POJ 2955-Brackets（括号匹配-区间DP）

Brackets

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu

Submit  Status

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 … a_n, 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, …, i_m where 1 ≤ i₁ < i₂ < … < i_m ≤ n, a_i1a_i2 … a_im 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

6
6
4
0
6

定义dp [ i ] [ j ] 为串中第 i 个到第 j 个括号的最大匹配数目

那么我们假如知道了 i 到 j 区间的最大匹配，那么i+1到 j+1区间的是不是就可以很简单的得到。

那么 假如第 i 个和第 j 个是一对匹配的括号那么 dp [ i ] [ j ] = dp [ i+1 ] [ j-1 ] + 2 ;

那么我们只需要从小到大枚举所有 i 和 j 中间的括号数目，然后满足匹配就用上面式子dp，然后每次更新dp [ i ] [ j ]为最大值即可。

更新最大值的方法是枚举 i 和 j 的中间值，然后让  dp[ i ] [ j ] = max ( dp [ i ] [ j ] , dp [ i ] [ f ] + dp [ f+1 ] [ j ] ) 。

这题很巧妙，首先要解决区间中从相邻的数开始的从内而外的局部最优扩展到全局最优的思想（类似于归并排序），过程中还要把局部最优加起来求全局最优（类似于线段树）

价值量很大

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 1100;
int dp[N][N];
char str[N];


int main()
{
    while(scanf("%s",str),strcmp("end",str)!=0)
    {
        memset(dp,0,sizeof(dp));
        int n=strlen(str);
        for(int i=1;i<n;i++)
        {
            for(int j=0, k=i;k<n;j++,k++)
            {
                if((str[j]=='('&&str[k]==')')||(str[j]=='['&&str[k]==']'))
                    dp[j][k]=max(dp[j][k],dp[j+1][k-1]+2);
                for(int l=j;l<k;l++)
                {
                    dp[j][k]=max(dp[j][l]+dp[l+1][k],dp[j][k]);
                }
            }
        }
        printf("%d\n",dp[0][n-1]);
    }
    return 0;
}

Brackets

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu

Submit  Status

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 … a_n, 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, …, i_m where 1 ≤ i₁ < i₂ < … < i_m ≤ n, a_i1a_i2 … a_im 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

6
6
4
0
6

定义dp [ i ] [ j ] 为串中第 i 个到第 j 个括号的最大匹配数目

那么我们假如知道了 i 到 j 区间的最大匹配，那么i+1到 j+1区间的是不是就可以很简单的得到。

那么 假如第 i 个和第 j 个是一对匹配的括号那么 dp [ i ] [ j ] = dp [ i+1 ] [ j-1 ] + 2 ;

那么我们只需要从小到大枚举所有 i 和 j 中间的括号数目，然后满足匹配就用上面式子dp，然后每次更新dp [ i ] [ j ]为最大值即可。

更新最大值的方法是枚举 i 和 j 的中间值，然后让  dp[ i ] [ j ] = max ( dp [ i ] [ j ] , dp [ i ] [ f ] + dp [ f+1 ] [ j ] ) 。

这题很巧妙，首先要解决区间中从相邻的数开始的从内而外的局部最优扩展到全局最优的思想（类似于归并排序），过程中还要把局部最优加起来求全局最优（类似于线段树）

价值量很大

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 1100;
int dp[N][N];
char str[N];


int main()
{
    while(scanf("%s",str),strcmp("end",str)!=0)
    {
        memset(dp,0,sizeof(dp));
        int n=strlen(str);
        for(int i=1;i<n;i++)
        {
            for(int j=0, k=i;k<n;j++,k++)
            {
                if((str[j]=='('&&str[k]==')')||(str[j]=='['&&str[k]==']'))
                    dp[j][k]=max(dp[j][k],dp[j+1][k-1]+2);
                for(int l=j;l<k;l++)
                {
                    dp[j][k]=max(dp[j][l]+dp[l+1][k],dp[j][k]);
                }
            }
        }
        printf("%d\n",dp[0][n-1]);
    }
    return 0;
}