codeforces 653B B. Bear and Compressing(dfs)
题目链接:
2 seconds
256 megabytes
standard input
standard output
Limak is a little polar bear. Polar bears hate long strings and thus they like to compress them. You should also know that Limak is so young that he knows only first six letters of the English alphabet: 'a', 'b', 'c', 'd', 'e' and 'f'.
You are given a set of q possible operations. Limak can perform them in any order, any operation may be applied any number of times. The i-th operation is described by a string ai of length two and a string bi of length one. No two of q possible operations have the same string ai.
When Limak has a string s he can perform the i-th operation on s if the first two letters of s match a two-letter string ai. Performing the i-th operation removes first two letters of s and inserts there a string bi. See the notes section for further clarification.
You may note that performing an operation decreases the length of a string s exactly by 1. Also, for some sets of operations there may be a string that cannot be compressed any further, because the first two letters don't match any ai.
Limak wants to start with a string of length n and perform n - 1 operations to finally get a one-letter string "a". In how many ways can he choose the starting string to be able to get "a"? Remember that Limak can use only letters he knows.
The first line contains two integers n and q (2 ≤ n ≤ 6, 1 ≤ q ≤ 36) — the length of the initial string and the number of available operations.
The next q lines describe the possible operations. The i-th of them contains two strings ai and bi (|ai| = 2, |bi| = 1). It's guaranteed thatai ≠ aj for i ≠ j and that all ai and bi consist of only first six lowercase English letters.
Print the number of strings of length n that Limak will be able to transform to string "a" by applying only operations given in the input.
3 5
ab a
cc c
ca a
ee c
ff d
4
2 8
af e
dc d
cc f
bc b
da b
eb a
bb b
ff c
1
6 2
bb a
ba a
0
In the first sample, we count initial strings of length 3 from which Limak can get a required string "a". There are 4 such strings: "abb", "cab", "cca", "eea". The first one Limak can compress using operation 1 two times (changing "ab" to a single "a"). The first operation would change "abb" to "ab" and the second operation would change "ab" to "a".
Other three strings may be compressed as follows:
- "cab" "ab" "a"
- "cca" "ca" "a"
- "eea" "ca" "a"
In the second sample, the only correct initial string is "eb" because it can be immediately compressed to "a".
题意:给一对string,前面两个字符匹配的话可以吧前边2个字符去掉换成这对string后面一个字符,问最后能得到a的长度为n的字符串有多少个;
思路:从a往前找,反过来操作;
AC代码:
#include <bits/stdc++.h>
using namespace std;
int a[],n,q,ans=;
char s[][];
int dfs(char x,int num)
{
if(num>=n-){ans+=a[x-'a'];return ;}
for(int i=;i<q;i++)
{
if(s[i][]==x)
{
char y=s[i][];
dfs(y,num+);
}
}
}
int main()
{
scanf("%d%d",&n,&q);
for(int i=;i<q;i++)
{
scanf("%s",s[i]+);
scanf("%s",s[i]+);
a[s[i][]-'a']++;
}
dfs('a',);
cout<<ans<<endl; return ;
}
codeforces 653B B. Bear and Compressing(dfs)的更多相关文章
- IndiaHacks 2016 - Online Edition (Div. 1 + Div. 2) B. Bear and Compressing 暴力
B. Bear and Compressing 题目连接: http://www.codeforces.com/contest/653/problem/B Description Limak is a ...
- IndiaHacks 2016 - Online Edition (Div. 1 + Div. 2) B. Bear and Compressing
B. Bear and Compressing 题目链接 Problem - B - Codeforces Limak is a little polar bear. Polar bears h ...
- 635B. Bear and Compressing
B. Bear and Compressing time limit per test 2 seconds memory limit per test 256 megabytes input stan ...
- Codeforces 653B Bear and Compressing【DFS】
题目链接: http://codeforces.com/problemset/problem/653/B 题意: 要求你构造一个长度为n的字符串使得通过使用m个操作,最终获得字符a.已知第i个操作将字 ...
- codeforces 680D D. Bear and Tower of Cubes(dfs+贪心)
题目链接: D. Bear and Tower of Cubes time limit per test 2 seconds memory limit per test 256 megabytes i ...
- codeforces Gym 100187J J. Deck Shuffling dfs
J. Deck Shuffling Time Limit: 2 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100187/pro ...
- CodeForces Gym 100500A A. Poetry Challenge DFS
Problem A. Poetry Challenge Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/10 ...
- codeforces 580C Kefa and Park(DFS)
题目链接:http://codeforces.com/contest/580/problem/C #include<cstdio> #include<vector> #incl ...
- Educational Codeforces Round 5 - C. The Labyrinth (dfs联通块操作)
题目链接:http://codeforces.com/contest/616/problem/C 题意就是 给你一个n行m列的图,让你求’*‘这个元素上下左右相连的连续的’.‘有多少(本身也算一个), ...
随机推荐
- lucas定理证明
Lucas 定理(证明) A.B是非负整数,p是质数.AB写成p进制:A=a[n]a[n-1]...a[0],B=b[n]b[n-1]...b[0]. 则组合数C(A,B)与C(a[n],b[n])* ...
- <%%>与<scriptrunat=server>,<%=%>与<%#%>的区别(转)
这些东西都是asp.net前台页面与后台代码交互过程中经常使用的,它们之间有的非常相似,又有一些不同.对比学习下,看看他们之间的联系与区别. 首先看<%%>与<scriptrunat ...
- 今天在网上查看了一个socket程序,运行的时候一直报错,经过队友解决?
1.首先是问题代码ip_port = ('192.168.12.2',8001)2.上边的代码本身没有问题,但是必须经过修改自己本机的局域网IP地址才能顺利链接请参考上一篇blog的地址,查看本机的i ...
- Delphi 对话框实现源码分析
Delphi 对话框实现源码分析 简介 在这篇文章中,我将大概的从Delphi XE2 的Dialogs单元入手,分析ShowMessage,MessageBox等对话框运行原理,希望能帮助你理解 ...
- Android系统移植与调试之------->如何修改Android默认字体大小和设置里面字体大小比例
因为我修改 ro.sf.lcd_density的值,将它从160修改 为120,所以导致整个系统的字体都变得很小.因此需要将整个字体变大,并且在设置-->显示-->字体大小的4个选项的值都 ...
- 使用asn1tools进行asn1编解码
最近在做3GPP的编解码,发现有两个第三方库比较好用.一个是ASN1C(c语言编译环境),一个是python第三方库asn1tools.这里介绍下asn1tools的使用方法: 1 第一步:生成asn ...
- 如何下载symfony
php -r "readfile('https://symfony.com/installer');" > symfony 可能无法下载,:那么你检查一下你的php.ini找 ...
- Openstack 架构简述
概述 在学习OpenStack的过程中,感觉对整个OpenStack的架构稍稍有些了解,所以将这些记录下来,一来防止自己忘记,二来也可以对有需要的人提供帮助 本文章相关的灵感/说明/图片来自于http ...
- 第11条:用zip函数同时遍历两个迭代器
核心知识点: (1)内置的zip函数可以平行地遍历多个迭代器. (2)python3中地zip相当于生成器,会在遍历过程中逐次产生元祖.而python2中地zip则是直接把这些元祖完全生成好,并一次性 ...
- 波浪分析数据转换:大智慧、钱龙、胜龙可用Advanced GET ToGet 数据转换器V3.05特别版
http://www.55188.com/thread-4185427-1-1.html Advanced GET ToGet 数据转换器V3.05特别版,大智慧可用软件数据类型选“分析家”源软件数据 ...