hihocoder1383 The Book List 字典树

时间限制:1000ms

单点时限:1000ms

内存限制:256MB

描述

The history of Peking University Library is as long as the history of Peking University. It was build in 1898. At the end of year 2015, it had about 11,000 thousand volumes of books, among which 8,000 thousand volumes were paper books and the others were digital ones. Chairman Mao Zedong worked in Peking University Library for a few months as an assistant during 1918 to 1919. He earned 8 Dayang per month there, while the salary of top professors in Peking University is about 280 Dayang per month.

Now Han Meimei just takes the position which Chairman Mao used to be in Peking University Library. Her first job is to rearrange a list of books. Every entry in the list is in the format shown below:

CATEGORY 1/CATEGORY 2/..../CATEGORY n/BOOKNAME

It means that the book BOOKNAME belongs to CATEGORY n, and CATEGORY n belongs to CATEGORY n-1, and CATEGORY n-1 belongs to CATEGORY n-2...... Each book belongs to some categories. Let's call CATEGORY1 "first class category", and CATEGORY 2 "second class category", ...ect. This is an example:

MATH/GRAPH THEORY
ART/HISTORY/JAPANESE HISTORY/JAPANESE ACIENT HISTORY
ART/HISTORY/CHINESE HISTORY/THREE KINDOM/RESEARCHES ON LIUBEI
ART/HISTORY/CHINESE HISTORY/CHINESE MORDEN HISTORY
ART/HISTORY/CHINESE HISTORY/THREE KINDOM/RESEARCHES ON CAOCAO

Han Meimei needs to make a new list on which the relationship between books and the categories is shown by indents. The rules are:

1) The n-th class category has an indent of 4×(n-1) spaces before it.
2) The book directly belongs to the n-th class category has an indent of 4×n spaces before it.
3) The categories and books which directly belong to a category X should be list below X in dictionary order. But all categories go before all books.
4) All first class categories are also list by dictionary order.

For example, the book list above should be changed into the new list shown below:

ARTHISTORYCHINESE HISTORYTHREE KINDOMRESEARCHES ON CAOCAORESEARCHES ON LIUBEICHINESE MORDEN HISTORYJAPANESE HISTORYJAPANESE ACIENT HISTORY
MATHGRAPH THEORY

Please help Han Meimei to write a program to deal with her job.

输入

There are no more than 10 test cases.
Each case is a list of no more than 30 books, ending by a line of "0".
The description of a book contains only uppercase letters, digits, '/' and spaces, and it's no more than 100 characters.
Please note that, a same book may be listed more than once in the original list, but in the new list, each book only can be listed once. If two books have the same name but belong to different categories, they are different books.

输出

For each test case, print "Case n:" first(n starts from 1), then print the new list as required.

样例输入

B/A
B/A
B/B
0
A1/B1/B32/B7
A1/B/B2/B4/C5
A1/B1/B2/B6/C5
A1/B1/B2/B5
A1/B1/B2/B1
A1/B3/B2
A3/B1
A0/A1
0

样例输出

Case 1:
BAB
Case 2:
A0A1
A1BB2B4C5B1B2B6C5B1B5B32B7B3B2
A3B1

自己做的一小撮测试用例（大小写不敏感），查看纯文本view plain，不然格式不对

input:

B/A
B/A
B/B
0
A1/B1/B32/B7
A1/B/B2/B4/C5
A1/B1/B2/B6/C5
A1/B1/B2/B5
A1/B1/B2/B1
A1/B3/B2
A3/B1
A0/A1
0
A/B/A/B
A/B/A/B
0
A/B
A/C
0
A
B
C
D
0
A
0
A1/B1/B1
A1/B1/B2
A1/B2/B1
A1/B3
A1/B4
B5
B6
C1/C1
C1/C2
C1/C3
0
ART/HISTORY/CHINESE HISTORY/THREE KINDOM/RESEARCHES ON CAOCAO
ART/HISTORY/CHINESE HISTORY/THREE KINDOM/RESEARCHES ON LIUBEI
ART/HISTORY/CHINESE HISTORY/CHINESE MORDEN HISTORY
ART/HISTORY/JAPANESE HISTORY/JAPANESE ACIENT HISTORY
MATH/GRAPH THEORY
0
a b/b c/d e
d e
a b/d e
0
MATH/GRAPH THEORY
ART/HISTORY/JAPANESE HISTORY/JAPANESE ACIENT HISTORY
ART/HISTORY/CHINESE HISTORY/THREE KINDOM/RESEARCHES ON LIUBEI
ART/HISTORY/CHINESE HISTORY/CHINESE MORDEN HISTORY
ART/HISTORY/CHINESE HISTORY/THREE KINDOM/RESEARCHES ON CAOCAO
0
0/0/0/0
0/0/0/1
0
1/2/3/4
1/2/3/2
1/2/3/2
0/ / / / / /a/ /
0/
0
0000/0000/ /0000
0 0 0/0000/ /0000
0 0 0/0000/0/0000
0
a/b/c
a/a
b/a
b/b
b/c/d
b/e/f/g
0/a
0
A/B/C
A/B
0

output:

Case 1:
BAB
Case 2:
A0A1
A1BB2B4C5B1B2B6C5B1B5B32B7B3B2
A3B1
Case 3:
ABAB
Case 4:
ABC
Case 5:
A
B
C
D
Case 6:
A
Case 7:
A1B1B1B2B2B1B3B4
C1C1C2C3
B5
B6
Case 8:
ARTHISTORYCHINESE HISTORYTHREE KINDOMRESEARCHES ON CAOCAORESEARCHES ON LIUBEICHINESE MORDEN HISTORYJAPANESE HISTORYJAPANESE ACIENT HISTORY
MATHGRAPH THEORY
Case 9:
a bb cd ed e
d e
Case 10:
ARTHISTORYCHINESE HISTORYTHREE KINDOMRESEARCHES ON CAOCAORESEARCHES ON LIUBEICHINESE MORDEN HISTORYJAPANESE HISTORYJAPANESE ACIENT HISTORY
MATHGRAPH THEORY
Case 11:
00001
Case 12:
12324
Case 13:a0Case 14:
0 0 00000000000000
000000000000
Case 15:
0a
abca
bcdefgab
Case 16:
ABCB

AC代码：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <map>
#include <string>
#include <algorithm>
#include <istream>
#include <cstdlib>using namespace std;const int maxnode = 300000, maxdep = 100000;
char a[maxdep + 5];
int lena;
map<string, int> ch[maxnode + 5][2];   //ch[父节点号][是不是叶子节点][节点名称] --> 子节点编号
map<int, string> m; //m[节点编号] --> 节点名称
int val[maxnode + 5]; //叶子节点标记struct Trie {int sz;Trie() {sz = 1;for (int i = 0; i < maxdep + 3; i++) {ch[i][0].clear();ch[i][1].clear();}m.clear();memset(val, 0, sizeof(val));}void clear() {sz = 1;for (int i = 0; i < maxdep + 3; i++) {ch[i][0].clear();ch[i][1].clear();}m.clear();memset(val, 0, sizeof(val));}void insert(char *s, int p) {int in, u = 0;while (1) {lena = 0;for (in = p; s[in] != '/' && s[in] != 0; in++) {a[lena++] = s[in];}a[lena] = 0;string ts = a;if (!ch[u][0][ts] && s[in] != 0) {ch[u][0][ts] = sz++;m[ch[u][0][ts]] = ts; //map from node id to string}if (!ch[u][1][ts] && s[in] == 0) {ch[u][1][ts] = sz++;m[ch[u][1][ts]] = ts; //map from node id to string}//debug// if (!ch[u][0][ts] && !ch[u][1][ts]) {//     if (s[in] == 0) {//       ch[u][1][ts] = sz++;//       m[ch[u][1][ts]] = ts; //map from node id to string//   }//     else {//        ch[u][0][ts] = sz++;//       m[ch[u][0][ts]] = ts; //map from node id to string//   }// }u = (s[in] ? ch[u][0][ts] : ch[u][1][ts]);if (s[in] == 0) {val[u] = 1;break;}p = in + 1;}}void output(int u, int dep) {//先输出目录for (map<string, int>::iterator i = ch[u][0].begin(); i != ch[u][0].end(); i++) {if (i->second != 0) {for (int j = 0; j < dep; j++) {printf("    ");}cout << m[i->second] << endl;output(i->second, dep + 1);}}//再输出书for (map<string, int>::iterator i = ch[u][1].begin(); i != ch[u][1].end(); i++) {if (val[i->second]) {for (int j = 0; j < dep; j++) {printf("    ");}cout << m[i->second] << endl;}}}
};char s[maxnode + 5];int main()
{/*freopen("D:\\input.txt", "r", stdin);freopen("D:\\output.txt", "w", stdout);*/Trie tree;int kase = 0;while (gets(s) != NULL) {if (s[0] == '0' && s[1] == '\0') {printf("Case %d:\n", ++kase);tree.output(0, 0);tree.clear();continue;}else {tree.insert(s, 0);}}return 0;
}

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hihocoder1383 The Book List 字典树

描述

输入

输出

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