[meet in the middle]String Coloring

最新推荐文章于 2022-08-02 17:49:00 发布

转载最新推荐文章于 2022-08-02 17:49:00 发布 · 155 阅读

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原文链接：http://www.cnblogs.com/lllxq/p/10365959.html

本文探讨了在给定长度为2N的字符串中，如何计算满足特定条件的字符串涂色方式数量。条件为从左至右读取的红色字符字符串等于从右至左读取的蓝色字符字符串。通过meet-in-the-middle搜索策略，文章提供了详细的算法实现，包括输入输出格式、示例和源代码。

题目描述

You are given a string S of length 2N consisting of lowercase English letters.
There are 22N ways to color each character in S red or blue. Among these ways, how many satisfy the following condition?
The string obtained by reading the characters painted red from left to right is equal to the string obtained by reading the characters painted blue from right to left.
Constraints
1≤N≤18
The length of S is 2N.
S consists of lowercase English letters.

输入

Input is given from Standard Input in the following format:
N
S

输出

Print the number of ways to paint the string that satisfy the condition.

样例输入

复制样例数据

4
cabaacba

样例输出

提示

There are four ways to paint the string, as follows:
cabaacba
cabaacba
cabaacba
cabaacba

思路：meet in the middle搜索

#include <iostream>
#include<cstdio>
#include<string>
#include<map>
typedef long long ll;
using namespace std;

char s[20];
map< pair<string,string>,ll > mp;

int main()
{
    ll n;scanf("%lld",&n);
    scanf("%s",s);
    for(ll i=0;i<=(1<<n)-1;i++){
        string a,b;
        for(ll j=0;j<n;j++){
            if((i>>j)&1) a+=s[j];
            else b+=s[j];
        }
        mp[make_pair(a,b)]++;
    }
    ll ans=0;
    for(ll i=0;i<=(1<<n)-1;i++){
        string a,b;
        for(ll j=0;j<n;j++){
            if((i>>j)&1) a+=s[2*n-1-j];
            else b+=s[2*n-1-j];
        }
        ans+=mp[make_pair(a,b)];
    }
    printf("%lld\n",ans);
    return 0;
}