【原】 POJ 1141 Brackets Sequence 动态规划 解题报告

http://poj.org/problem?id=1141

方法：
对角线方向求解DP。

设c[i][j]表示原串str[i...j]补齐后的最短长度。a[i][j]表示原串str[i...j]补齐后的字符串。
c[1][n]和a[1][n]即为所求结果。n为原串的长度。
 
初始状态：
c[i][i] = 2
a[i][i] = "()"，if str[i]="(" or ")"
        = "[]"，if str[i]="[" or "]"
c[i][j] = 0，if i>j

根据性质（3）一个序列如果是AB形式的话，我们可以划分为A，B两个子问题，我们可以得到递归式：
枚举[i,j)范围内的k
c[i][j] = min{ c[i][k] + c[k+1][j]，for all k in [i,j) }
a[i][j] = a[i][k] + a[k+1][j]
此递归式为“最优二分检索树问题”形式：C[i,j]=w[i,j]+opt{C[i,k-1]+C[k,j]}.

根据性质（2）一个序列如果是[A]或者(A)的形式，我们可以把它降为向下分解为A即可
c[i][j] = min{ tmp , c[i+1][j-1]+2 } ,其中tmp就是由性质3得到的c[i][j]
a[i][j] = str[i] + c[i+1][j-1]+ str[j]

由上面可以看出c[N][N]矩阵的对角线都为2，左下部分全为0。
c[i][j]依赖于c[i][k]、c[k+1][j]和c[i+1][j-1]。所以需要以对角线为边界，沿对角线方向求解。

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

()[()]

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

using namespace std ;

const int N = 101 ;
const int INF = 0x7fffffff ;

int c[N][N] ;    //c[i][j]表示原串str[i...j]补齐后的最短长度
string a[N][N] ; //a[i][j]表示原串str[i...j]补齐后的字符串
char str[N] ;    //原串，从索引1开始

//返回原串长n
int Initialize( )
{
    int i,j ;
    int n ;

    scanf( "%s", (str+1) ) ;  //从索引1开始
    n = strlen(str+1) ;

    memset( c , 0 , sizeof(c) ) ;

    for( i=1 ; i<=n ; ++i )  //设置对角线初始值
    {
        for( j=1 ; j<=n ; ++j )
            a[i][j] = "" ;
        c[i][i] = 2 ;
        if( str[i]=='(' || str[i]==')' )
            a[i][i] = "()" ;
        else
            a[i][i] = "[]" ;
    }

    return n ;
}

inline
bool Match( char a, char b )
{
    if( a=='(' && b==')' )
        return true ;
    else if( a=='[' && b==']' )
        return true ;
    else
        return false ;
}

void DP( int n )
{
    int i,j,k ;
    int b,e ;

    //精妙的双循环使得按照对角线方向求解
    for( j=1 ; j<n ; ++j )  //j为纵向比横向多出的距离
    {
        for( i=1 ; i+j<=n ; ++i )  //i+j即为纵向坐标
        {
            b = i ;
            e = i+j ;
            c[b][e] = INF ;

            for( k=b ; k<e ; ++k )  //性质3
            {
                if( c[b][e] > c[b][k]+c[k+1][e] )
                {
                    c[b][e] = c[b][k] + c[k+1][e] ;
                    a[b][e] = a[b][k] + a[k+1][e] ;
                }
            }

            if( Match( str[b] , str[e] ) )  //性质2
            {
                if( c[b][e] > c[b+1][e-1]+2 )
                {
                    c[b][e] = c[b+1][e-1]+2 ;
                    a[b][e] = str[b] + a[b+1][e-1] + str[e] ;
                }
            }
        }
    }
}

void run1141()
{
    int n ;
    n = Initialize() ;
    DP(n) ;
    cout<<a[1][n]<<endl ;
}