程序師世界是廣大編程愛好者互助、分享、學習的平台,程序師世界有你更精彩!
首頁
編程語言
C語言|JAVA編程
Python編程
網頁編程
ASP編程|PHP編程
JSP編程
數據庫知識
MYSQL數據庫|SqlServer數據庫
Oracle數據庫|DB2數據庫
 程式師世界 >> 編程語言 >> C語言 >> C++ >> C++入門知識 >> POJ 1731 Orders

POJ 1731 Orders

編輯:C++入門知識

Orders Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 8268 Accepted: 5168 Description The stores manager has sorted all kinds of goods in an alphabetical order of their labels. All the kinds having labels starting with the same letter are stored in the same warehouse (i.e. in the same building) labelled with this letter. During the day the stores manager receives and books the orders of goods which are to be delivered from the store. Each order requires only one kind of goods. The stores manager processes the requests in the order of their booking.   You know in advance all the orders which will have to be processed by the stores manager today, but you do not know their booking order. Compute all possible ways of the visits of warehouses for the stores manager to settle all the demands piece after piece during the day. Input Input contains a single line with all labels of the requested goods (in random order). Each kind of goods is represented by the starting letter of its label. Only small letters of the English alphabet are used. The number of orders doesn't exceed 200. Output Output will contain all possible orderings in which the stores manager may visit his warehouses. Every warehouse is represented by a single small letter of the English alphabet -- the starting letter of the label of the goods. Each ordering of warehouses is written in the output file only once on a separate line and all the lines containing orderings have to be sorted in an alphabetical order (see the example). No output will exceed 2 megabytes. Sample Input bbjd Sample Output bbdj bbjd bdbj bdjb bjbd bjdb dbbj dbjb djbb jbbd jbdb jdbb Source CEOI 1999 解題思路:會用STL的next_permutation函數即可。 [cpp]   #include<iostream>   #include<algorithm>   #include<cstring>   #include<cstdio>   using namespace std;   int main()   www.2cto.com {       char a[210];       gets(a);       int l=strlen(a);       sort(a,a+l);       do       {           puts(a);       }       while(next_permutation(a,a+l));       return 0;   }    

  1. 上一頁:
  2. 下一頁:
Copyright © 程式師世界 All Rights Reserved