poj1141 Brackets Sequence

Brackets Sequence

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 24949 Accepted: 7018 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

([(]

Brackets Sequence

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 24949 Accepted: 7018 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

()[()]

题目大意.

给出一组括号，添加最少的括号个数使其完全匹配.

1.空串是匹配的

2.如果s是匹配的，那(s),[s]是匹配的.

3.如果a匹配，b匹配.那ab是匹配的.

注意:输入数据只有一个，且可能有多余空格（用gets()读取就好了）.

4.输出注意是有换行的（这和其他题目类似）

我的想法.

1.将原题的匹配问题看成是,从0，到n-1的匹配.问题的定义是从l到r的匹配.dp[l][r]

2.只有一个字符（l==r）是一定不匹配的,(当匹配后,一定是输出() 或者 [] )

3.如果l<r；那麽dp[l][r]=min(dp[l][k]+dp[k+1][r]);但是要处理直接有l于r匹配的情况.否则无限递归.

具体如下.

<pre name="code" class="cpp">    /*Source Code    Problem: 1141User:     Memory: 8188KTime: 94MS    Language: GCCResult: Accepted    Source Code*/    #include <stdio.h>    #include <string.h>    #define maxn 1001          //开大了哈    #define maxv ( (1 << 30) -1 )    int dp[maxn][maxn]={-1};    int path[maxn][maxn]={-1};    int p[maxn]={0};    int min(int x,int y){    return x<y?x:y;    }    int isMatch(char left,char right){    if((left=='('&&right==')')||(left=='['&&right==']'))          return 1;    else           return 0;    }    int Match(char str[],int l,int r){    if(l>r) return 0;                       if(dp[l][r]!=-1) return dp[l][r]; //计算过的    if(l==r){                         //一定会添加一个    return (dp[l][r]=1);    }else{    int k;    int m=maxv;    if(isMatch(str[l],str[r])){ //如果可以匹配先记录当前匹配的个数    m=Match(str,l+1,r-1);    path[l][r]=-2;      //嫖妓l,r的匹配为直接匹配.    }    for(k=l;k+1<=r;k++){    if(m==maxv||(m>Match(str,l,k)+Match(str,k+1,r))){ //没有计算或者，存在分为两段的更好方法.          m=Match(str,l,k)+Match(str,k+1,r);          path[l][r]=k;               //记录l,r，是经过分为l,path[l][k], 和 path[l][k]+1到r这样分割的.    }    }    return dp[l][r]=m;    }    }    void print(char str[],int l,int r){    if(l>r) return;    else if(l==r){    if(str[l]==')') printf("(%c",str[l]);    if(str[l]==']') printf("[%c",str[l]);    if(str[l]=='(') printf("%c)",str[l]);    if(str[l]=='[') printf("%c]",str[l]);    }else{    if(path[l][r]==-2){    printf("%c",str[l]);    print(str,l+1,r-1);    printf("%c",str[r]);    }else{    print(str,l,path[l][r]);    print(str,path[l][r]+1,r);    }    }    }    int main(){    char str[maxn]="";    int l=0,r,m=maxv;    int i;    memset(dp,-1,sizeof(dp));    memset(path,-1,sizeof(path));    if (gets(str) == NULL)    return 0;    r=strlen(str)-1;    Match(str,l,r);    print(str,l,r);    printf("\n");    return 0;    }