题目传送门
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
", dp[1][len]);
82 }
83
84 return 0;
85 }