[ACM] poj 1141 Brackets Sequence （动态规划）

Brackets Sequence

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 23884		Accepted: 6727		Special Judge

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

参考资料：

http://www.cnblogs.com/android-html5/archive/2010/05/30/2534035.html

http://blog.csdn.net/lijiecsu/article/details/7589877

dp[i][j]更新的顺序

假设字符串的长度为6

（0,1） （1,2） （2,3） （3,4） （4,5）

（0,2） （1,3） （2,4） （3,5）

（0,3） （1,4） （2,5）

（0,4） （1,5）

（0,5）

任何可能的区间都在里面，而且区间长度由小到大，这也符合动规把大问题转化为小问题的思想。

代码：

#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
using namespace std;
const int maxn=250;
const int inf=0x7fffffff;
char str[maxn];
int dp[maxn][maxn],ps[maxn][maxn];
//dp[i][j]代表串从i到j需要添加几个字符才能合法，ps[i][j]，记录的从i到j在哪里断成两部分
int len;

void print(int i,int j)//递归输出
{
    if(i>j)
        return;
    if(i==j)
    {
        if(str[i]=='('||str[i]==')')
            cout<<"()";
        else
            cout<<"[]";
    }
    else if(ps[i][j]==-1)//从i到j括号匹配
    {
        cout<<str[i];
        print(i+1,j-1);
        cout<<str[j];
    }
    else//不匹配分成两部分输出
    {
        print(i,ps[i][j]);
        print(ps[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++)//区间从小到大
        for(int i=0;i+k<len;i++)
    {
        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];
            ps[i][j]=-1;//括号匹配，不用添加另外的括号
        }
        for(int mid=i;mid<j;mid++)//看是否能划分为两部分
        {
            if(dp[i][j]>(dp[i][mid]+dp[mid+1][j]))
            {
                dp[i][j]=dp[i][mid]+dp[mid+1][j];
                ps[i][j]=mid;//mid为串i到j区间断开的位置
            }
        }
    }
    print(0,len-1);
    cout<<endl;
    return 0;
}