记忆化搜索(DP+DFS) URAL 1183 Brackets Sequence

题目传送门

/*
    记忆化搜索(DP+DFS)：dp[i][j] 表示第i到第j个字符，最少要加多少个括号
        dp[x][x] = 1 一定要加一个括号；dp[x][y] = 0, x > y;
        当s[x] 与 s[y] 匹配，则搜索 (x+1, y-1); 否则在x~y-1枚举找到相匹配的括号，更新最小值
*/
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <iostream>
#include <cstring>
using namespace std;

const int MAXN = 1e2 + 10;
const int INF = 0x3f3f3f3f;
int dp[MAXN][MAXN];
int pos[MAXN][MAXN];
char s[MAXN];

void DFS(int x, int y)
{
    if (dp[x][y] != -1)    return ;
    if (x > y)        {dp[x][y] = 0;    return ;}
    if (x == y)    {dp[x][y] = 1;    return ;}

    dp[x][y] = INF;
    if ((s[x]=='(' && s[y]==')') || (s[x]=='[' && s[y]==']'))
    {
        pos[x][y] = -1;
        DFS (x+1, y-1);
        dp[x][y] = dp[x+1][y-1];
    }
    for (int i=x; i<y; ++i)
    {
        DFS (x, i);    DFS (i+1, y);
        int cur = dp[x][i] + dp[i+1][y];
        if (cur < dp[x][y])
        {
            dp[x][y] = cur;        pos[x][y] = i;
        }
    }
}

void print(int x, int y)
{
    if (x > y)    return ;
    if (x == y)
    {
        if (s[x] == '(' || s[y] == ')')    printf ("()");
        else    printf ("[]");
        return ;
    }
    if (pos[x][y] == -1)
    {
        printf ("%c", s[x]);
        print (x+1, y-1);
        printf ("%c", s[y]);
    }
    else
    {
        print (x, pos[x][y]);    print (pos[x][y]+1, y);
    }
}

int main(void)        //URAL 1183 Brackets Sequence
{
    //freopen ("O.in", "r", stdin);

    while (scanf ("%s", s+1) == 1)
    {
        int len = strlen (s+1);
        memset (dp, -1, sizeof (dp));
        memset (pos, 0, sizeof (pos));

        DFS (1, len);
        print (1, len);    puts ("");
    }

    return 0;
}