Brackets POJ

枚举起点和终点，当起点和终点匹配时，dp[start][end]=dp[start+1][end-1]+2
然后枚举中间点

#include<iostream>
#include<cstring>
using namespace std;
const int N=1010;
int dp[N][N];
string str;
int main()
{
	while(cin>>str)
	{
		if(str[0]=='e')
			break;
		memset(dp,0,sizeof dp);
		for(int len=1;len<=str.size();len++)
			for(int start=0;start<str.size();start++)
			{
				int end=start+len-1;
				if(end>=str.size())
					continue;
				if(str[start]=='('&&str[end]==')' || str[start]=='['&&str[end]==']')
					dp[start][end]=dp[start+1][end-1]+2;
				for(int k=start;k<=end;k++)
					dp[start][end]=max(dp[start][end],dp[start][k]+dp[k+1][end]);
			}
			cout<<dp[0][str.size()-1]<<endl;
	}
	return 0;
}