poj 2955 Brackets（区间DP）

Brackets

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 … an, 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, …, im where 1 ≤ i1 < i2 < … < im ≤ n, ai1ai2 … aim 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

6
6
4
0
6

思路：区间DP
如果((s[i]==’(‘&&s[j]==’)’)||(s[i]==’[‘&&s[j]==’]’))， 很明显这时侯
dp[i][j]=max(dp[i][j],dp[i+1][j-1]+2);

其他情况下状态转移方程：dp[i][j]=max(dp[i][j],dp[i][k]+dp[k+1][j]);

代码：

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;char s[110];int dp[110][110];int judge(char a,char b){    if((a=='('&&b==')')||(a=='['&&b==']'))        return 1;    return 0;}int main(){    while(~scanf("%s",s+1))    {        memset(dp,0,sizeof(dp));        if(!strcmp(s+1,"end"))            break;        int len=strlen(s+1);        int ans=0;        for(int l=1;l<len;++l)            for(int i=1;i+l<=len;++i)        {            int j=i+l;            if(judge(s[i],s[j]))                dp[i][j]=max(dp[i][j],dp[i+1][j-1]+2);            for(int k=i;k<j;++k)                dp[i][j]=max(dp[i][j],dp[i][k]+dp[k+1][j]);            ans=max(ans,dp[i][j]);        }        printf("%d\n",ans);    }    return 0;}