区间dp - 括号匹配并输出方案

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：给你一个不完整的括号序列，要求你添加最少的括号，使其可以构成一个左右互相匹配的完整的序列。
思路分析：开始想了个贪心，但是是不对
　　　　正解是区间 dp，dp[i][j]表示区间 i - j 内添加最小数量的括号可以使其匹配，然后在转移的过程中判断一下当前区间的括号是否可以匹配上，如果可以此时的值则等于其内部区间的值，否则则在加一层 for去判断
　　　　输出的地方采用递归的方式去输出，比较经典的一个题
代码示例：

char s[105];
int dp[105][105], path[105][105];

void print(int l, int r){
    if (l > r) return;
    
    if (l == r){
        if (s[l] == '(' || s[l] == ')') printf("()");
        if (s[l] == '[' || s[l] == ']') printf("[]");
        return;
    }
    
    if (path[l][r] == -1){
        putchar(s[l]);
        print(l+1, r-1);
        putchar(s[r]);
    }
    else{
        print(l, path[l][r]);
        print(path[l][r]+1, r);
    }
}

int main() {
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    
    while(gets(s+1) != NULL){
        int n = strlen(s+1);
        memset(dp, 0, sizeof(dp));
        for(int i = 1; i <= n; i++) dp[i][i] = 1;
    
        for(int len = 2; len <= n; len++){ // 区间长度
            for(int i = 1; i <= n; i++){
                int j = i+len-1;
                if (j > n) break;
                dp[i][j] = inf;
                if ((s[i]=='('&&s[j]==')') || (s[i]=='['&&s[j]==']')){
                    dp[i][j] = dp[i+1][j-1];
                    path[i][j] = -1;
                }
                for(int k = i; k < j; k++){
                    if (dp[i][k]+dp[k+1][j] < dp[i][j]){
                        dp[i][j] = dp[i][k]+dp[k+1][j];
                        path[i][j] = k;
                    }
                }
          //      printf("+++ %d %d %d \n", i, j, dp[i][j]);
            }
        }        
        print(1, n);
        //printf("%d\n", dp[1][n]);
        printf("\n");
    }
    return 0;
}
/*
([(] 
*/