poj2955（简单区间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：给出一串括号字符，求最多的匹配个数。

思路：找到转移方程就好。

max0=max(max0,dp[i][k]+dp[k+1][j]);

if((s[i]=='('&&s[j]==')')||(s[i]=='['&&s[j]==']'))

dp[i][j]=max(max0,dp[i+1][j-1]+1);

else

dp[i][j]=max0;

#include <iostream>#include <stdio.h>#include <stdlib.h>#include<string.h>#include<algorithm>#include<math.h>#include<queue>#include<stack>using namespace std;typedef long long ll;int dp[110][110];char s[110];int fdp(int a,int b){    if(a>=b)return 0;    if(dp[a][b]!=-1)return dp[a][b];    int max0=0;    for(int i=a; i<b; i++)        max0=max(max0,fdp(a,i)+fdp(i+1,b));    if((s[a]=='('&&s[b]==')')||(s[a]=='['&&s[b]==']'))        return dp[a][b]=max(max0,fdp(a+1,b-1)+1);    else        return dp[a][b]=max0;}int main(){    while(~scanf("%s",s))    {        if(strcmp(s,"end")==0)            break;        int l=strlen(s);        memset(dp,-1,sizeof(dp));        printf("%d\n",fdp(0,l-1)<<1);    }    return 0;}