pku 2955 Brackets 区间DP

http://poj.org/problem?id=2955

题意：

给定一个只包含'(' , ')' , '[', ']'的字符串，求满足括号匹配的最长子串。

思路：

区间DP，只要找到满足（）或者  []  匹配的， dp[i][j] = dp[i +１][j - 1] + 2;然后再枚举i到j之间一点求最大值。

 

记忆化搜索：

//#pragma comment(linker,"/STACK:327680000,327680000")
#include <iostream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <cstring>
#include <algorithm>
#include <string>
#include <set>
#include <functional>
#include <numeric>
#include <sstream>
#include <stack>
#include <map>
#include <queue>

#define CL(arr, val)    memset(arr, val, sizeof(arr))

#define ll long long
#define inf 0x7f7f7f7f
#define lc l,m,rt<<1
#define rc m + 1,r,rt<<1|1
#define pi acos(-1.0)
#define ll long long
#define L(x)    (x) << 1
#define R(x)    (x) << 1 | 1
#define MID(l, r)   (l + r) >> 1
#define Min(x, y)   (x) < (y) ? (x) : (y)
#define Max(x, y)   (x) < (y) ? (y) : (x)
#define E(x)        (1 << (x))
#define iabs(x)     (x) < 0 ? -(x) : (x)
#define OUT(x)  printf("%I64d\n", x)
#define lowbit(x)   (x)&(-x)
#define Read()  freopen("din.txt", "r", stdin)
#define Write() freopen("dout.txt", "w", stdout);


#define N 1007
using namespace std;

int dp[N][N];
bool vt[N][N];
char str[N];
int n;

int DP(int s,int e)
{
    if (s == e) return 0;
    if (vt[s][e]) return dp[s][e];
    if ((str[s] == '(' && str[e] == ')') || (str[s] == '[' && str[e] == ']'))
    dp[s][e] = DP(s + 1,e - 1) + 2;
    for (int t = s + 1; t < e; ++t)
    {
        dp[s][e] = max(dp[s][e],DP(s,t) + DP(t,e));
    }
     vt[s][e] = true;
    return dp[s][e];
}
int main()
{
    //Read();
    while (~scanf("%s",str))
    {
        if (str[0] == 'e') break;
        CL(dp,0); CL(vt,false);
        n = strlen(str);
        int ans = DP(0,n - 1);
        printf("%d\n",ans);
    }
    return 0;

}

　　

DP:

//#pragma comment(linker,"/STACK:327680000,327680000")
#include <iostream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <cstring>
#include <algorithm>
#include <string>
#include <set>
#include <functional>
#include <numeric>
#include <sstream>
#include <stack>
#include <map>
#include <queue>

#define CL(arr, val)    memset(arr, val, sizeof(arr))

#define ll long long
#define inf 0x7f7f7f7f
#define lc l,m,rt<<1
#define rc m + 1,r,rt<<1|1
#define pi acos(-1.0)
#define ll long long
#define L(x)    (x) << 1
#define R(x)    (x) << 1 | 1
#define MID(l, r)   (l + r) >> 1
#define Min(x, y)   (x) < (y) ? (x) : (y)
#define Max(x, y)   (x) < (y) ? (y) : (x)
#define E(x)        (1 << (x))
#define iabs(x)     (x) < 0 ? -(x) : (x)
#define OUT(x)  printf("%I64d\n", x)
#define lowbit(x)   (x)&(-x)
#define Read()  freopen("din.txt", "r", stdin)
#define Write() freopen("dout.txt", "w", stdout);


#define N 1007
using namespace std;

int dp[N][N];

char str[N];
int n;


int main()
{
    int i,j,len,t;
   
    while (~scanf("%s",str))
    {
        if (str[0] == 'e') break;
        n = strlen(str);
        CL(dp,0);
        for (len = 1; len < n; ++len)
        {
            for (i = 0; i < n - len; ++i)
            {
                //printf(">>%d %d\n",i,i + len);
                if ((str[i] == '(' && str[i + len] == ')') || (str[i] == '[' && str[i + len] == ']'))
                dp[i][i + len] = dp[i + 1][i + len - 1] + 2;
                for (t = i + 1; t < i + len; ++t)
                {
                    dp[i][i + len] = max(dp[i][i + len],dp[i][t] + dp[t][i + len]);
                }
            }
        }
        printf("%d\n",dp[0][n - 1]);
    }
    return 0;

}