Description  ``Accordian'' Patience  You are to simulate the playing of games of ``Accordian'' patience, the rules for which are as follows: Deal cards one by one in a row from left to right, not overlapping. Whenever the card matches its immediate n…

题目:52张扑克,从左到右在平面上排列,按着如下规则处理: 1.按照从左到右的顺序,如果一张牌和左边的第一张或者第三张匹配,就把它放到对应的牌上面. 2.如果可以移动到多个位置,移动到最左端的牌上面.(匹配:花色或者数值相同) 分析:数据结构.栈.模拟.对于每叠牌建立一个栈,进行模拟即可. 注意:每次只移动每叠牌的最顶上的牌. #include <iostream> #include <cstdlib> #include <cstdio> using namespace…

Input Input data to the program specifies the order in which cards are dealt from the pack. The input contains pairs of lines, each line containing 26 cards separated by single space characters. The final line of the input file contains a # as its fi…

题意:52张牌排一行,一旦出现任何一张牌与它左边的第一张或第三张"匹配",即花色或点数相同,则须立即将其移动到那张牌上面,将其覆盖.能执行以上移动的只有压在最上面的牌.直到最后没有牌能向左移动. 注意细则:如果同时有多张牌都可以移动,你应该采取的策略是移动最左边可移动的牌.当一张牌既可以移动到左边第一张,又可以移动到左边第三张时,应移动到左边第三张上面. 代码:(Accepted,0.100s) //UVa127 - "Accordian" Patience //A…

"Accordian" Patience You are to simulate the playing of games of ``Accordian'' patience, the rules for which are as follows: Deal cards one by one in a row from left to right, not overlapping. Whenever the card matches its immediate neighbour on…

Description  ``Accordian'' Patience  You are to simulate the playing of games of ``Accordian'' patience, the rules for which are as follows: Deal cards one by one in a row from left to right, not overlapping. Whenever the card matches its immediate n…

UVa和POJ都有这道题. 不同的是UVa要求区分单复数,而POJ不要求. 使用STL做会比較简单,这里纯粹使用指针做了,很麻烦的指针操作,一不小心就错. 调试起来还是很费力的 本题理解起来也是挺费力的,要搞清楚怎样模拟也不easy啊,读题要非常细致. 纯指针的操作挺快的吧. 只是POJ 0ms,而UVa就0.2左右了. 三相链表: 1 仅仅要有叠起来的牌.那么就使用一个down指针指向以下的牌就能够了. 2 使用双向链表,能够方便前后遍历. 3 记得有了更新牌之后.又要又一次開始检查是否须要更…

[链接] 我是链接,点我呀:) [题意] 在这里输入题意 [题解] 链表模拟即可. 1pile不能加s... [代码] #include <bits/stdc++.h> using namespace std; const int N = 60; string s[N]; int l[N], r[N]; vector <string> v[N]; int main() { #ifdef LOCAL_DEFINE freopen("F:\\c++source\\rush_i…

题目大意:一种纸牌游戏,将52张扑克牌排成一列,每次操作可将一张扑克牌移到它的前一张或前面第三张上当牌的点数或花色匹配时.每次都移动最靠左的扑克牌,并且能移动三格就移动三格.求最终扑克牌状态. 题目分析:利用栈这种数据结构模拟,以为会超时,没想到AC了. 代码如下: # include<iostream> # include<cstdio> # include<map> # include<stack> # include<cstring> #…

题目: 把52张牌从左到右排好,每张牌自成一个牌堆.当某张牌与它左边那张牌或者左边第三张牌匹配时(花色或者点数相同)时,就把这张牌移到那张牌上面. 移动之后还要查看是否可以进行其他移动.只有位于牌堆顶部的牌才能移动或者参与匹配.当牌堆之间出现空隙时要立刻把右边的所有牌堆左移一格来填 补空隙.如果有多张牌可以移动,先移动最左边的那张牌:如果即可以移一格也可以移3格时,移3格.按顺序输入52张牌,输出最后牌堆数以及各牌堆的牌数. 思路: 看完之后知道要用vector和stack来结合解题,但是没有想…