描述

Clickomania is a puzzle in which one starts with a rectangular grid of cells of different colours. In each step, a player selects ("clicks") a cell. All connected cells of the same colour as the selected cell (including itself) are removed if the selected cell is connected to at least one other cell of the same colour. The resulting "hole" is filled in by adjacent cells based on some rule, and the object of the game is to remove all cells in the grid. In this problem, we are interested in the one-dimensional version of the problem. The starting point of the puzzle is a string of colours (each represented by an uppercase letter).
At any point, one may select (click) a letter provided that the same letter occurs before or after the one selected. The substring of the same letter containing the selected letter is removed, and the string is shortened to remove the hole created. To solve the puzzle, the player has to remove all letters and obtain the empty string. If the player obtains a non-empty string in which no letter can be selected, then the player loses. For example, if one starts with the string "ABBAABBAAB", selecting the first "B" gives "AAABBAAB". Next, selecting the last "A" gives "AAABBB". Selecting an "A" followed by a "B" gives the empty string. On the other hand, if one selects the third "B" first, the string "ABBAAAAB" is obtained. One may verify that regardless of the next selections, we obtain either the string "A" or the string "B" in which no letter can be selected. Thus, one must be careful in the sequence of selections chosen in order to solve a puzzle. Furthermore,
there are some puzzles that cannot be solved regardless of the choice of selections. For example, "ABBAAAAB" is not a solvable puzzle. Some facts are known about solvable puzzles: The empty string is solvable. If x and y are solvable puzzles, so are xy, AxA, and AxAyA for any uppercase letter
A. All other puzzles not covered by the rules above are unsolvable.
Given a puzzle, your task is to determine whether it can be solved or not.

输入

Each case of input is specified by a single line. Each line contains a string of uppercase letters. Each string has at least one but no more than 150 characters. The input is terminated by the end of file.

输出

For each input case, print solvable on a single line if there is a sequence of selections that solves the puzzle. Otherwise, print unsolvable on a line.

样例输入

ABBAABBAAB
ABBAAAAB

样例输出

solvable
unsolvable

题目大意:

每次去掉一段字符相同(两个以上)的去掉,问最后能不能去完。

dp[i][j]代表区间[i,j]能不能去完。然后就是讨论AxA,AA,AxAyA,xy几种情况。

#include <bits/stdc++.h>
using namespace std;
char s[];
int dp[][];
int main()
{
while(~scanf("%s",s+))
{
memset(dp,,sizeof dp);
int len=strlen(s+);
for(int l=;l<=len;l++)
{
for(int i=;i+l<=len;i++)
{
int j=i+l;
if(s[i]==s[j])
{
if(dp[i+][j-]||l==)///AxA||AA
dp[i][j]=;
for(int k=i;k<=j;k++)///AxAyA
if(s[i]==s[k])
if((k-<i+||dp[i+][k-])&&(k+>j-||dp[k+][j-]))///AAyA||AxAA||AAA
dp[i][j]=;
}
for(int k=i;k<=j;k++)///xy
if(dp[i][k]&&dp[k+][j]) dp[i][j]=;
}
}
if(dp[][len]) printf("solvable\n");
else printf("unsolvable\n");
}
return ;
}

Clickomania(区间DP)的更多相关文章

  1. 【BZOJ-4380】Myjnie 区间DP

    4380: [POI2015]Myjnie Time Limit: 40 Sec  Memory Limit: 256 MBSec  Special JudgeSubmit: 162  Solved: ...

  2. 【POJ-1390】Blocks 区间DP

    Blocks Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 5252   Accepted: 2165 Descriptio ...

  3. 区间DP LightOJ 1422 Halloween Costumes

    http://lightoj.com/volume_showproblem.php?problem=1422 做的第一道区间DP的题目,试水. 参考解题报告: http://www.cnblogs.c ...

  4. BZOJ1055: [HAOI2008]玩具取名[区间DP]

    1055: [HAOI2008]玩具取名 Time Limit: 10 Sec  Memory Limit: 162 MBSubmit: 1588  Solved: 925[Submit][Statu ...

  5. poj2955 Brackets (区间dp)

    题目链接:http://poj.org/problem?id=2955 题意:给定字符串 求括号匹配最多时的子串长度. 区间dp,状态转移方程: dp[i][j]=max ( dp[i][j] , 2 ...

  6. HDU5900 QSC and Master(区间DP + 最小费用最大流)

    题目 Source http://acm.hdu.edu.cn/showproblem.php?pid=5900 Description Every school has some legends, ...

  7. BZOJ 1260&UVa 4394 区间DP

    题意: 给一段字符串成段染色,问染成目标串最少次数. SOL: 区间DP... DP[i][j]表示从i染到j最小代价 转移:dp[i][j]=min(dp[i][j],dp[i+1][k]+dp[k ...

  8. 区间dp总结篇

    前言:这两天没有写什么题目,把前两周做的有些意思的背包题和最长递增.公共子序列写了个总结.反过去写总结,总能让自己有一番收获......就区间dp来说,一开始我完全不明白它是怎么应用的,甚至于看解题报 ...

  9. Uva 10891 经典博弈区间DP

    经典博弈区间DP 题目链接:https://uva.onlinejudge.org/external/108/p10891.pdf 题意: 给定n个数字,A和B可以从这串数字的两端任意选数字,一次只能 ...

  10. 2016 年沈阳网络赛---QSC and Master(区间DP)

    题目链接 http://acm.hdu.edu.cn/showproblem.php?pid=5900 Problem Description Every school has some legend ...

随机推荐

  1. netty-socketio即时通讯

    jar包和依赖包在360云盘中:所有文件 > 学习 > jar包 > netty-socketio-1.7.10以及依赖 原文链接:http://www.cnblogs.com/al ...

  2. 装饰者模式--Java篇

    装饰者模式(Decorator):动态地给一个对象添加一些额外的职责,就增加功能来说,装饰者模式比生成子类更为灵活. 1.定义接口,可以动态的给对象添加职责. package com.lujie; p ...

  3. iOS 画圆图片的几种方法

    方法一: self.cycleImv= [[UIImageView alloc]initWithFrame:CGRectMake(100, 100, 50, 50)]; [self.view addS ...

  4. ARM指令解析

    今天我来总结一下arm指令的学习,今天我不会对所有的arm指令进行一一的解析,在这里希望大家去看arm汇编手册,这个手册的中文版我放在了http://download.csdn.net/detail/ ...

  5. Window10 开启传统启动界面

    Windows 10沿袭了Windows 8的快速启动,导致在启动过程中无法通过按F8进入启动选项,这样当系统遇到问题无法进入时根本无法通过进入安全模式等方式进行处理(当然通过其他一些工具还是能够引导 ...

  6. Eclipse Java类编辑器里出现乱码的解决方案

    如图:在Java Class编辑器里出现的这种乱码,非常烦人. 解决方案:Windows->Preference->General->Appearance, 在里面将Theme设置成 ...

  7. 技术大众化--10款无需编程的App DIY开发工具

    你有一个很棒的创意但不会编程怎么办?外包.合伙开发还是从零学编程?这里提供另外一种方式--使用无需编程的App DIY开发工具.DIY开发工具不仅节省了开发时间和资金,更为那些创意无限热爱应用的人提供 ...

  8. ant design table td 文字显示过长添加省略号、ant 文字过长时添加tootip提示

    方法1: overflow: hidden; text-overflow: ellipsis; display: -webkit-box; -webkit-line-clamp: 2; -webkit ...

  9. 如何查看 JAR 包的源代码

    ava 项目的编译文件经常被打包成 JAR(Java Archive,Java 归档文件)文件,当然,作为学习,有时候也非常想看到这个 JAR 被打包前的源代码是怎么样的. 下面提供几种查看 JAR ...

  10. 选择法数组排序参考(Java)

    package com.swift; public class Xuanze { public static void main(String[] args) { int[] arr= {28,2,3 ...