Poj-1159-Palindrome-learning scrolling Array

Question: Here is a string that can contain at least a few characters to make it a return string. Practice: Set a [I] to this string, and B [I] to the reverse string of this string. Then, the longest common subsequence of a [I] and B [I] is the largest input string in the string. Then, the length of the string minus the maximum length of the input string is required. Formula for finding the longest common subsequence: dp [I] [j] = max (dp [I-1] [j], dp [I] [J-1]) if (a [I] = B [I]) dp [I] [j] = max (dp [I] [j], dp [I-1] [J-1] + 1); If you directly find it, it is bound to open a 5001*5001 array, set MLE. There are two solutions: 1. Open the short int type array. This is the result returned by poj: 2. Use a dynamic array. According to the dp rolling process, we can know that the value of dp [I] [j] is not related to the value of dp [I-2] [0... n. Then the value of dp [I] [j] can be overwritten on dp [I-2] [j. That is, dp [I] [j] Is dp [I % 2] [j]. The result returned by poj is as follows: Compared with the above two methods, it is obvious that the method 2 is very space-saving, it takes a little longer. 1. short int array [html] # include <iostream> # include <stdio. h> # include <string. h> # define max (a, B) (a> B? A: B) using namespace std; short int dp [5001] [5001]; int main () {int a [5001]; int B [5001]; int I, j; int n; char str; cin> n; getchar (); for (I = 1; I <= n; I ++) {scanf ("% c ", & str); a [I] = str; B [n-I + 1] = str ;}for (I = 0; I <= n; I ++) {dp [I] [0] = 0; dp [0] [I] = 0 ;}for (I = 1; I <= n; I ++) {for (j = 1; j <= n; j ++) {dp [I] [j] = max (dp [I] [J-1], dp [I-1] [j]); if (a [I] = B [j]) {dp [I] [j] = max (dp [I] [j], dp [I-1] [J-1] + 1) ;}} int len; len = dp [n] [n]; printf ("% d \ n", n-len); return 0;} 2, rolling array [html] # include <iostream> # include <stdio. h> # include <string. h> using namespace std; int main () {int a [5001]; int B [5001]; int dp [10] [5001]; int I, j; int n; char str; cin> n; getchar (); for (I = 1; I <= n; I ++) {scanf ("% c", & str ); a [I] = str; B [n-I + 1] = str;} dp [1] [0] = dp [0] [0] = 0; for (I = 0; I <= n; I ++) {dp [0] [I] = 0;} for (I = 1; I <= n; I ++) {for (j = 1; j <= n; j ++) {dp [I % 2] [j] = max (dp [I % 2] [J-1], dp [(I-1) % 2] [j]); if (a [I] = B [j]) {dp [I % 2] [j] = max (dp [I % 2] [j], dp [(I-1) % 2] [J-1] + 1) ;}} int len; len = dp [n % 2] [n]; printf ("% d \ n", n-len); return 0 ;}