[ACM] POJ 1141 Brackets Sequence (区间动态规划）

Brackets Sequence

Description

Let us define a regular brackets sequence in the following way:

1. Empty sequence is a regular sequence.

2. If S is a regular sequence, then (S) and [S] are both regular sequences.

3. If A and B are regular sequences, then AB is a regular sequence.

For example, all of the following sequences of characters are regular brackets sequences:

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

And all of the following character sequences are not:

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

Some sequence of characters ‘(‘, ‘)‘, ‘[‘, and ‘]‘ is given. You are to find the shortest possible regular brackets sequence, that contains the given character sequence as a subsequence. Here, a string a1 a2 ... an is called a subsequence of the string b1 b2
... bm, if there exist such indices 1 = i1 < i2 < ... < in = m, that aj = bij for all 1 = j = n.

Input

The input file contains at most 100 brackets (characters ‘(‘, ‘)‘, ‘[‘ and ‘]‘) that are situated on a single line without any other characters among them.

Output

Write to the output file a single line that contains some regular brackets sequence that has the minimal possible length and contains the given sequence as a subsequence.

Sample Input

([(]

Sample Output

()[()]

Source

Northeastern Europe 2001

解题思路：

dp[i][j] 代表从i到j位置中至少添加几个括号使得括号匹配，pos[i][j]= -1,说明 str[i]　str[j]是一对匹配的括号，否则记录的是从哪个位置把str[i].....str[j]分为两部分，输出答案采用递归的形式。

代码：

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
const int maxn=220;
const int inf=0x7fffffff;
int pos[maxn][maxn];//从i到j在哪里分开
int dp[maxn][maxn];//从i到j至少添几个符号
char str[maxn];
int len;

void print(int i,int j)
{
    if(i>j)
        return ;//递归出口
    if(i==j)
    {
        if(str[i]=='('||str[i]==')')
            cout<<"()";
        else
            cout<<"[]";
    }
    else if(pos[i][j]==-1)//两边是对称的
    {
        cout<<str[i];
        print(i+1,j-1);
        cout<<str[j];
    }
    else//可分割
    {
        print(i,pos[i][j]);
        print(pos[i][j]+1,j);
    }
}

int main()
{
    cin>>str;
    len=strlen(str);
    memset(dp,0,sizeof(dp));
    for(int i=0;i<len;i++)
        dp[i][i]=1;
    for(int k=1;k<len;k++)//长度
        for(int i=0;i+k<len;i++)//起点
        {
            int j=i+k;
            dp[i][j]=inf;
            if((str[i]=='('&&str[j]==')')||(str[i]=='['&&str[j]==']'))
            {
                dp[i][j]=dp[i+1][j-1];
                pos[i][j]=-1;//暂时让它等于-1
            }
            for(int mid=i;mid<j;mid++)//这个必须要执行的。
            {
                if(dp[i][j]>(dp[i][mid]+dp[mid+1][j]))
                {
                    dp[i][j]=dp[i][mid]+dp[mid+1][j];
                    pos[i][j]=mid;
                }
            }
        }
    print(0,len-1);
    cout<<endl;
    return 0;
}

[ACM] POJ 1141 Brackets Sequence (区间动态规划）