codeforces 1385D（dfs）

题目描述

You are given a string s[1…n] consisting of lowercase Latin letters. It is guaranteed that n=2k for some integer k≥0.

The string s[1…n] is called c-good if at least one of the following three conditions is satisfied:

The length of s is 1, and it consists of the character c (i.e. s1=c); The length of s is greater than 1, the first half of the string consists of only the character c (i.e. s1=s2=⋯=sn2=c) and the second half of the string (i.e. the string sn2+1sn2+2…sn) is a (c+1)-good string; The length of s is greater than 1, the second half of the string consists of only the character c (i.e. sn2+1=sn2+2=⋯=sn=c) and the first half of the string (i.e. the string s1s2…sn2) is a (c+1)-good string. For example: “aabc” is ‘a’-good, “ffgheeee” is ‘e’-good.

In one move, you can choose one index i from 1 to n and replace si with any lowercase Latin letter (any character from ‘a’ to ‘z’).

Your task is to find the minimum number of moves required to obtain an ‘a’-good string from s (i.e. c-good string for c= ‘a’). It is guaranteed that the answer always exists.

You have to answer t independent test cases.

Another example of an ‘a’-good string is as follows. Consider the string s=“cdbbaaaa”. It is an ‘a’-good string, because:

the second half of the string (“aaaa”) consists of only the character ‘a’; the first half of the string (“cdbb”) is ‘b’-good string, because: the second half of the string (“bb”) consists of only the character ‘b’; the first half of the string (“cd”) is ‘c’-good string, because: the first half of the string (“c”) consists of only the character ‘c’; the second half of the string (“d”) is ‘d’-good string.

给定一个字符串，求把字符串变成a-good的最小次数

思路

每次都可以分成两半来搜，所以这道题可以分治然后dfs，复杂度为nlogn。

AC代码

#include<bits/stdc++.h>
#define x first
#define y second
#define IOS ios::sync_with_stdio(false);cin.tie(0);
using namespace std;
typedef unsigned long long ULL;
typedef pair<int,int> PII;
typedef pair<long,long> PLL;
typedef pair<char,char> PCC;
typedef long long LL;
const int N=2*1e5+10;
const int M=150;
const int INF=0x3f3f3f3f;
const int MOD=998244353;
string s;
int n;
int check(string str,char ch){
    int cnt=0;
    for(int i=0;i<str.size();i++){
        if(str[i]!=ch) cnt++;
    }
    return cnt;
}
int dfs(string str,char ch,int res){
	//如果剩下一个字母，说明已经改变完毕，结束递归
    if(str.size()==1){
        if(str[0]!=ch) return res+1;
        return res;
    }
    int len=str.size();
    string s1=str.substr(0,len/2);
    string s2=str.substr(len/2);
    int cnt1=check(s1,ch);//改变s1需要的次数
    int cnt2=check(s2,ch);//改变s2需要的次数

    return min(res+dfs(s1,ch+1,res)+cnt2,res+dfs(s2,ch+1,res)+cnt1);
    //比较先改变s1为good串和先改变s2为good串的次数，选最小
}
void solve(){
    cin>>n>>s;
    cout<<dfs(s,'a',0)<<endl;
}
int main(){
    IOS;
    int t;cin>>t;
    while(t--){
        solve();
    }
    return 0;
}