poj1141 Brackets Sequence

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

()[()]

题意：给你一个含括号的字符串，问你最小添加多少括号能使这个字符串的括号完全匹配。可以记录dp[i][j]为区间[i,j]中所有括号全部匹配需要的最少括号数，c[i][j]表示区间[i,j]中断开的坐标，便于待会递归输出。当i==j时，c[i][j]=-1,dp[i][j]=1;当str[i]==str[j]时，dp[i][j]=dp[i+1][j-1];再进行状态转移方程dp[i][j]=min(dp[i][j],dp[i][k]+dp[k+1][j]).同时更新c[i][j].

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<string>
#include<algorithm>
using namespace std;
#define inf 99999999
int dp[106][106],c[106][106];
char str[106];
int ok(char a,char b){
    if( (a=='('&&b==')') || (a=='[' && b==']'))return 1;
    return 0;
}

void shuchu(int l,int r)
{
    int i,j;
    if(l>r)return;
    if(l==r){
        if(str[l]=='(' || str[l]==')')printf("()");
        else printf("[]");
        return;
    }
    if(c[l][r]!=-1){
        shuchu(l,c[l][r]);
        shuchu(c[l][r]+1,r);
        return;
    }
    else{
        if(str[l]=='('){
            printf("(");
            shuchu(l+1,r-1);
            printf(")");
        }
        else{
            printf("[");
            shuchu(l+1,r-1);
            printf("]");
        }
        return;
    }
}

int main()
{
    int n,m,i,j,len,len1,k,l,r;
    while(gets(str)>0)
    {
        memset(c,-1,sizeof(c));
        len1=strlen(str);
        for(i=0;i<len1;i++){
            dp[i][i]=1;
        }
        for(i=0;i<len1-1;i++){
            if(ok(str[i],str[i+1]))dp[i][i+1]=0;
            else {
                dp[i][i+1]=2;
                c[i][i+1]=i;
            }
        }
        for(len=3;len<=len1;len++){
            for(i=0;i+len-1<len1;i++){
                j=i+len-1;
                dp[i][j]=inf;
                if(ok(str[i],str[j])){
                    dp[i][j]=dp[i+1][j-1];
                    c[i][j]=-1;
                }

                for(k=i;k<j;k++){
                    if(dp[i][j]>dp[i][k]+dp[k+1][j]){
                        dp[i][j]=dp[i][k]+dp[k+1][j];
                        c[i][j]=k;
                    }
                }
            }
        }
        shuchu(0,len1-1);
        printf("\n");
    }
    return 0;
}