POJ 1141(Bracket Sequence)-区间DP

Brackets Sequence

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 23158		Accepted: 6508		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

 

 

经典区间DP。。。

/************************************************
* author:crazy_石头
* Pro:POJ 1141 Bracket Sequence
* algorithm:区间DP
* Time:110ms
* Judge Status:Accepted
* 解题思路：
* dp[i][j]表示i~j间最少添加的字符能使所有括号匹配
* 那么当i和j匹配时，i~j的匹配数(dp[i][j])等于i+1~j-1
* 间的匹配数，即dp[i][j]=dp[i+1][j-1];
* 对于一般情况下，dp[i][j]=min(dp[i][j],dp[i][k]+dp[k+1][j]
* 不断枚举k即可；找出最小值，用path记录路径，path=-1
* 表示i~j间括号得到匹配，输出递归输出即可；
**************************************************/
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>

using namespace std;

#define LL long long
#define rep(i,h,n) for(int i=(h);i<=(n);i++)
#define ms(a,b) memset((a),(b),sizeof(a))
#define INF 1<<29

const int maxn=1500+5;
const int maxm=32000+5;
const LL mod=10000007 ;

int dp[maxn][maxn];//i，j间最少添加几个字符串才匹配；
int path[maxn][maxn];
char str[maxn];

inline bool match(int i,int j)
{
    if((str[i]=='['&&str[j]==']')||(str[i]=='('&&str[j]==')'))
    return true;
    return false;
}

inline void output(int i,int j)
{
    if(i>j)return;
    if(i==j)
    {
        if(str[i]=='['||str[i]==']')printf("[]");
        else printf("()");
        return ;
    }
    if(path[i][j]==-1)//path[i][j]==-1时表示匹配；
    {
        printf("%c",str[i]);
        output(i+1,j-1);
        printf("%c",str[j]);
        return ;
    }

    output(i,path[i][j]);
    output(path[i][j]+1,j);

}

inline void solve()
{
    int len=strlen(str);
    rep(i,0,len-1)
    dp[i][i]=1;
    for(int i=1;i<len;i++)
    {
        for(int j=0;j+i<len;j++)
        {
            int t=j+i;
            dp[j][t]=INF;
            if(match(j,t))
            {
                dp[j][t]=dp[j+1][t-1];
                path[j][t]=-1;
            }
            for(int k=j;k<t;k++)
            {
                if(dp[j][t]>dp[j][k]+dp[k+1][t])
                {
                    dp[j][t]=dp[j][k]+dp[k+1][t];
                    path[j][t]=k;
                }
            }
        }
    }
}

int main()
{
    while(gets(str))
    {
        ms(dp,0);
        ms(path,0);
        int len=strlen(str)-1;
        solve();
        output(0,len);
        printf("\n");
    }
    return 0;
}