uva208 - Firetruck
The Center City fire department collaborates with the transportation department to maintain maps of the city which reflects the current status of the city streets. On any given day, several streets are closed for repairs or construction. Firefighters need to be able to select routes from the firestations to fires that do not use closed streets.
Central City is divided into non-overlapping fire districts, each containing a single firestation. When a fire is reported, a central dispatcher alerts the firestation of the district where the fire is located and gives a list of possible routes from the firestation to the fire. You must write a program that the central dispatcher can use to generate routes from the district firestations to the fires.
Input
The city has a separate map for each fire district. Streetcorners of each map are identified by positive integers less than 21, with the firestation always on corner #1. The input file contains several test cases representing different fires in different districts.
- The first line of a test case consists of a single integer which is the number of the streetcorner closest to the fire.
- The next several lines consist of pairs of positive integers separated by blanks which are the adjacent streetcorners of open streets. (For example, if the pair 4 7 is on a line in the file, then the street between streetcorners 4 and 7 is open. There are no other streetcorners between 4 and 7 on that section of the street.)
- The final line of each test case consists of a pair of 0's.
Output
For each test case, your output must identify the case by number (CASE #1, CASE #2, etc). It must list each route on a separate line, with the streetcorners written in the order in which they appear on the route. And it must give the total number routes from firestation to the fire. Include only routes which do not pass through any streetcorner more than once. (For obvious reasons, the fire department doesn't want its trucks driving around in circles.)
Output from separate cases must appear on separate lines.
The following sample input and corresponding correct output represents two test cases.
Sample Input
6
1 2
1 3
3 4
3 5
4 6
5 6
2 3
2 4
0 0
4
2 3
3 4
5 1
1 6
7 8
8 9
2 5
5 7
3 1
1 8
4 6
6 9
0 0
Sample Output
CASE 1:
1 2 3 4 6
1 2 3 5 6
1 2 4 3 5 6
1 2 4 6
1 3 2 4 6
1 3 4 6
1 3 5 6
There are 7 routes from the firestation to streetcorner 6.
CASE 2:
1 3 2 5 7 8 9 6 4
1 3 4
1 5 2 3 4
1 5 7 8 9 6 4
1 6 4
1 6 9 8 7 5 2 3 4
1 8 7 5 2 3 4
1 8 9 6 4
There are 8 routes from the firestation to streetcorner 4.
// 题意:给一个无向图,输出从结点1到给定结点的所有路径,要求结点不能重复经过
// 算法:数据不难,直接回溯即可。但是需要注意两点:
// 1. 要事先判断路径是否存在,否则会超时
// 2. 必须按照字典序从小到大输出各路径。本程序的解决方法是给每个点的相邻点编号排序
预判dfs复杂度:
回溯dfs复杂度:O(b^n) b为分支数
算法耗时 9 ms
#include<cstdio>
#include<cstring>
#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
const int maxn=21+5;
int target, cnt;
int route[maxn], vis[maxn];
vector<int> G[maxn]; void dfs(int d, int v)
{
if(v==target) {
cnt++;
for(int i=0;i<d-1;i++) printf("%d ", route[i]);
printf("%d\n", route[d-1]);
return;
}
for(int i=0;i<G[v].size();i++)
{
int u=G[v][i];
if(!vis[u]) {
route[d]=u;
vis[u]=1; dfs(d+1, u); vis[u]=0;
}
}
} bool can_reach_target(int u)
{
if(u==target) return true;
for(int i=0;i<G[u].size();i++)
{
int v=G[u][i];
if(!vis[v]) {
vis[v]=1;
if(can_reach_target(v)) return true;
}
}
return false;
} int main()
{
int kase=0;
while(scanf("%d", &target)==1) {
int u,v;
for(int i=0;i<maxn;i++) G[i].clear();
cnt=0;
while(scanf("%d%d", &u, &v)==2&&(u||v)) {
G[u].push_back(v);
G[v].push_back(u);
}
for(int i=0;i<maxn;i++) sort(G[i].begin(), G[i].end()); printf("CASE %d:\n", ++kase);
memset(vis, 0, sizeof(vis));
if(vis[1]=1, can_reach_target(1)) {
memset(vis, 0, sizeof(vis));
route[0]=1;vis[1]=1;
dfs(1, 1);
}
printf("There are %d routes from the firestation to streetcorner %d.\n", cnt, target);
} return 0;
}
解法二:双向搜索进行剪枝
从起点开始搜索之前,很有必要先确定一下有那些路是可以到达目标的。
如何确定那些路可以到目标呢? 我们只需要先从目标点开始进行搜索,把所有搜索到得路径都进行标记。
然后,再从起点处进行搜索,在搜索之前,要先判断一下这个路径是否有被之前标记过,如果没有被标记,那么说明它是不可能
走到目标处的。这样的话,就不会盲目地去走了,也大大提高了效率。
下面代码加入一个mark数组,表示终点可以到达的点。mark_can_reach_target复杂度也是O(E)啦。 算法耗时 9 ms
#include<cstdio>
#include<cstring>
#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
const int maxn=21+5;
int target, cnt;
int route[maxn], vis[maxn], mark[maxn];
vector<int> G[maxn]; void dfs(int d, int v)
{
if(v==target) {
cnt++;
for(int i=0;i<d-1;i++) printf("%d ", route[i]);
printf("%d\n", route[d-1]);
return;
}
for(int i=0;i<G[v].size();i++)
{
int u=G[v][i];
if(!vis[u] && mark[u]) {
route[d]=u;
vis[u]=1; dfs(d+1, u); vis[u]=0;
}
}
} bool mark_can_reach_target(int u)
{
bool reach_start=false;
if(mark[u])
return false;
mark[u]=1;
if(u==1)
reach_start=true;
for(int i=0;i<G[u].size();i++)
{
int v=G[u][i];
if(mark_can_reach_target(v))
reach_start=true;
}
return reach_start;
} int main()
{
int kase=0;
while(scanf("%d", &target)==1) {
int u,v;
for(int i=0;i<maxn;i++) G[i].clear();
cnt=0;
while(scanf("%d%d", &u, &v)==2&&(u||v)) {
G[u].push_back(v);
G[v].push_back(u);
}
for(int i=0;i<maxn;i++) sort(G[i].begin(), G[i].end()); printf("CASE %d:\n", ++kase);
memset(mark, 0, sizeof(mark));
if(mark_can_reach_target(target)) {
memset(vis, 0, sizeof(vis));
route[0]=1;vis[1]=1;
dfs(1, 1);
}
printf("There are %d routes from the firestation to streetcorner %d.\n", cnt, target);
} return 0;
}
uva208 - Firetruck的更多相关文章
- UVa-208 Firetruck (图的DFS)
UVA-208 天道好轮回.UVA饶过谁. 就是一个图的DFS. 不过这个图的边太多,要事先判一下起点和终点是否联通(我喜欢用并查集),否则会TLE. #include <iostream> ...
- UVA-208 Firetruck (回溯)
题目大意:给一张无向图,节点编号从1到n(n<=20),按字典序输出所有从1到n的路径. 题目分析:先判断从1是否能到n,然后再回溯. 注意:这道题有坑,按样例输出会PE. 代码如下: # in ...
- UVA208 Firetruck 消防车(并查集,dfs)
要输出所有路径,又要字典序,dfs最适合了,用并查集判断1和目的地是否连通即可 #include<bits/stdc++.h> using namespace std; ; int p[m ...
- 7-1 FireTruck 消防车 uva208
题意: 输入一个n <=20 个结点的无向图以及某个结点k 按照字典序从小到大顺序输出从结点1到结点k的所有路径 要求结点不能重复经过 标准回溯法 要实现从小到大字典序 现在数组中排序好即 ...
- 【习题 7-1 UVA-208】Firetruck
[链接] 我是链接,点我呀:) [题意] 在这里输入题意 [题解] 预处理一下终点能到达哪些点. 暴力就好. 输出结果的时候,数字之间一个空格.. [代码] /* 1.Shoud it use lon ...
- Uva 208 - Firetruck
[题目链接]http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=24&p ...
- UVa 208 - Firetruck 回溯+剪枝 数据
题意:构造出一张图,给出一个点,字典序输出所有从1到该点的路径. 裸搜会超时的题目,其实题目的数据特地设计得让图稠密但起点和终点却不相连,所以直接搜索过去会超时. 只要判断下起点和终点能不能相连就行了 ...
- UVA - 208 Firetruck(消防车)(并查集+回溯)
题意:输入着火点n,求结点1到结点n的所有路径,按字典序输出,要求结点不能重复经过. 分析:用并查集事先判断结点1是否可以到达结点k,否则会超时.dfs即可. #pragma comment(link ...
- uva208
一道简单的路径打印,首先需要一次dfs判断能否从1到达目标点,否则可能会超时.还有一点就是那个格式需要注意下,每条路径前没有空格(看起来好像有3个空格)-. AC代码: #include<cst ...
随机推荐
- Hibernate查询出现java.lang.IllegalArgumentException异常解决方法
Hibernate查询出现java.lang.IllegalArgumentException. 异常信息如下:java.lang.IllegalArgumentException at ...
- Memcached 内存级缓存
Memcached在大型网站中应用 memcached是一个高性能的分布式的内存对象缓存系统,通过在内存里维护一个统一的巨大的hash表,它能够用来存储各种格式的数据,包括图像.视 频.文件以及 ...
- DataTable添加列和行的三种方法
JRoger 原文 #region 方法一: DataTable tblDatas =new DataTable("Datas"); DataColumn dc =null; dc ...
- C++模板知识小结
模板定义:模板就是实现代码重用机制的一种工具,它可以实现类型参数化,即把类型定义为参数, 从而实现了真正的代码可重用性.模版可以分为两类,一个是函数模版,另外一个是类模版. 由于类模板包含类型参数,因 ...
- Delphi 提示在Delphi的IDE中,按Ctrl+Shift+G键可以为一个接口生成一个新的GUID。
对于Object Pascal语言来说,最近一段时间最有意义的改进就是从Delphi3开始支持接口(interface),接口定义了能够与一个对象进行交互操作的一组过程和函数.对一个接口进行定义包含两 ...
- SAS、R以及SPSS的比较__统计语言大战
- c++数组、字符串操作
一.数组操作 1.数组初始化1-1一维数组初始化:标准方式一: int value[100]; // value[i]的值不定,没有初始化标准方式二: int value[100] = {1,2}; ...
- ubuntu 14.04 root破解
Advanced Programmable interrupt controller 高级可编程中断控制;Advanced create table `users` (`id` int(11) not ...
- javascript里面的闭包,作用域,预解析
函数的作用域 1.全局变量=公用卫生间 2.局部变量=次卧卫生间 局部变量 全局无法使用 局部声明变量不加var的话就变成全局变量(不推荐使用) 3.闭包=次卧的可以用自己的卫生间 ...
- 【125】固定IP的电脑配置无线路由
标题所提到的情况即为有些电脑是用静态IP上网的,与普通的宽带连接稍微有些不同,例如我的电脑的静态IP设置是这样的: 只有按照上面的设置才可以正常上网,因此在配置无线路由器的时候也要用到上面的内容,废话 ...