UVA1351 ----- String Compression ----- interval DP (memory-based search implementation)

According to his idea, the algorithm is self-implemented to give a string, which can be abbreviated to k (S) form for the same continuous part, and S is a string, k indicates that there are consecutive identical S. For example, abgogogogo can be abbreviated as ab4 (go ). you can also nest abbreviations, such as "nowletsgogogoletsgogogo", abbreviated as "now2 (lets3 (go)". idea: a range of dp, but this question is not really f (I, j) represents the string I ~ Then f (I, j) = min {f (I, k) + f (k + 1, j ), I <= k <j}, min {digitNum (k) + f [l] [l + k-1] + 2, if the string can be composed of the first k strings,} digitNum (k) indicates the number of digits of the number k to determine the interval (I, j) whether there are consecutive strings consisting of k consecutive strings, it is easy to use the O (n) Time to determine the interval DP to search by memory, you don't have to think about iterative writing, so you can use the code for memorizing search to write the interval DP in the future: # include <cstdio> # include <cstring> # include <iostream> # include <algorithm> # include <string. h> using namespace std; const int maxn = 300; int dp [maxn] [maxn]; char s [maxn]; const Int INF = 0x3f3f3f; bool check (int l, int r, int k) {int I; int len = r-l + 1; I = 0; while (I <k) {int p; for (p = 1; l + p * k + I <= r; p ++) {if (s [l + I]! = S [l + p * k + I]) return false;} I ++;} return true;} int min (int a, int B) {return a <B? A: B;} int digitnum (int k) {int len = 0; while (k> 0) {len ++; k/= 10;} return len ;} int DP (int l, int r) {if (dp [l] [r]! =-1) return dp [l] [r]; int len = r-l + 1; int d; dp [l] [r] = INF; for (int k = l; k <r; k ++) dp [l] [r] = min (dp [l] [r], DP (l, k) + DP (k + 1, r); for (d = 1; d <= len/2; d ++) {if (len % d! = 0) continue; if (check (l, r, d) {dp [l] [r] = min (dp [l] [r], digitnum (len/d) + DP (l, l + D-1) + 2);} return dp [l] [r];} int main () {int t; scanf ("% d", & t); while (t --) {scanf ("% s", s); memset (dp,-1, sizeof (dp )); int len = strlen (s); int I; for (I = 0; I <len; I ++) dp [I] [I] = 1; cout <DP (0, len-1) <endl;} return 0 ;}