Brackets Sequence

1183. Brackets Sequence Time limit: 1.0 second Memory limit: 64 MB Let us define a regular brackets sequence in the following way: Empty sequence is a regular sequence. If S is a regular sequence, then (S) and [S] are both regular sequences. 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 a1a2...an is called a subsequence of the string b1b2...bm, if there exist such indices 1 ≤ i1 < i2 < ... < in ≤ m, that aj=bij for all 1 ≤ j ≤ n. Input The input contains at most 100 brackets (characters '(', ')', '[' and ']') that are situated on a single line without any other characters among them. Output Write 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 output ([(] ()[()]

#include <map>
#include <set>
#include <list>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <vector>
#include <bitset>
#include <cstdio>
#include <string>
#include <numeric>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
typedef long long  ll;
typedef unsigned long long ull;

int dx[4]= {-1,1,0,0};
int dy[4]= {0,0,-1,1}; //up down left right
bool inmap(int x,int y,int n,int m)
{
    if(x<1||x>n||y<1||y>m)return false;
    return true;
}
int hashmap(int x,int y,int m)
{
    return (x-1)*m+y;
}

#define eps 1e-8
#define inf 0x7fffffff
#define debug puts("BUG");
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define Read freopen("in.txt","r",stdin)
#define Write freopen("out.txt","w",stdout)

char str[200]= {0};
int dp[110][110]= {0},len;
int tag[111][111]= {0};

void read()
{
    scanf("%s",str);
}
void print(int s,int e)
{
    if (s>e)return;
    if (s==e)
    {
        if (str[s]=='(' || str[s]==')')
            printf("()");
        else
            printf("[]");
        return ;
    }
    if (tag[s][e]==-1)
    {
        printf("%c",str[s]);
        print(s+1,e-1);
        printf("%c",str[e]);
    }
    else
    {
        print(s,tag[s][e]);
        print(tag[s][e]+1,e);
    }
}
void DP()
{
    len = strlen(str);
    for (int i=0; i<len; i++)
    {
        dp[i][i]=1;//初始
        dp[i+1][i]=0;//下面在进行DP时，i=j+1时，i=i+1、j=j-1会造成i=j+1
    }
    for (int st=1; st<len; st++)
    {
        for (int i=0; i+st<len; i++)
        {
            int j=i+st;
            int temp=9999999;
            if ((str[i]=='(' && str[j]==')')||(str[i]=='[' && str[j]==']'))
            {
                temp=dp[i+1][j-1];//第一种情况（s）或[s]
            }
            tag[i][j]=-1;
            for (int k=i; k<j; k++) //第二种情况，枚举i到j之间的k 即AB
            {
                int res=dp[i][k]+dp[k+1][j];
                if (res<temp)
                {
                    temp=res;
                    tag[i][j]=k;
                }
            }
            dp[i][j]=temp;
        }
    }
    print(0,len-1);//根据标记情况进行回溯
    printf("\n");
}
int main()
{
    read();
    DP();
    return 0;
}