Bracket Matching (ii)--classic dynamic programming

The parentheses here match, and if two are the same then execute the following statement

if (CMP (Str[i],str[j]))                       = Min (dp[i][j],dp[i+1][j-1]);

each time you determine the parentheses that need to be filled from I to J, this value is

1#include <stdio.h>2#include <string.h>3#include <math.h>4#include <iostream>5#include <limits.h>6#include <algorithm>7#include <queue>8#include <vector>9#include <Set>Ten#include <stack> One#include <string> A#include <sstream> -#include <map> -#include <cctype> the using namespacestd; - intdp[ the][ the]; - BOOLcmpintNintm) - { +     if(n = ='('&&m = =')')|| (n = ='['&&m = =']')) -         return true; +     return false; A } at intMainvoid) - {  -     intn,m,i,j,k; -     Charstr[101]; -scanf"%d",&n); -      while(n--) in     { -scanf"%s", str); to         intLength =strlen (str); +Memset (DP,0,sizeof(DP)); -          for(i =0; i < length; i++) theDp[i][i] =1; *          for(M =1; m < length; m++)//let's first calculate the difference between the two numbers. $         {Panax Notoginseng              for(i =0; i < length-m; i++) -             { thej = i + M;//J and I difference M +DP[I][J] = the;//the default between I and J requires 105 parentheses to fill A                 if(CMP (STR[I],STR[J]))//See if I and J are compatible theDp[i][j] = min (dp[i][j],dp[i+1][j-1]);//the matching words from the last and +                  for(k = i; k < J; k++) -                 { $Dp[i][j] = min (dp[i][j],dp[i][k]+dp[k+1][j]); $                 } -             } -         } theprintf"%d\n", dp[0][length-1]); -     }Wuyi     return 0; the}

Bracket Matching (ii)--classic dynamic programming