Brackets Sequence （POJ-1141）

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：给你一行括号，只有小括号（）和中括号【】，令其匹配好，即左括号一定有个右括号匹配，然后把最短的匹配好的括号输出；

思路：区域DP，但是比较难的地方是输出，用pos把需要添加的地方记录下来，然后二分查找输出；

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#define INF 0x3f3f3f3fusing namespace std;int f[105][105],pos[105][105];char s[105];void fun(int x,int y){    if(x>y) return;    if(x==y){        if(s[x]=='('||s[x]==')') printf("()");        if(s[x]=='['||s[x]==']') printf("[]");    }    else if(x<y){        if(pos[x][y]==-1){            if(s[x]=='('){                printf("(");                fun(x+1,y-1);                printf(")");            }            else{                printf("[");                fun(x+1,y-1);                printf("]");            }        }        else{            fun(x,pos[x][y]);            fun(pos[x][y]+1,y);        }    }}int main(){    scanf("%s",s);    memset(f,0,sizeof(f));    int len=strlen(s);    for(int i=len;i>0;i--){        s[i]=s[i-1];        f[i][i]=1;    }    for(int l=1;l<=len;l++){        for(int i=1;i<=len-l;i++){            int j=l+i;            f[i][j]=INF;            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];            }            pos[i][j]=-1;            for(int k=i;k<j;k++){                if(f[i][k]+f[k+1][j]<f[i][j]){                    f[i][j]=f[i][k]+f[k+1][j];                    pos[i][j]=k;                }            }        }    }    fun(1,len);    printf("\n");    return 0;}