POJ2955 Brackets 区间DP

               

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 ofs. That is, you wish to find the largest m such that for indicesi₁, i₂, …, i_m where 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

题意：求出互相匹配的括号的总数

思路：一道区间DP，dp[i][j]存的是i~j区间内匹配的个数

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

           

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