区间DP UVA 1351 String Compression

题目传送门

 /*
     题意：给一个字符串，连续相同的段落可以合并，gogogo->3(go)，问最小表示的长度
     区间DP：dp[i][j]表示[i,j]的区间最小表示长度，那么dp[i][j] = min (dp[j][k] + dp[k+1][i+j-1]),
             digit (i / k) + dp[j][j+k-1] + 2)后者表示可以压缩成k长度连续相同的字符串
 4.5　　详细解释 */
 /************************************************
 * Author        :Running_Time
 * Created Time  :2015-8-12 15:28:11
 * File Name     :UVA_1351.cpp
  ************************************************/

 #include <cstdio>
 #include <algorithm>
 #include <iostream>
 #include <sstream>
 #include <cstring>
 #include <cmath>
 #include <string>
 #include <vector>
 #include <queue>
 #include <deque>
 #include <stack>
 #include <list>
 #include <map>
 #include <set>
 #include <bitset>
 #include <cstdlib>
 #include <ctime>
 using namespace std;

 #define lson l, mid, rt << 1
 #define rson mid + 1, r, rt << 1 | 1
 typedef long long ll;
 const int MAXN = 2e2 + ;
 const int INF = 0x3f3f3f3f;
 const int MOD = 1e9 + ;
 char str[MAXN];
 int dp[MAXN][MAXN];

 bool check(int l, int r, int k)   {
     int i = ;
     while (i < k)   {
         for (int p=; l+p*k+i<=r; ++p)  {
             if (str[l+i] != str[l+p*k+i])   return  false;
         }
         ++i;
     }
     return true;
 }

 int digit(int x)    {
     int ret = ;
     while (x)   {
         ret++;  x /= ;
     }
     return ret;
 }

 int main(void)    {     //UVA 1351 String Compression
     int T;  scanf ("%d", &T);
     while (T--) {
         scanf ("%s", str + );
         int len = strlen (str + );
         for (int i=; i<=len; ++i)  dp[i][i] = ;
         for (int i=; i<=len; ++i)    {
             for (int j=; j+i-<=len; ++j)  {
                 dp[j][j+i-] = INF;
                 int &x = dp[j][j+i-];
                 for (int k=j; k<i+j-; ++k) {
                     x = min (x, dp[j][k] + dp[k+][i+j-]);
                 }
                 for (int k=; k<=i/; ++k)  {
                     if (i % k != ) continue;
                     if (check (j, i + j - , k))    {
                         x = min (x, digit (i / k) + dp[j][j+k-] + );
                     }
                 }
             }
         }

         printf ("%d\n", dp[][len]);
     }

     return ;
 }