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poj2955 Brackets 区间dp

2014-05-06 23:30 363 查看

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Brackets

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 2686		Accepted: 1394

Description

We give the following inductive definition of a “regular brackets” sequence:

the empty sequence is a regular brackets sequence,
if s is a regular brackets sequence, then (s) and [s] are regular brackets sequences, and
if a and b are regular brackets sequences, then ab is a regular brackets sequence.
no other sequence is a regular brackets sequence

For instance, all of the following character sequences are regular brackets sequences:

(), [], (()), ()[], ()[()]

while the following character sequences are not:

(, ], )(, ([)], ([(]

Given a brackets sequence of characters a1a2 … an, your goal is to find the length of the longest regular brackets sequence that is a subsequence of s. That is, you wish to find the largest m such
that for indices i1, i2, …, imwhere 1 ≤ i1 < i2 < … < im ≤ n, ai1ai2 … aim is
a regular brackets sequence.

Given the initial sequence

([([]])]

, the longest regular brackets subsequence is

[([])]

.

Input

The input test file will contain multiple test cases. Each input test case consists of a single line containing only the characters

, and

; each input test will have length between 1 and 100, inclusive.
The end-of-file is marked by a line containing the word “end” and should not be processed.

Output

For each input case, the program should print the length of the longest possible regular brackets subsequence on a single line.

Sample Input

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

Sample Output

Source

Stanford Local 2004

题意：给定一个括号序，求找到一个序列，里面的括号两两配对，求序列的最长长度。

思路：设dp[i][j]为区间[i,j]中的最大括号数,考虑第i个括号，那么我们可以分为两种情况,(1)不考虑i,直接看dp[i+1][j]; (2)i为左括号，找到与i相匹配的右括号,设为k，分两边加起来,状态方程为：dp[i][j]=max(dp[i][j],dp[i+1][k-1]+dp[k+1][j]+2);详见程序：

记忆化搜索：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAXN=100+100;
const int inf=0x3fffffff;
char s[MAXN];
int dp[MAXN][MAXN];
void dfs(int i,int j)
{
    if(j<=i)
    {
        dp[i][j]=0;
        return ;
    }
    if(dp[i][j]!=-1)
        return ;
    if(j-i==1)
    {
        if((s[i]=='(' && s[j]==')')||(s[i]=='[' && s[j]==']'))
            dp[i][j]=2;
        else
            dp[i][j]=0;
        return ;
    }
    dfs(i+1,j);
    dp[i][j]=dp[i+1][j];
    for(int k=i+1;k<=j;k++)
         if((s[i]=='('&&s[k]==')')||s[i]=='['&&s[k]==']')
            {
                dfs(i+1,k-1);  dfs(k+1,j);
                dp[i][j]=max(dp[i][j],dp[i+1][k-1]+dp[k+1][j]+2);
            }
}
int main()
{
    //freopen("text.txt","r",stdin);
    while(~scanf("%s",s) && s[0]!='e')
    {
        int n=strlen(s);
        memset(dp,-1,sizeof(dp));
        for(int i=0;i<n;i++)
            dp[i][i]=0;
        dfs(0,n-1);
        printf("%d\n",dp[0][n-1]);
    }
    return 0;
}

线性：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAXN=100+10;
char s[MAXN];
int dp[MAXN][MAXN];
int main()
{
    //freopen("text.txt","r",stdin);
    while(scanf("%s",s)!=EOF&&s[0]!='e')
    {
        int n=strlen(s);
        memset(dp,0,sizeof(dp));
        for(int i=n-1;i>=0;i--)
            for(int j=i+1;j<n;j++)
            {
                dp[i][j]=dp[i+1][j];
                for(int k=i+1;k<=j;k++)
                   if((s[i]=='('&&s[k]==')')||s[i]=='['&&s[k]==']')
                        dp[i][j]=max(dp[i][j],dp[i+1][k-1]+dp[k+1][j]+2);
            }
        printf("%d\n",dp[0][n-1]);
    }
    return 0;
}

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