Compress String

Problem

Now we have a string and try to find its shortest compressed character length (brackets count).
For example, AAAAAABCABC = 6(A)BCABC, which has a compressed length of 9.

Solution

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;

const int maxn = 210;
int f[maxn][maxn];  // minium number of division
int len;            // the length of the string
char s[maxn];   

int get(int x) {    //calculate the digits
    int sum = 0;
    while (x) {
        sum++;
        x /= 10;
    }
    return sum;
}

int main() {
    scanf("%s", s + 1);
    memset(f, 0x3f, sizeof(f));  // initialize
    len = strlen(s + 1);

    // 
    for (int i = 1; i <= len; ++i) { //Pre-processed sub-intervals of length 1
        f[i][i] = 1;
    }

    // expand the length of sub-intervals
    for (int le = 2; le <= len; ++le) {  // the length of the sub-intervals
        for (int l = 1; l + le - 1 <= len; ++l) {  // the initial position of the sub-intervals
            f[l][l + le - 1] = min(le, f[l][l + le - 1]);  // Treat the whole interval as a subinterval, initialised to the length of the substring
            // Enumerated intervals have several complete subintervals inside them with the same first character as the interval
            for (int t = 1;; t++) {
                bool flag = true;
                int st = l + t * le;
                if (st + le - 1 > len) break;

                // Determine if the internal subintervals are identical
                for (int p = 1; p <= le; p++) {
                    if (s[l + p - 1] != s[st + p - 1]) {
                        flag = false;
                        break;
                    }
                }

                // If the internal subintervals are not identical, end the enumeration
                if (!flag) break;

                // Update the minimum number of divisions of a subinterval
                f[l][l + (t + 1) * le - 1] = min(min(le + get(t + 1) + 2, (t + 1) * le), f[l][l + (t + 1) * le - 1]);
                // where le + get(t + 1) + 2 means that the number of internal subintervals is expressed as the required length of the string 
                // (t + 1) * le denotes the number of partitions of the interval in terms of t+1 identical substrings
            }

            // Enumerate from the middle position and update the minimum number of divisions of the subinterval
            int j = l + le - 1;
            for (int k = l; k < j; k++) {
                f[l][j] = min(f[l][j], f[l][k] + f[k + 1][j]);
            }
        }
    }

    cout << f[1][len];