[Minimum number of characters inserted when DP generates a return string] poj 1159

Note that the space complexity is optimized. The first 49776 K of the scrolling array is not used, and the last 188 K is used. This is the difference !!!

Transfer equation:

If (STR [I] = STR [J])
DP [I] [J] = DP [I-1] [J-1];
Else
DP [I] [J] = 1 + min (DP [I + 1] [J], DP [I] [J-1]);

# Include <map> # include <set> # include <list> # include <queue> # include <deque> # include <stack> # include <string> # include <time. h> # include <cstdio> # include <math. h> # include <iomanip> # include <cstdlib> # include <limits. h> # include <string. h> # include <iostream> # include <fstream> # include <algorithm> using namespace STD; # define ll long # define min int_min # define Max int_max # define PI ACOs (-1.0) # define fre Freopen ("input.txt", "r", stdin) # define FF freopen ("output.txt", "W", stdout) # define n 5005 /*********************************** * ********* if (STR [I] = STR [J]) DP [I] [J] = DP [I-1] [J-1]; else DP [I] [J] = 1 + min (DP [I + 1] [J], DP [I] [J-1]); **************************************** * *****/INT min (int, int B) {return A> B? B: A;} Char STR [N]; short DP [2] [N]; // short integer optimized space complexity int main () {int N; while (scanf ("% d % s", & N, STR + 1 )! =-1) {int I, j; memset (DP, 0, sizeof (DP); int K = 0; for (I = N; I> = 1; I --) {for (j = I + 1; j <= N; j ++) {if (I = J) {continue ;} if (STR [I] = STR [J]) {DP [k] [J] = DP [1-k] [J-1];} else DP [k] [J] = 1 + min (DP [1-k] [J], DP [k] [J-1]);} For (j = 1; j <= N; j ++) DP [1-k] [J] = DP [k] [J]; // The space complexity of rolling Array Optimization K ^ = 1 ;} printf ("% d \ n", DP [1-k] [N]);} return 0 ;}