每k个字符划分一个组，该组内字符顺序可以任意重排，定义块为最长的连续的字符子串，求长度为m*k的字符串中最少的块的数目
设\(dp[i][j]\)：前\(i\)组中第\(i\)组结尾为\(j\)的最优解
然后分情况转移即可
吐槽：机子开夜间模式不小心把输出放到循环里（看不见匹配），样例居然过了...
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<string>
#include<vector>
#include<stack>
#include<queue>
#include<set>
#include<map>
#define rep(i,j,k) for(register int i=j;i<=k;i++)
#define rrep(i,j,k) for(register int i=j;i>=k;i--)
#define erep(i,u) for(register int i=head[u];~i;i=nxt[i])
#define iin(a) scanf("%d",&a)
#define lin(a) scanf("%lld",&a)
#define din(a) scanf("%lf",&a)
#define s0(a) scanf("%s",a)
#define s1(a) scanf("%s",a+1)
#define print(a) printf("%lld",(ll)a)
#define enter putchar(‘\n‘)
#define blank putchar(‘ ‘)
#define println(a) printf("%lld\n",(ll)a)
#define IOS ios::sync_with_stdio(0)
using namespace std;
const int maxn = 1e4+11;
const double eps = 1e-10;
typedef long long ll;
const int oo = 0x3f3f3f3f;
const double ERR = -2.3333;
ll read(){
ll x=0,f=1;register char ch=getchar();
while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
return x*f;
}
int dp[maxn][27];
char str[maxn];
int have[29],kind;
int main(){
int T=read();
while(T--){
int k=read();
s1(str);
memset(dp,0x3f,sizeof dp);
int n=strlen(str+1);
int m=n/k;
rep(i,1,m){
int left=1+(i-1)*k,right=i*k;
kind=0;memset(have,0,sizeof have);
rep(j,left,right){
have[str[j]-‘a‘]++;
if(have[str[j]-‘a‘]==1) kind++;
}
if(i==1) rep(j,0,‘z‘-‘a‘){
if(have[j]) dp[1][j]=kind;
}
if(i==1)continue;
rep(j,0,‘z‘-‘a‘){
if(!have[j])continue;
rep(k,0,‘z‘-‘a‘){
if(j==k){
if(kind==1){
dp[i][j]=min(dp[i][j],dp[i-1][k]);//all j
}else{
dp[i][j]=min(dp[i][j],dp[i-1][k]+kind);
}
}else if(have[k]){
dp[i][j]=min(dp[i][j],dp[i-1][k]+kind-1);
}else{
dp[i][j]=min(dp[i][j],dp[i-1][k]+kind);
}
}
}
}
int ans=oo;
rep(i,0,‘z‘-‘a‘){
ans=min(ans,dp[m][i]);
}
println(ans);
}
return 0;
}