String Compression LA4256

这道题转移方程不太好想，table[i][j]表示从i到j压缩后最小的字符数，comp[i][mid][j]表示str[i~j]中从str[i]开始str[i~(mid-1)]的最长循环串次数-1，例如letsgogogo  comp[5][7][10] = 2，

table[i][j] = min    table(bit(comp[i][k][j])+2)+table[k+(k-i+1)*(comp[i][k][j]+1))][j])

   table[i][k]+table[k+1][j]

   (i <= k < j)

由于要计算comp[i][k][j]，所以复杂度实际为O(n^4)，所以要先把com[i][k][j]预处理出来，我想到的是先用O(n^3)计算出comp[i][k][j] =1的情况，再用O(n^3)递推出其他的情况，所以最后的复杂度为O(n^3)+O(n^3)+O(n^3)=O(n^3)

#include <iostream>#include <cstdio>#include <cstdlib>#include <cmath>#include <queue>#include <algorithm>#include <vector>#include <cstring>#include <stack>#include <cctype>#include <utility>   #include <map>#include <string>  #include <climits> #include <set>#include <string> #include <sstream>#include <utility>#include <ctime> using std::priority_queue;using std::vector;using std::swap;using std::stack;using std::sort;using std::max;using std::min;using std::pair;using std::map;using std::string;using std::cin;using std::cout;using std::set;using std::queue;using std::string;using std::istringstream;using std::make_pair;using std::greater;using std::endl;const int MAXN(210);int comp[MAXN][MAXN][MAXN];int table[MAXN][MAXN];char str[MAXN];int main(){int T;scanf("%d", &T);while(T--){scanf("%s", str+1);int len = strlen(str+1);memset(comp+1, 0, sizeof(comp[0])*len);for(int i = 2; i <= len; ++i){int l = i-1, r = i, tl = 1;while(l >= 1 && r <= len){if(memcmp(str+l, str+i, tl) == 0){for(int j = 0; j < tl; ++j)comp[l][i][r+j] = 1;}--l;++r;++tl;}}for(int i = 2; i <= len; ++i){for(int l = i-1; l >= 1; --l){int temp = i-l;if(comp[l][i][i+temp-1]){int count = 2;for(int r = i+temp*2-1; r <= len; r+= temp, ++count){if(!comp[r-temp+1-temp][r-temp+1][r])break;for(int k = 0; k < temp; ++k)comp[l][i][r+k] = count;}}}}for(int i = 1; i <= len; ++i)table[i][i] = 1;for(int l = 2; l <= len; ++l){int tl1 = len-l+1;for(int i = 1; i <= tl1; ++i){int j = i+l-1;int &cur = table[i][j];cur = l;for(int k = i; k < j; ++k){int temp = table[i][k]+table[k+1][j];int t1 = comp[i][k+1][j];if(t1){++t1;temp = min(temp, (t1 >= 10? (t1 >= 100? 3: 2): 1)+table[i][k]+2+table[k+1+(t1-1)*(k-i+1)][j]);}cur = min(cur, temp);}}}printf("%d\n", table[1][len]);}return 0;}