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;
}
时间: 2024-10-04 06:35:16