POJ 1141 Brackets Sequence

括号序列，刘汝佳黑书上的经典例题。但是这道题要输出我们最后得到的添加括号最少的序列，

输出序列确实很麻烦，参考了题解，才勉勉强强写出来，以后还得把这道题敲一遍。

/*Accepted    288K    32MS    C++    1862B    2012-07-23 11:46:07*/

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
using namespace std;
#define o -2
#define lr -3

const int MAXN = 105;
int f[MAXN][MAXN], record[MAXN][MAXN];
char s[MAXN];
const int inf = 0x3f3f3f3f;

bool judge( int l, int r)
{
    if( s[l] == '(' && s[r] == ')' || s[l] == '[' && s[r] == ']')
        return true;
    return false;
}

int dps(int l, int r)
{
    int &ans = f[l][r], i;
    if(ans != inf)
        return ans;
    if(l > r)
        return 0;
    if(l == r)
        return record[l][r] = o, ans = 1;
    for( i = l; i < r; i ++)
        if( dps(l, i) + dps(i + 1, r) < ans)
            ans = f[l][i] + f[i + 1][r], record[l][r] = i;
    if( judge(l, r) && dps(l + 1, r - 1) < ans)
        ans = f[l + 1][r - 1], record[l][r] = lr;
    return ans;
}

void print(int l, int r)
{
    switch( record[l][r])
    {
        case o :{
            if( s[l] == '(' || s[l] == '[')
            {
                printf( "%c", s[l]);
                print( l + 1, r);
                printf( s[l] == '(' ? ")" : "]");
            }
            else
            {
                printf(s[l] == ')' ? "(" : "[");
                printf("%c", s[l]);
                print(l + 1, r);
            }
            break;
        }
        case lr:{
            printf( "%c", s[l]);
            print(l + 1, r - 1);
            printf( "%c", s[r]);
            break;
        }
        case -1:
            return;
        default:
            print( l, record[l][r]), print(record[l][r] + 1, r);
            break;
    }
}

int main()
{
    while( scanf( "%s", s + 1) == 1)
    {
        memset(f, 0x3f, sizeof f);
        memset(record, -1, sizeof record);
        dps(1, strlen(s + 1));
        print(1, strlen(s + 1));
        printf( "\n");
    }
    return 0;
}