UVA10562 Undraw the Trees【樹】

Professor Homer has been reported missing. We suspect that his recent research works might have had something to with this. But we really don’t know much about what he was working on! The detectives tried to hack into his computer, but after hours of failed efforts they realized that the professor had been lot more intelligent than them. If only they could realize that the professor must have been absent minded they could get the clue rightaway. We at the crime lab were not at all surprised when the professor’s works were found in a 3.5」 floppy disk left inside the drive.
    The disk contained only one text file in which the professor had drawn many trees with ASCII characters. Before we can proceed to the next level of investigation we would like to match the trees drawn with the ones that we have in our database. Now you are the computer geek —we leave this trivial task for you. Convert professor’s trees to our trees.
Input
The first line of the input file (which you can assume comes from standard input) contains the number of trees, T (1 ≤ T ≤ 20) drawn in the file. Then you would have T trees, each ending with a single hash (‘#’) mark at the beginning of the line. All the trees drawn here are drawn vertically in top down fashion. The labels of each of node can be any printable character except for the symbols ‘-’, ‘|’, ‘ ’ (space) and ‘#’. Every node that has children has a ‘|’ symbol drawn just below itself. And in the next line there would be a series of ‘-’ marks at least as long as to cover all the immediate children. The sample input section will hopefully clarify your doubts if there is any. No tree drawn here requires more than 200 lines, and none of them has more than 200 characters in one line.
Output
Our trees are drawn with parenthesis and the node labels. Every subtree starts with an opening parenthesis and ends with a closing parenthesis; inside the parenthesis the node labels are listed. The sub trees are always listed from left to right. In our database each tree is written as a single string in one line, and they do not contain any character except for the node labels and the parenthesis pair. The node labels can be repeated if the original tree had such repetitions.
Sample Input
2
A
|
--------
B C D
| |
----- -
E F G

e
|
----
f g

Sample Output
(A(B()C(E()F())D(G())))
(e(f()g()))

問題連接：UVA10562 Undraw the Trees
問題簡述：（略）
問題分析：
    輸入的樹是字符圖形，將其轉換爲表達式形式的樹，是一種從文本到文本的處理。在字符界面時代，這種處理是常見的。
    用遞歸程序來處理是建樹的一種好辦法。遞歸過程就是樹形的結構的。實際上不須要建樹，只須要創建遞歸過程，而後輸出層次結構便可。
程序說明：（略）
參考連接：（略）
題記：（略）

AC的C++程序以下：

/* UVA10562 Undraw the Trees */

#include <bits/stdc++.h>

using namespace std;

const int N = 200;
char s[N + 1][N + 1];
int lcnt;

void dfs(int x, int y)
{
    printf("%c(", s[x][y]);
    if(x < lcnt - 1 && s[x + 1][y] == '|') {        // 出現「|」有子樹
        int i = y;
        while(i >= 1 && s[x + 2][i - 1] == '-') i--;  // 查找最左邊' - '
        while(s[x + 2][i] == '-' && s[x + 3][i] != '\0') {
            if(s[x + 3][i] != ' ')    //查找子樹根（相差3行）
                dfs(x + 3, i);
            i++;
        }
    }
    printf(")");
}

int main()
{
    int t;
    scanf("%d", &t);
    getchar();
    while(t--) {
        lcnt = 0;
        do {
            gets(s[lcnt++]);
        } while(strcmp(s[lcnt - 1], "#") != 0);
        lcnt--;

        printf("(");
        if(lcnt) {
            int len = strlen(s[0]);
            for(int i = 0; i < len; i++)        // 查找樹根結點
                if(s[0][i] != ' ') {
                    dfs(0, i);
                    break;
                }
        }
        printf(")\n");
    }
    return 0;
}