poj2955 水区间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 … 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

这个题的转移其实就是dp[i][j] =max(dp[i+k][j-k-i] )
如果碰到了左括号，那么无论如何不会出现新的匹配，那么就把dp[1-i][j]全部 赋值为dp[1-i][j-1]
如果碰到了右括号，那么就进行一遍dp[i][j] =max(dp[i+k][j-k-i] )
注意！！！
说一下为什么碰到了右中括号不能直接赋值的原因
当a为右中括号的时候。
b的位置可能为左小括号….
这样的话就要漏情况了….
所以这样不行，也要进行一遍转移

#include<iostream>#include<cmath>#include<queue>#include<stack>#include<algorithm>#include<cstdio>#include<string>#include<memory.h>using namespace std;int dp[500][500];int main(){    string q;    while(cin>>q)    {        if(q=="end")break;        memset(dp,0,sizeof(dp));        for(int a=0;a<q.size();a++)        {            int sum=0;            if(q[a]=='('||q[a]=='[')            {                for(int c=0;c<a;c++)                {                    dp[c+1][a+1]=dp[c+1][a-1+1];                }            }            else if(q[a]==']')            {                for(int b=a-1;b>=0;b--)                {                    if(q[b]!='[')                    {                        dp[b+1][a+1]=dp[b+1+1][a+1];                        //sum=max(dp[1][b]+dp[b+1][a+1],sum);                        for(int c=a-1;c>=b;c--)                        {                            dp[b+1][a+1]=max(dp[b+1][a+1],dp[b+1][c+1]+dp[c+1][a+1]);                        }                    }                    else                    {                        dp[b+1][a+1]=dp[b+1+1][a-1+1]+2;                        for(int c=a-1;c>=b;c--)                        {                            dp[b+1][a+1]=max(dp[b+1][a+1],dp[b+1][c+1]+dp[c+1][a+1]);                        }                    //  sum=max(dp[1][b]+dp[b+1][a+1],sum);                    }                }            }            else if(q[a]==')')            {                for(int b=a-1;b>=0;b--)                {                    if(q[b]!='(')                    {                        dp[b+1][a+1]=dp[b+1+1][a+1];                    //  sum=max(dp[1][b]+dp[b+1][a+1],sum);                        for(int c=a-1;c>=b;c--)                        {                            dp[b+1][a+1]=max(dp[b+1][a+1],dp[b+1][c+1]+dp[c+1][a+1]);                        }                    }                    else                    {                        dp[b+1][a+1]=dp[b+1+1][a-1+1]+2;                        for(int c=a-1;c>=b;c--)                        {                            dp[b+1][a+1]=max(dp[b+1][a+1],dp[b+1][c+1]+dp[c+1][a+1]);                        }                    }                }            }        }        cout<<dp[1][q.size()]<<endl;    }       return 0;}