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 ;
}
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