POJ 1141 Brackets Sequence(区间DP)

Brackets Sequence

 

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 27286		Accepted: 7727		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

 

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define INF 1000000000
const int N = 110;
int dp[N][N],path[N][N];
char s[N];
void print(int i,int j)
{
    if(i>j)return;
    if(i==j)
    {
        if(s[i]=='['||s[i]==']')
            printf("[]");
        else
            printf("()");
    }
    else if(path[i][j] == -1)
    {
        printf("%c",s[i]);
        print(i+1,j-1);
        printf("%c",s[j]);
    }
    else
    {
        print(i,path[i][j]);
        print(path[i][j]+1,j);
    }
}
int main()
{
    while(gets(s))
    {
        int len = strlen(s);
        if(len == 0)
        {
            printf("\n");
            continue;
        }
        memset(dp,0,sizeof(dp));
        for(int i=0; i<len; i++)
            dp[i][i] = 1;
        for(int i=len-1; i>=0; i--)
            for(int j=i+1; j<len; j++)
            {
                dp[i][j]=INF;
                if((s[i]=='['&&s[j]==']')||(s[i]=='('&&s[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,len-1);
        printf("\n");
    }
    return 0;
}