UVA 1626 - Brackets sequence


http://www.cnblogs.com/20143605--pcx/p/4791122.html
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 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 begins with a single positive integer on a line by itself indicating the number of the cases following, each of them as described below. This line is followed by a blank line, and there is also a blank line between two consecutive inputs.


The input file contains at most 100 brackets (characters '(', ')', '[' and ']') that are situated on a single line without any other characters among them.

Output 

For each test case, the output must follow the description below. The outputs of two consecutive cases will be separated by a blank line.


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 

1

([(]

Sample Output 

()[()]

这是《入门经典》上的一道例题。如果仅让求最短序列是极简单的,定义dp(i,j)表示将区间 i~j 变为合法添加的最小字符数.
      则 dp(i,j)=dp(i+1,j-1)   (i与j能匹配时),    dp(i,j)=min(dp(i,k)+dp(k+1,j))。
但是,输出的时候就比较慢了。从左往右通过比较选择最优方案递归输出(看的书上代码)。
char p[105];
int dp[105][105];
const int INF=100000;
bool match(int x,int y)
{
    if(p[x]=='('&&p[y]==')')
        return true;
    if(p[x]=='['&&p[y]==']')
        return true;
    return false;
}
void DP()
{
    int len=strlen(p);
    for(int l=1;l<=len;++l){
        for(int i=0;i+l-1<len;++i){
            int r=i+l-1;
            if(l==1){
                dp[i][r]=1;
                continue;
            }
            if(l==2){
                if(match(i,r))
                    dp[i][r]=0;
                else
                    dp[i][r]=2;
                continue;
            }
            dp[i][r]=INF;
            if(match(i,r))
                dp[i][r]=dp[i+1][r-1];
            for(int k=i;k<r;++k){
                dp[i][r]=min(dp[i][r],dp[i][k]+dp[k+1][r]);
            }
        }
    }
}
void print(int i,int j)
{
    if(i>j)
        return ;
    if(i==j){
        if(p[i]=='('||p[i]==')')
            printf("()");
        else
            printf("[]");
        return ;
    }
    int ans=dp[i][j];
    if(match(i,j)&&ans==dp[i+1][j-1]){
        printf("%c",p[i]);
        print(i+1,j-1);
        printf("%c",p[j]);
        return ;
    }
    for(int k=i;k<j;++k){
        if(ans==dp[i][k]+dp[k+1][j]){
            print(i,k);
            print(k+1,j);
            return ;
        }
    }
}
int main()
{
    int T;
    scanf("%d",&T);
    getchar();
    while(T--)
    {
        getchar();
        gets(p);
        int len=strlen(p);
        if(len>0){
            DP();
            print(0,len-1);
        }
        printf("\n");
        if(T)
            printf("\n");
    }
    return 0;
}

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