toj 1705 Scavenger Hunt

最新推荐文章于 2022-03-24 12:38:32 发布

weixin_30289831

最新推荐文章于 2022-03-24 12:38:32 发布

阅读量153

点赞数

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/forever4444/archive/2009/05/16/1458415.html

本文介绍了一个算法问题，旨在通过一系列步骤提示来重建完整的路线。输入包括多个场景及其对应的连续步骤对，输出则是每个场景的完整步骤顺序。该算法利用了字符串映射和路径追踪的技术。

1705. Scavenger Hunt

Time Limit: 1.0 Seconds Memory Limit: 65536K
Total Runs: 399 Accepted Runs: 214 Multiple test files

Background

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

The first line contains the number of scenarios. Each scenario describes one route and its first line tells you how many steps (3 ≤ S ≤ 333) the route has. The next S-1 lines each contain one consecutive pair of the steps on the route separated by a single space. The name of each step is always a single string of letters.

Output

The output for every scenario begins with a line containing "Scenario #i:", where i is the number of the scenario starting at 1. Then print S lines containing the steps of the route in correct order. Terminate the output for the scenario with a blank line.

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: TUD Programming Contest 2005

Submit List Runs Forum Statistics

#include < iostream >
#include < string >
#include < map >
#define MAX 700
using namespace std;
int path[MAX];
map < string , int > M;
map < string , int > ::iterator p;
string str[MAX];
int t,n;
int main()
{
    cin >> t;
     int zz = 1 ;
     while (t -- )
    {
        cin >> n;
         string a,b;
        M.clear();
         int k,i,j;
         int numa,numb;
         for (i = 1 ;i <= n;i ++ )
        {
            path[i] =- 1 ;
        }
        cin >> a >> b;
        M[a] = 1 ;
        M[b] = 2 ;
        str[ 1 ] = a;
        str[ 2 ] = b;
        path[ 2 ] = 1 ;
        k = 3 ;
         for (i = 2 ;i < n;i ++ )
        {
            cin >> a >> b;
            p = M.find(a);
             if (p == M.end())
            {
                M[a] = k;
                str[k] = a;
                numa = k;
                k ++ ;
            }
             else
            {
                numa = p -> second;
            }
            p = M.find(b);
             if (p == M.end())
            {
                M[b] = k;
                str[k] = b;
                numb = k;
                k ++ ;
            }
             else
            {
                numb = p -> second;
            }
            path[numb] = numa;
        }
         int start;
        printf( " Scenario #%d:\n " ,zz ++ );
         for (i = 1 ;i <= n;i ++ )
        {
             if (path[i] ==- 1 )
            {
                start = i;
                cout << str[i] << endl;
                 break ;
            }
        }
         int kk = 0 ;
         int next = start;
         while ( 1 )
        {
             if (kk >= n)
                 break ;
             for (i = 1 ;i <= n;i ++ )
            {
                 if (path[i] == next)
                {
                    cout << str[i] << endl;
                     break ;
                }
            }
            next = i;
            kk ++ ;
        }
        cout << endl;
    }
     return 0 ;
}

转载于:https://www.cnblogs.com/forever4444/archive/2009/05/16/1458415.html