codeforces1385D （递归+分治）

题意描述

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.

思路

根据题意描述，第一次操作时，我们可以选择 s [ 0 ] s[0] s[0]到 [ l e n / 2 ] [len/2] [len/2]来进行改变，第二次操作时，我们可以选择 s [ l e n / 2 + 1 ] s[len/2+1] s[len/2+1]到 s [ l e n ] s[len] s[len]来进行改变，之后的每次改变都会将字符串分为一半，取两次操作的最小值即可

AC代码

#include<bits/stdc++.h>
#define x first
#define y second
#define PB push_back
#define mst(x,a) memset(x,a,sizeof(x))
#define all(a) begin(a),end(a)
#define rep(x,l,u) for(ll x=l;x<u;x++)
#define rrep(x,l,u) for(ll x=l;x>=u;x--)
#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=1e6+10;
const int INF=0x3f3f3f3f;
const int MOD=1e9+7;
int n;
string s;
int check(string str,char ba){
    int cnt=0;
    rep(i,0,str.size()) if(str[i]!=ba) cnt++;
    return cnt;
}
int dfs(string str,char ba,int res){
    if(str.size()==1){
        if(str[0]==ba) return res;
        return res+1;
    }
    int len=str.size();
    string s1=str.substr(0,len/2);
    string s2=str.substr(len/2,len/2);
    int cnt1=check(s1,ba);//对s1进行改变的结果    
    int cnt2=check(s2,ba);//对s2进行改变的结果
    //取两次操作的最小值
    return min(res+dfs(s2,ba+1,res)+cnt1,res+dfs(s1,ba+1,res)+cnt2);
}
void solve(){
    cin>>n>>s;
    cout<<dfs(s,'a',0)<<endl;
}
int main(){
    IOS;
    int t;cin>>t;
    while(t--){
        solve();
    }
    return 0;
}