题意：给你一串()[]括号，要你求出这串括号的最大匹配个数，如'('与')'匹配，为2个，'['与']'匹配，为2个，其他不能匹配.......

思路：dp[i][j]代表从区间i到区间j所匹配的括号的最大个数，首先，假设不匹配，那么dp[i][j]=dp[i+1][j]；然后查找i+1~~j有木有与第i个括号匹配的

有的话，dp[i][j]=max(dp[i][j],dp[i+1][k-1]+dp[k][j]+2).....

代码：

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#define INF 0x3fffffff
using namespace std;
char a[105];
int dp[105][105];
int main(){
    char b[6] = "end";
    while(scanf("%s",a) != EOF){
        if(!strcmp(b,a))break;
        int n = strlen(a);
        memset(dp,0,sizeof(dp));
        for(int i = 2; i <= n; i++){
            for(int j = 0; j < n; j++){
                int theend = j+i-1;
                dp[j][theend] = dp[j][theend-1];
                if(a[theend] == ')'){
                    for(int k = theend-1; k >= j; k--){
                        if(a[k] == '(')
                            dp[j][theend] = max(dp[j][theend],dp[j][k-1]+dp[k+1][theend-1]+2);
                    }
                }
                if(a[theend] == ']'){
                    for(int k = theend-1; k >= j; k--){
                        if(a[k] == '[')
                            dp[j][theend] = max(dp[j][theend],dp[j][k-1]+dp[k+1][theend-1]+2);
                    }
                }
            }
        }
        cout << dp[0][n-1] << endl;
    }
}