poj 2955 区间dp

题意：求最大括号匹配

链接：点我

区间dp水题，但也是很经典，区间dp都是由此延伸出来的，越是基础题越是需要勤加练习

Sample Input

((()))
()()()
([]])
)[)(
([][][)
end

Sample Output

6
6
4
0
6

2015-07-19：专题复习纪念

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<queue>
#include<map>
using namespace std;
#define MOD 1000000007
const int INF=0x3f3f3f3f;
const double eps=1e-5;
typedef long long ll;
#define cl(a) memset(a,0,sizeof(a))
#define ts printf("*****
");
const int MAXN=1005;
int n,m,tt;
char s[MAXN];
int dp[MAXN][MAXN];
int flag[MAXN][MAXN];
int main()
{
    int i,j,k;
    #ifndef ONLINE_JUDGE
    freopen("1.in","r",stdin);
    #endif
    while(gets(s))
    {
        if(s[0]=='e')   break;
        int len=strlen(s);
        for(i=0;i<len;i++)  dp[i][i]=0;
        for(int d=1;d<len;d++)
        {
            for(i=0;i+d<len;i++)
            {
                j=i+d;
                dp[i][j]=-999999;
                if((s[i]=='('&&s[j]==')')||(s[i]=='['&&s[j]==']'))
                {
                    dp[i][j]=dp[i+1][j-1]+1;
                }
                for(k=i;k<j;k++)
                {
                    if(dp[i][j]<dp[i][k]+dp[k+1][j])
                    {
                        dp[i][j]=dp[i][k]+dp[k+1][j];
                    }
                }
            }
        }
        printf("%d
",dp[0][len-1]*2);
    }
}

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<queue>
#include<map>
using namespace std;
#define MOD 1000000007
const int INF=0x3f3f3f3f;
const double eps=1e-5;
typedef long long ll;
#define cl(a) memset(a,0,sizeof(a))
#define ts printf("*****
");
const int MAXN=1005;
int n,m,tt;
int dp[MAXN][MAXN];
int main()
{
    int i,j,k;
    #ifndef ONLINE_JUDGE
    freopen("1.in","r",stdin);
    #endif
    char s[MAXN];
    while(scanf("%s",s)!=EOF)
    {
        if(s[0]=='e')   break;
        int len=strlen(s);
        cl(dp);
        for(int d=1;d<len;d++)
        {
            for(i=0;d+i<len;i++)
            {
                j=d+i;
                if((s[i]=='('&&s[j]==')')||(s[i]=='['&&s[j]==']'))  dp[i][j]=dp[i+1][j-1]+2;
                for(k=i;k<=j;k++)
                {
                    dp[i][j]=max(dp[i][j],dp[i][k]+dp[k+1][j]);
                }
            }
        }
        printf("%d
",dp[0][len-1]);
    }
}