hdu 4041

题意：如下图

遍历树的方法是粉色箭头所示
此图的输入是”walpurgis(charlotte(patricia,gertrud),elly,gisela)”
要求输出经过的结点名称，和每一步的起点和终点。
题解：用栈维护，遇到’(‘把’(‘前面的数字输出并压栈，如果遇到’,’，输出当前点和栈顶，如果遇到’)’，输出当前点和栈顶并把栈顶数字弹出。

#include <stdio.h>
#include <stack>
#include <string.h>
using namespace std;
const int N = 1000005;
const int M = 50005;
char str[N], room[M][15];
int res[N];
stack<int> s;

int main() {
    int t;
    scanf("%d", &t);
    while (t--) {
        while (!s.empty())
            s.pop();
        scanf("%s", str);
        int len = strlen(str);
        int k = 0, ss = 0, cnt = 0;
        for (int i = 0; i < len; i++) {
            if (str[i] <= ‘z‘ && str[i] >= ‘a‘)
                room[k][ss++] = str[i];
            else if (str[i] == ‘(‘) {
                room[k][ss] = 0;
                ss = 0;
                k++;
                s.push(k);
                res[cnt++] = k;
            }
            else if (str[i] == ‘)‘) {
                if (str[i - 1] != ‘)‘) {
                    room[k][ss] = 0;
                    k++;
                }
                else {
                    s.pop();
                    res[cnt++] = s.top();
                    ss = 0;
                    continue;
                }
                ss = 0;
                res[cnt++] = k;
                res[cnt++] = s.top();
            }
            else if (str[i] == ‘,‘) {
                if (str[i - 1] != ‘)‘) {
                    room[k][ss] = 0;
                    k++;
                }
                else {
                    s.pop();
                    res[cnt++] = s.top();
                    ss = 0;
                    continue;
                }
                ss = 0;
                res[cnt++] = k;
                res[cnt++] = s.top();
            }
        }
        if (k == 0) {
            printf("1\n");
            for (int i = 0; i < len; i++)
                if (str[i] <= ‘z‘ && str[i] >= ‘a‘)
                    printf("%c", str[i]);
            printf("\n\n");
            continue;
        }
        printf("%d\n", k);
        for (int i = 0; i < k; i++)
            printf("%s\n", room[i]);
        for (int i = 0; i < cnt - 1; i++)
            printf("%d %d\n", res[i], res[i + 1]);
        printf("\n");
    }
    return 0;
}