G - Brackets Sequence POJ - 1141 (动态规划）

G - Brackets Sequence

 POJ - 1141 

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

()[()]

 动态规划，用dp[i][j]保存从第i个字符（从零开始算）到第j个字符间形成完整regular sequences需要添加的字符个数，pos[i][j]表示形成第i个字符（从零开始算）到第j个字符间形成完整regular sequences需要断开而形成的题目所说的AB的形式的分界点（该点为A的右端），枚举过程i到j的长度是从小到大的，因此计算第i个字符到第j个字符时，在（i,j）位置的dp值是可知的，具体看代码，不难理解

#include<cstdio>
#include<stack>
#include<set>
#include<vector>
#include<queue>
#include<algorithm>
#include<cstring>
#include<string>
#include<map>
#include<iostream>
#include<cmath>
using namespace std;
#define inf 0x3f3f3f3f
typedef long long ll;
const int N=300;
const double esp = 1e-9;
const double PI=3.1415926;
char ch[N];
int dp[N][N],pos[N][N];
void solve(int i,int j)
{
    if(i>j)
        return;
    if(i==j)
    {
        if(ch[i]=='('||ch[j]==')')
            printf("()");
        else
            printf("[]");
    }
    else if(pos[i][j]==-1)
    {
        printf("%c",ch[i]);
        solve(i+1,j-1);  //递归中间部分的过程
        printf("%c",ch[j]);
    }
    else
    {
        solve(i,pos[i][j]);
        solve(pos[i][j]+1,j);
    }
}
int main()
{
    while(gets(ch)!=NULL)  //不能用scanf输入，scanf不能输入带空格字符串
    {
        int len=strlen(ch);
        memset(dp,0,sizeof(dp));
        for(int i=0; i<len; i++)
        {
            dp[i][i]=1;
        }
        for(int t=1; t<len; t++)
        {
            for(int i=0; i+t<len; i++)
            {
                int j=i+t;
                dp[i][j]=inf;
                if((ch[i]=='('&&ch[j]==')')||(ch[i]=='['&&ch[j]==']'))
                {
                    dp[i][j]=dp[i+1][j-1];
                    pos[i][j]=-1;
                }
                for(int mid=i; mid<j; mid++) //计算是否有更优解
                {
                    int ans=dp[i][mid]+dp[mid+1][j];
                    if(dp[i][j]>ans)
                    {
                        pos[i][j]=mid;
                        dp[i][j]=ans;
                    }
                }
            }
        }
        solve(0,len-1);
        printf("\n");
    }
    return 0;
}