POJ 2955 Brackets（括号最大匹配，区间DP）

原题地址

Brackets

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 6711 Accepted: 3612

Description

We give the following inductive definition of a “regular brackets” sequence:

• 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, then ab is a regular brackets sequence.
• no other sequence is a regular brackets sequence

For instance, all of the following character sequences are regular brackets sequences:

(), [], (()), ()[], ()[()]

while the following character sequences are not:

(, ], )(, ([)], ([(]

Given a brackets sequence of characters a₁a₂ … a_n, your goal is to find the length of the longest regular brackets sequence that is a subsequence of s. That is, you wish to find the largest m such that for indices i₁, i₂, …, i_mwhere 1 ≤ i₁ < i₂ < … < i_m ≤ n, a_i1a_i2 … a_im is a regular brackets sequence.

Given the initial sequence ([([]])], the longest regular brackets subsequence is [([])].

Input

The input test file will contain multiple test cases. Each input test case consists of a single line containing only the characters (, ), [, and ]; each input test will have length between 1 and 100, inclusive. The end-of-file is marked by a line containing the word “end” and should not be processed.

Output

For each input case, the program should print the length of the longest possible regular brackets subsequence on a single line.

Sample Input

((()))()()()([]]))[)(([][][)end

Sample Output

66406

求能匹配上的括号的总数

用dp[ i ][ j ]代表区间[ i , j ]内能匹配上的括号的总数

AC代码：

#include<cstdio>#include<cstdlib>#include<cstring>#include<cmath>#include<iostream>#include<algorithm>using namespace std;char ch[200];int dp[106][106];int main(){    while(~scanf("%s",&ch))    {        if(strcmp(ch,"end")==0)        {            break;        }        int i,j,k,l;        memset(dp,0,sizeof(dp));        l=strlen(ch);        for(i=1;i<l;i++)        {            for(j=0,k=i;k<l;j++,k++)            {                if(ch[j]=='('&&ch[k]==')'||ch[j]=='['&&ch[k]==']')                    dp[j][k]=dp[j+1][k-1]+2;                for(int p=j;p<k;p++)                {                    if(dp[j][k]<dp[j][p]+dp[p+1][k])                    {                        dp[j][k]=dp[j][p]+dp[p+1][k];                    }                }            }        }        printf("%d\n",dp[0][l-1]);    }    return 0;}