題目大意:

摺疊的定義如下：

1. 一個字串可以看成它自身的摺疊。記作S = S
2. X(S)是X(X>1)個S連線在一起的串的摺疊。記作X(S) = SSSS…S(X個S)。
3. 如果A = A’, B = B’，則AB = A’B’ 例如，因為3(A) = AAA, 2(B) = BB，所以3(A)C2(B) = AAACBB，而2(3(A)C)2(B) = AAACAAACBB

給一個字串，求它的最短折疊。例如AAAAAAAAAABABABCCD的最短折疊為：9(A)3(AB)CCD。

解題思路:

　一道挺不錯的區間dp,適合我這種入門級選手做.定義f[i][j]為區間[i,j]壓縮後的dp值,那麼有兩種狀態轉移:一,從[i,k],[k,j]直接轉移;二,從[i,k]進行壓縮轉移,$f[i,j]=f[i,k]+2+m[(j-i+1)/(k-i+1)]$,其中需滿足$(j-i+1)%(k-i+1)==0$.

code:

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#include<queue>
#define R register
#define next exnt
#define debug puts("mlg")
using namespace std;
typedef long long ll;
typedef unsigned 
 
long long ull;
typedef long double ld;
inline ll read();
inline void write(ll x);
inline void writesp(ll x);
inline void writeln(ll x);
ll n;
char wn,c[1000];
ll f[150][150],m[120];
inline bool check(ll l,ll r,ll len){
    for(R ll i=l+len;i<=r;i++){
        if(c[(i-l)%len+l]!=c[i]) return false; 
    }
    
 
return true;
}
int main(){    
    while(!(('A'<=wn&&wn<='Z')||('a'<=wn&&wn<='z'))) wn=getchar();
    while((('A'<=wn&&wn<='Z')||('a'<=wn&&wn<='z'))) c[++n]=wn,wn=getchar();
    for(R ll i=1;i<=9;i++) m[i]=1;
    for(R ll j=10;j<=99;j++) m[j]=2;
    m[100]=3;
    memset(f,0x3f,sizeof f);
    for(R ll i=1;i<=n;i++) f[i][i]=1;
    for(R ll l=2;l<=n;l++){
        for(R ll i=1,j=i+l-1;j<=n;j++,i++){
            for(R ll k=i;k<j;k++) f[i][j]=min(f[i][j],f[i][k]+f[k+1][j]);
            for(R ll k=i;k<j;k++){
                ll len=k-i+1;
                if(l%len) continue;
                if(check(i,j,len)) f[i][j]=min(f[i][j],f[i][k]+2+m[l/len]);
            }
        }
    }
    writeln(f[1][n]);
}
inline ll read(){
    ll x=0,t=1;char ch=getchar();
    while(ch<'0'||ch>'9'){
        if(ch=='-') t=-1;
        ch=getchar();
    }
    while(ch>='0'&&ch<='9'){
        x=x*10+ch-'0';
        ch=getchar();
    }
    return x*t;
}
inline void write(ll x){
    if(x<0){putchar('-');x=-x;}
    if(x<=9){putchar(x+'0');return;}
    write(x/10);putchar(x%10+'0');
}
inline void writesp(ll x){
    write(x);putchar(' ');
}
inline void writeln(ll x){
    write(x);putchar('\n');
}