pku 2955 Brackets 区间DP

最新推荐文章于 2016-08-02 09:45:17 发布

转载最新推荐文章于 2016-08-02 09:45:17 发布 · 58 阅读

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原文链接：http://www.cnblogs.com/E-star/archive/2013/01/28/2879385.html

本文介绍了一种使用区间动态规划解决给定字符串中满足括号匹配的最长子串问题的方法。通过递归地寻找并匹配'()'和'[]'，并利用记忆化搜索优化算法性能。

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;

}

转载于:https://www.cnblogs.com/E-star/archive/2013/01/28/2879385.html