POJ 1141 Brackets Sequence

黑书动态规划第一题

                                            Brackets Sequence

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 21666		Accepted: 6068		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

#include <iostream>
#include <cstring>
#include <cstdio>

using namespace std;

const int maxn=111;

int dp[maxn][maxn],path[maxn][maxn];
char str[maxn];

void print(int i,int j)
{
    if(i>j) return ;
    else if(i==j)
    {
        if(str[i]=='['||str[i]==']') printf("[]");
        else printf("()");
    }
    else if(path[i][j]==-1)
    {
        printf("%c",str[i]);
        print(i+1,j-1);
        printf("%c",str[j]);
    }
    else
    {
        int k=path[i][j];
        print(i,k);
        print(k+1,j);
    }

}

int main()
{
    while(gets(str))
    {
        int n=strlen(str);

        if(n==0) { putchar(10);continue; }

        memset(dp,0,sizeof(dp));

        for(int i=0;i<n;i++)  dp[i][i]=1;

        for(int r=1;r<n;r++)
        for(int i=0;i<n-r;i++)
        {
            int j=i+r;
            dp[i][j]=999999999;
            if((str[i]=='['&&str[j]==']')||(str[i]=='('&&str[j]==')'))
                if(dp[i][j]>dp[i+1][j-1])
                    dp[i][j]=dp[i+1][j-1],path[i][j]=-1;

            for(int k=i;k<j;k++)
                if(dp[i][j]>(dp[i][k]+dp[k+1][j]))
                dp[i][j]=dp[i][k]+dp[k+1][j],path[i][j]=k;
        }

        print(0,n-1);
        putchar(10);
    }

    return 0;
}