poj 2955 Brackets(区间DP，经典问题)求有规律的括号的最大长度

1、http://poj.org/problem?id=2955

2、题目大意

给出一个只包含()[]的字符序列，求出该字符序列中有规律的符号序列的最长长度

有规律的序列要求如下：

• the empty sequence is a regular brackets sequence,
• if s is a regular brackets sequence, then (s) and [s] are regular brackets sequences, and
• if a and b are regular brackets sequences, thenab is a regular brackets sequence.
• no other sequence is a regular brackets sequence

dp[i][j]表示i到j区间有规律字符串的最大长度

dp[i][j]=max(dp[i][j],dp[i][k]+dp[k][j])其中i=<k<j

3、AC代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define N 105
char str[N];
int dp[N][N];
int check(char a,char b)
{
    if((a=='(' && b==')') || (a=='[' && b==']'))
    return 1;
    return 0;
}
int main()
{
    while(scanf("%s",str)!=EOF)
    {
        if(strcmp(str,"end")==0)
        break;
        int len=strlen(str);
        memset(dp,0,sizeof(dp));
        for(int i=0;i<len-1;i++)
        {
            if(check(str[i],str[i+1]))
            dp[i][i+1]=2;
        }
        for(int i=3;i<=len;i++)
        {
            for(int j=0;i+j-1<len;j++)
            {
                //dp[j][i+j-1]=0;
                if(check(str[j],str[i+j-1]))
                dp[j][i+j-1]=dp[j+1][i+j-2]+2;
                for(int k=j;k<i+j-1;k++)
                {
                    dp[j][i+j-1]=max(dp[j][i+j-1],dp[j][k]+dp[k][i+j-1]);
                }
                //printf("*%d %d %d\n",j,j+i-1,dp[j][i+j-1]);
            }
        }
        printf("%d\n",dp[0][len-1]);
    }
    return 0;
}