SDUT 2022 Winter Individual Contest - D(K)

Posted 2023-02-04 佐鼬Jun

链接: link.

K - Dishonest Driver

题意：

给定一个长度为$N$的字符串，现在需要将字符串进行压缩，压缩的规则就是相同的子串的可以缩写，例如$aaabaaab$ 缩写成$((a^3)b{)}^2$,现在问给定的字符串能压缩成的最短长度

思路：

区间$dp$
定义$dp(i,j)$为从字符串位置从$i到j$可以压缩成的最短长度
那么此时$s_{i,j}$的最短循环节长度为$mlen$,如果$le{n}_{i,j}$是$mlen$的整数倍，且$le{n}_{i,k}$也是$mlen$的整数倍的话，那么转移方程就是$dp(i,j)=min(dp(i,j),dp(i,k))$
否则就是把两个字符串合并
$dp(i,j)=min(dp(i,j),dp(i,k)+dp(k+1,j))$
而最短循环节的长度就直接每次都跑$kmp$求出字符串的对应的$next$数组，然后最短循环节的长度就是$n-next[n]$，或者提前跑个二维$kmp$，预处理二维的$next$数组,最终输出$dp(1,n)$即可

#include <bits/stdc++.h>
using namespace std;
const int inf = 0x3f3f3f3f;
const int N = 777;
char s[N];
int n;
int dp[N][N];
int get_next(int l, int r) 
    char str[N];
    int Next[N] = 0;
    for (int i = 1; i <= r - l + 1; i++) 
        str[i] = s[i + l - 1];
    
    int n = r - l + 1;
    for (int i = 2, j = 0; i <= n; i++) 
        while (j && str[i] != str[j + 1]) j = Next[j];
        if (str[i] == str[j + 1]) j++;
        Next[i] = j;
    
    return n - Next[n];


int main() 
    cin >> n;
    cin >> s + 1;
    memset(dp, 0x3f, sizeof(dp));
    for (int i = 0; i <= n; i++) 
        dp[i][i] = 1;
    

    for (int len = 2; len <= n; len++) 
        for (int i = 1; i <= n - len + 1; i++) 
            int j = i + len - 1;
            dp[i][j] = inf;
            int minlen = get_next(i, j);
            for (int k = i; k <= j; k++) 
                int dex = k - i + 1;
                if (len % minlen == 0 && dex % minlen == 0) 
                    dp[i][j] = min(dp[i][j], dp[i][k]);
                 else 
                    dp[i][j] = min(dp[i][j], dp[i][k] + dp[k + 1][j]);
                
            
        
    
    cout << dp[1][n] << endl;