POJ 1141 Brackets Sequence(DP)

Brackets Sequence

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 18313		Accepted: 5020		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

 

 

 

 

题目描述：
     给出一个由(、)、[、]组成的字符串，添加最少的括号使得所有的括号匹配，输出最后得到的字符串。

解题思路：
      用DP求最少需要括号数：以p从1到n（字符串长度），记录下从i到i+p需要添加的最少括号数f[i][j],同时记录下中间需要添加括号的位置pos[i][j]——为-1表示不需要添加。

 

 

DP

 

 

#include<stdio.h>
#include<string.h>

#define MAXN 120
const int INF=0x7fffffff;

int f[MAXN][MAXN],pos[MAXN][MAXN];
char s[MAXN];

int n;

int DP()
{
    n=strlen(s);
    memset(f,0,sizeof(f));
    
    for(int i=n;i>0;i--)
    {
        s[i]=s[i-1];
        f[i][i]=1;
    }    
    int tmp;
    for(int p=1;p<=n;p++)
    {
        for(int i=1;i<=n-p;i++)
        {
            int j=i+p;
            f[i][j]=INF;
            if((s[i]=='('&&s[j]==')')||(s[i]=='['&&s[j]==']'))
            {
                tmp=f[i+1][j-1];
                if(tmp<f[i][j])
                   f[i][j]=tmp;
            }    
            pos[i][j]=-1;
            for(int k=i;k<j;k++)
            {
                tmp=f[i][k]+f[k+1][j];
                if(tmp<f[i][j])
                {
                    f[i][j]=tmp;
                    pos[i][j]=k;
                }    
            }    
        }    
    }  
    return f[1][n];  
} 

void print(int beg,int end)
{
    if(beg>end)return ;
    if(beg==end)
    {
        if(s[beg]=='('||s[beg]==')')
          printf("()");
        else  printf("[]");
    }    
    else
    {
        if(pos[beg][end]==-1)
        {
            if(s[beg]=='(')
            {
                printf("(");
                print(beg+1,end-1);
                printf(")");
            }    
            else
            {
                printf("[");
                print(beg+1,end-1);
                printf("]");
            }    
        } 
        else
        {
            print(beg,pos[beg][end]);
            print(pos[beg][end]+1,end);
        }       
    }    
}       
int main()
{
   // while(scanf("%s",&s)!=EOF)  //这样输入会 WR 
    scanf("%s",&s);
    
        DP();
        print(1,n);
        printf("\n");
    
    return 0;
}