区间DP UVA 1351 String Compression

 1 /*
 2     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
 3     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]), 
 4             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
 5 */
 6 /************************************************
 7 * Author        :Running_Time
 8 * Created Time  :2015-8-12 15:28:11
 9 * File Name     :UVA_1351.cpp
10  ************************************************/
11 
12 #include <cstdio>
13 #include <algorithm>
14 #include <iostream>
15 #include <sstream>
16 #include <cstring>
17 #include <cmath>
18 #include <string>
19 #include <vector>
20 #include <queue>
21 #include <deque>
22 #include <stack>
23 #include <list>
24 #include <map>
25 #include <set>
26 #include <bitset>
27 #include <cstdlib>
28 #include <ctime>
29 using namespace std;
30 
31 #define lson l, mid, rt << 1
32 #define rson mid + 1, r, rt << 1 | 1
33 typedef long long ll;
34 const int MAXN = 2e2 + 10;
35 const int INF = 0x3f3f3f3f;
36 const int MOD = 1e9 + 7;
37 char str[MAXN];
38 int dp[MAXN][MAXN];
39 
40 bool check(int l, int r, int k)   {
41     int i = 0;
42     while (i < k)   {
43         for (int p=1; l+p*k+i<=r; ++p)  {
44             if (str[l+i] != str[l+p*k+i])   return  false;
45         }
46         ++i;
47     }
48     return true;
49 }
50 
51 int digit(int x)    {
52     int ret = 0;
53     while (x)   {
54         ret++;  x /= 10;
55     }
56     return ret;
57 }
58 
59 int main(void)    {     //UVA 1351 String Compression
60     int T;  scanf ("%d", &T);
61     while (T--) {
62         scanf ("%s", str + 1);
63         int len = strlen (str + 1);
64         for (int i=1; i<=len; ++i)  dp[i][i] = 1;
65         for (int i=2; i<=len; ++i)    {
66             for (int j=1; j+i-1<=len; ++j)  {
67                 dp[j][j+i-1] = INF;
68                 int &x = dp[j][j+i-1];
69                 for (int k=j; k<i+j-1; ++k) {
70                     x = min (x, dp[j][k] + dp[k+1][i+j-1]);
71                 }
72                 for (int k=1; k<=i/2; ++k)  {
73                     if (i % k != 0) continue;
74                     if (check (j, i + j - 1, k))    {
75                         x = min (x, digit (i / k) + dp[j][j+k-1] + 2);
76                     }
77                 }
78             }
79         }
80 
81         printf ("%d\n", dp[1][len]);
82     }
83 
84     return 0;
85 }