(i=0; i<len; i++)
{
c=str[i];
if(str[i+1]==':')
{
pre=c-'A'+1;
hav[pre]++;
}
else if(c>='A'&&c<='Z')
{
now=c-'A'+1;
hav[now]++;
maps[now][pre]=maps[pre][now]=1;
}
}
for(i=0; i<27; i++)
{
if(hav[i])
p[n++]=i;
}
dfs(1);
for(i=1; i<n; i++)
printf("%c ",ans[i]+'A'-1);
printf("-> %d",pmax);
printf("\n");
}
return 0;
}
UVA140 ——bandwidth(搜索)的更多相关文章
- uva140 - Bandwidth
Bandwidth Given a graph (V,E) where V is a set of nodes and E is a set of arcs in VxV, and an orderi ...
- UVa140 Bandwidth 小剪枝+双射小技巧+枚举全排列+字符串的小处理
给出一个图,找出其中的最小带宽的排列.具体要求见传送门:UVa140 这题有些小技巧可以简化代码的编写. 本题的实现参考了刘汝佳老师的源码,的确给了我许多启发,感谢刘老师. 思路: 建立双射关系:从字 ...
- Uva140 Bandwidth 全排列+生成测试法+剪枝
参考过仰望高端玩家的小清新的代码... 思路:1.按字典序对输入的字符串抽取字符,id[字母]=编号,id[编号]=字母,形成双射 2.邻接表用两个vector存储,存储相邻关系 ...
- UVa140 Bandwidth 【最优性剪枝】
题目链接:https://vjudge.net/contest/210334#problem/F 转载于:https://www.cnblogs.com/luruiyuan/p/5847706.ht ...
- 递归回溯 UVa140 Bandwidth宽带
本题题意:寻找一个排列,在此排序中,带宽的长度最小(带宽是指:任意一点v与其距离最远的且与v有边相连的顶点与v的距离的最大值),若有多个,按照字典序输出最小的哪一个. 解题思路: 方法一:由于题目说结 ...
- UVA-140 Bandwidth (回溯+剪枝)
题目大意:求一个使带宽最小的排列和最小带宽.带宽是指一个字母到其相邻字母的距离最大值. 题目分析:在递归生成全排列的过程中剪枝,剪枝方案还是两个.一.当前解不如最优解优时,减去:二.预测的理想解不必最 ...
- 7-6 Bandwidth UVA140
没有清空向量导致debug了好久 这题难以下手 不知道怎么dfs 原来是用排序函数. letter[n]=i; id[i]=n++; 用来储存与设置标记十分巧妙 for(;;) { while(s[ ...
- uva 140 bandwidth (好题) ——yhx
Bandwidth Given a graph (V,E) where V is a set of nodes and E is a set of arcs in VxV, and an orde ...
- UVa OJ 140 - Bandwidth (带宽)
Time limit: 3.000 seconds限时3.000秒 Problem问题 Given a graph (V,E) where V is a set of nodes and E is a ...
随机推荐
- JAVA反射机制基础概念
反射机制:所谓的反射机制就是java语言在运行时拥有一项自观的能力.通过这种能力可以彻底的了解自身的情况为下一步的动作做准备.下面具体介绍一下java的反射机制.这里你将颠覆原来对java的理解. J ...
- Linux进程调度分析
原文:http://www.2cto.com/os/201112/113229.html 操作系统要实现多进程,进程调度必不可少. 有人说,进程调度是操作系统中最为重要的一个部分.我觉得这种说法说得太 ...
- Tornado 网站demo 一
web服务器的工作过程 创建 listen socket, 在指定的监听端口, 等待客户端请求的到来 listen socket 接受客户端的请求, 得到 client socket, 接下来通过 c ...
- RxSwift:ReactiveX for Swift 翻译
RxSwift:ReactiveX for Swift 翻译 字数1787 阅读269 评论3 喜欢3 图片发自简书App RxSwift | |-LICENSE.md |-README.md |-R ...
- 关于tomcat部署应用的三种方式
关于tomcat部署应用虽然不是一个经常的操作,因为一旦选择了一种部署方式,我们其他的应用就会不经大脑的使用这种既定模式, 如果不使用这种部署方式,但是对于其他的部署方式不是很清楚的话,很容易抓瞎,所 ...
- LR回放https协议脚本失败: 错误 -27778: 在尝试与主机“www.baidu.com”connect 时发生 SSL 协议错误
今天用LR录制脚本协议为https协议,回放脚本时出现报错: Action.c(14): 错误 -27778: 在尝试与主机"www.baidu.com"connect 时发生 S ...
- Document Object Model
什么是DOM W3C制定的书写HTML分析器的标准接口规范 全称 Document Object Model 文档对象模型DOM为HTML文档提供的一个API(接口) 可以操作HTML文档 <! ...
- Java8-如何构建一个Stream
Stream的创建方式有很多种,除了最常见的集合创建,还有其他几种方式. List转Stream List继承自Collection接口,而Collection提供了stream()方法. List& ...
- Java面试题(二)
系统整理了一下有关Java的面试题,包括基础篇,javaweb篇,框架篇,数据库篇,多线程篇,并发篇,算法篇等等,陆续更新中.其他方面如前端后端等等的面试题也在整理中,都会有的. 注:文末有福利! 1 ...
- [UWP] Custom Capability的使用
Custom Capability 是uwp开发中普通开发者较为不常用的内容,但是在一些OEM和驱动厂商,使用频率比较高 Custom Capability 有两种用户: 1.普通应用程序开发者: 2 ...