POJ 1141 Brackets Sequence

Brackets Sequence

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 29502		Accepted: 8402		Special Judge

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

()[()]

Source

Northeastern Europe 2001

题目大意: 给你一贯括号序列(只包含小括号和中括号)，让你找出长度最小的regular brackets sequence包含此子序列.其中的regular brackets sequence定义如下:

1)空序列是一个regular brackets sequence;

2)如果s是一个regular brackets sequence,那么[s] 也是一个regular brackets sequence,(s)也是一个regular brackets sequence.

3)如果A,B都是regular brackets sequence,那么AB也是一个regular brackets sequence.

例如:()、[]、()[] 、([]) 、([])()[()]都是regular brackets sequence。

而[[[、 (((((、 ([)] 则都不是regular brackets sequence。其中以“([)]”为例，包含它最小的regular brackets sequence有两个:()[()]或者是([])[].而你只要输出其中一个就行。

//f[i][j]表示区间i~j内需要最少的字符数能够匹配，path[i][j]表示到达该状态是哪种情况，  
//  -1表示第一个和最后一个，其他表示中间的分段点，然后递归输出(递归改变次序)   
#include<cstdio>
#include<cstring>
using namespace std;
#define N 1010
char s[N];
int f[N][N],path[N][N];
void out(int l,int r){
    if(l>r) return ;
    if(l==r){//到达了最后
        if(s[l]=='('||s[l]==')')
            printf("()");
        else
            printf("[]");
        return ;
    }
    if(path[l][r]==-1){//首尾，先输出开始，然后递归输出中间，最后输出结尾
        putchar(s[l]);
        out(l+1,r-1);
        putchar(s[r]);
    }
    else{
        out(l,path[l][r]);
        out(path[l][r]+1,r);
    }
}
int main(){
    while(gets(s+1)){//有空串，scanf("%s"),不能读空串，然后少一个回车，会出错
        int n=strlen(s+1);
        for(int i=1;i<=n;i++) f[i][i]=1;//一个的话只需一个就可以匹配  
        for(int x=1;x<n;x++){//枚举区间长度
            for(int i=1;i<=n-x;i++){//枚举区间开始位置
                int j=i+x;
                f[i][j]=0x3f3f3f3f;
                if((s[i]=='('&&s[j]==')')||(s[i]=='['&&s[j]==']')){
                    if(f[i+1][j-1]<f[i][j]){
                        f[i][j]=f[i+1][j-1];path[i][j]=-1;
                    }
                }
                for(int k=i;k<j;k++){//中间分隔情况
                    if(f[i][k]+f[k+1][j]<f[i][j]){
                        f[i][j]=f[i][k]+f[k+1][j];path[i][j]=k;
                    }
                }
            }
        }
        out(1,n);
        putchar('
');
    }
    return 0;
} 

//update 2.0
#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
const int N=105;
int n,f[N][N];char s[N];
bool match(char a,char b){
    if(a=='('&&b==')') return 1;
    if(a=='['&&b==']') return 1;
    return 0;
}
void print(int i,int j){
    if(i>j) return ;
    if(i==j){
        if(s[i]=='('||s[i]==')') printf("()");
        else printf("[]");
        return ;
    }
    int ans=f[i][j];
    if(match(s[i],s[j])&&ans==f[i+1][j-1]){
        printf("%c",s[i]);
        print(i+1,j-1);
        printf("%c",s[j]);
        return ;
    }
    for(int k=i;k<j;k++){
        if(ans==f[i][k]+f[k+1][j]){
            print(i,k);
            print(k+1,j);
            return ;
        }
    }
} 
int main(){
    while(gets(s+1)){
        n=strlen(s+1);
        memset(f,0x3f3f3f3f,sizeof f);
        for(int i=1;i<=n;i++){
            f[i+1][i]=0;
            f[i][i]=1;
        }
        for(int i=n-1;i>=1;i--){
            for(int j=i+1;j<=n;j++){
                f[i][j]=n;
                if(match(s[i],s[j])) f[i][j]=min(f[i][j],f[i+1][j-1]);
                for(int k=i;k<j;k++){
                    f[i][j]=min(f[i][j],f[i][k]+f[k+1][j]);
                }
            }
        }
        //printf("%d
",f[1][n]);
        print(1,n);
        putchar('
');
    }
    return 0;
}