POJ 2491 Scavenger Hunt map
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 2848 | Accepted: 1553 |
Description
Bill has been the greatest boy scout in America and has become quite a superstar because he always organized the most wonderful scavenger hunts (you know, where the kids have to find a certain route following certain hints). Bill has retired now, but a nationwide election quickly found a successor for him, a guy called George. He does a poor job, though, and wants to learn from Bill's routes. Unfortunately Bill has left only a few notes for his successor.
Problem
Bill never wrote down his routes completely, he only left lots of little sheets on which he had written two consecutive steps of the routes. He then mixed these sheets and memorized his routes similarly to how some people learn for exams: practicing again and again, always reading the first step and trying to remember the following. This made much sense, since one step always required something from the previous step.
George however would like to have a route written down as one long sequence of all the steps in the correct order. Please help him make the nation happy again by reconstructing the routes.
Input
Output
Sample Input
2
4
SwimmingPool OldTree
BirdsNest Garage
Garage SwimmingPool
3
Toilet Hospital
VideoGame Toilet
Sample Output
Scenario #1:
BirdsNest
Garage
SwimmingPool
OldTree Scenario #2:
VideoGame
Toilet
Hospital
Source
SwimmingPool OldTree(SwimmingPool在OldTree前面)
BirdsNest Garage
Garage SwimmingPool
由此可知 BirdsNest -> Garage -> SwimmingPool -> OldTree.
然后再按这顺序每行一个名词输出.
这个题目运用的方法是用两个map保存前序和后序
然后找到第一个,再按顺序输出
#include <iostream>
#include <cstring>
#include <cstdio>
#include <map>
#include <string>
using namespace std;
int main()
{
int t;
cin >> t;
for(int i=; i<=t; i++)
{
map<string,string>next,pre;//定义两个保存前序,后序的map
int n;
cin >> n;
n=n-;
string tmp,s,s1;
while(n--)
{
cin >> s >> s1;
next[s] = s1;
pre[s1] = s;
}
while(pre[s]!="")
s= pre[s];//当前序为空的时候,说明找到第一个
printf("Scenario #%d:\n",i);
cout << s << endl;//先输出第一个
while(next[s]!="")//当后面不为空的时候输出下一个,最后一个不输出
{
cout << next[s] << endl;
s = next[s];//让s到下一个
}
cout << endl;
}
return ;
}
POJ 2491 Scavenger Hunt map的更多相关文章
- POJ 1066 Treasure Hunt(线段相交判断)
Treasure Hunt Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 4797 Accepted: 1998 Des ...
- poj 2418 Hardwood Species (map)
题目:http://poj.org/problem?id=2418 在poj 上交题总是有各种错误,再次感叹各个编译器. c++ AC代码,G++为超时,上代码: #include<cstdio ...
- poj 2503 Babelfish (查找 map)
题目:http://poj.org/problem?id=2503 不知道为什么 poj 的 数据好像不是100000,跟周赛的不一样 2000MS的代码: #include <iostrea ...
- POJ 1066 Treasure Hunt (线段相交)
题意:给你一个100*100的正方形,再给你n条线(墙),保证线段一定在正方形内且端点在正方形边界(外墙),最后给你一个正方形内的点(保证不再墙上) 告诉你墙之间(包括外墙)围成了一些小房间,在小房间 ...
- POJ 1840 Eqs 二分+map/hash
Description Consider equations having the following form: a1x13+ a2x23+ a3x33+ a4x43+ a5x53=0 The co ...
- 简单几何(线段相交) POJ 1066 Treasure Hunt
题目传送门 题意:从四面任意点出发,有若干障碍门,问最少要轰掉几扇门才能到达终点 分析:枚举入口点,也就是线段的两个端点,然后选取与其他线段相交点数最少的 + 1就是答案.特判一下n == 0的时候 ...
- POJ 1066 Treasure Hunt(计算几何)
题意:给出一个100*100的正方形区域,通过若干连接区域边界的线段将正方形区域分割为多个不规则多边形小区域,然后给出宝藏位置,要求从区域外部开辟到宝藏所在位置的一条路径,使得开辟路径所需要打通的墙壁 ...
- poj 1066 Treasure Hunt
http://poj.org/problem?id=1066 #include <cstdio> #include <cstring> #include <cmath&g ...
- POJ 1066 Treasure Hunt(相交线段&&更改)
Treasure Hunt 大意:在一个矩形区域内.有n条线段,线段的端点是在矩形边上的,有一个特殊点,问从这个点到矩形边的最少经过的线段条数最少的书目,穿越仅仅能在中点穿越. 思路:须要巧妙的转换一 ...
随机推荐
- 3、K-近邻算法
K最近邻(k-Nearest Neighbor,KNN)分类算法 1.定义:如果一个样本在特征空间中的k个最近似(即特征空间中最临近)的样本中大多数属于某一类别,则该样本也属于这个类别. 2.计算公式 ...
- 08_代码块丶继承和final
Day07笔记 课程内容 1.封装 2.静态 3.工具类 4.Arrays工具类 封装 概述 1.封装:隐藏事物的属性和实现细节,对外提供公共的访问方式 2.封装的好处: 隐藏了事物的实现细节 提高了 ...
- Spring系列(一):Spring核心概念
一.Spring概念 Spring是一种多层的J2EE应用程序框架,其核心就是管理资源组件以及依赖关系,Spring框架为现代基于java的企业应用程序提供了一个全面的编程和配置模型. 二.Sprin ...
- redhat linux 5.3安装activeMQ
安装环境:linux redhat enterprise 5.3 activemq版本:5.9.01.从http://activemq.apache.org/download.html地址下载apac ...
- 洛谷 P3628 特别行动队
洛谷题目页面传送门 题意见洛谷. 这题一看就是DP... 设\(dp_i\)表示前\(i\)个士兵的最大战斗力.显然,最终答案为\(dp_n\),DP边界为\(dp_0=0\),状态转移方程为\(dp ...
- leetcode bug free
---不包含jiuzhang ladders中出现过的题.如出现多个方法,则最后一个方法是最优解. 目录: 1 String 2 Two pointers 3 Array 4 DFS &&am ...
- 基于Visual C#的AutoCAD开发——一些网址
https://blog.csdn.net/xwebsite/article/details/5578446 http://www.cadgj.com/?p=1504
- Javascript中,实现十大排序方法之一(冒泡排序及其优化设想)
冒泡排序的Javascript实现 首先定义一个取值范围在(0~100000)之间的随机值的长度为10万的数组, function bubbleSort(arr) { console.time('冒泡 ...
- 物联网时代-跟着Thingsboard学IOT架构-HTTP设备协议及API相关限制
thingsboard官网: https://thingsboard.io/ thingsboard GitHub: https://github.com/thingsboard/thingsboar ...
- python(自用手册)三
第三章 基础 3.1编码初识 ascii 256字母没有中文 一个字节 8位 gbk 中国 中文2字节 16位 英文1字节8位 unicode 万国码 前期 2字节 8位 后期变成4个字节 32位 u ...