【题目传送门】
1383 : The Book List
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:
ART
HISTORY
CHINESE HISTORY
THREE KINDOM
RESEARCHES ON CAOCAO
RESEARCHES ON LIUBEI
CHINESE MORDEN HISTORY
JAPANESE HISTORY
JAPANESE ACIENT HISTORY
MATH
GRAPH 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:
B
A
B
Case 2:
A0
A1
A1
B
B2
B4
C5
B1
B2
B6
C5
B1
B5
B32
B7
B3
B2
A3
B1
字典树大模拟。因为输出要按字典序输出，所以输入的时候需要给字符串排个序。而且输出的时候在相同层有后继的优先输出（题目没说清QAQ），所以在输出上要一点小处理。
1 #include<cstdio> 2 #include<iostream> 3 #include<cstring> 4 #include<algorithm> 5 #define clr(x) memset(x,0,sizeof(x)) 6 #define maxnode 3010 7 using namespace std; 8 struct node 9 { 10 int lt,rt; 11 char *str; 12 int val; 13 }; 14 struct trie 15 { 16 node poi[maxnode]; 17 int nodelen; 18 int head; 19 trie() {nodelen=1; head=0; clr(poi);} 20 void clear() {nodelen=1;head=0; clr(poi);} 21 void insert(int fa,int now,char *stri) 22 { 23 // printf("fat:%d p:%d nodelen:%d head:%d char:%s ",fa,now,nodelen,head,stri); 24 int end=false,i=0; 25 while(stri[i] && stri[i]!='/') 26 i++; 27 if(stri[i]==0) 28 end=true; 29 stri[i]=0; 30 if(!now) 31 { 32 now=newnode(stri); 33 poi[fa].lt=now; 34 if(!end) 35 insert(now,0,stri+i+1); 36 else 37 poi[now].val++; 38 return ; 39 } 40 int q; 41 while(now && strcmp(poi[now].str,stri)!=0) 42 { 43 q=now; 44 now=poi[now].rt; 45 } 46 if(!now) 47 { 48 now=newnode(stri); 49 poi[q].rt=now; 50 } 51 if(!end) 52 { 53 insert(now,poi[now].lt,stri+i+1); 54 } 55 else 56 { 57 poi[now].val++; 58 } 59 return ; 60 } 61 int newnode(char *stri) 62 { 63 if(!head) 64 { 65 head=nodelen; 66 } 67 poi[nodelen].str=stri; 68 return nodelen++; 69 } 70 void output(int node,int dep) 71 { 72 if(node==0) 73 { 74 return ; 75 } 76 int q=node; 77 while(q) 78 { 79 if(poi[q].lt!=0) 80 { 81 for(int i=0;i<dep;i++) 82 printf(" "); 83 printf("%s",poi[q].str); 84 printf(" "); 85 output(poi[q].lt,dep+1); 86 } 87 q=poi[q].rt; 88 } 89 q=node; 90 while(q) 91 { 92 if(poi[q].val) 93 { 94 for(int i=0;i<dep;i++) 95 printf(" "); 96 printf("%s",poi[q].str); 97 printf(" "); 98 } 99 q=poi[q].rt; 100 } 101 return ; 102 } 103 104 105 }tried; 106 bool cmp(char *a,char *b) 107 { 108 return strcmp(a,b)<0; 109 } 110 char s[110],deal[maxnode][110]; 111 char *dir[maxnode]; 112 int main() 113 { 114 int n,kase=0; 115 while(fgets(s,110,stdin)!=NULL) 116 { 117 clr(deal); 118 n=0; 119 s[strlen(s)-1]=' '; 120 tried.clear(); 121 strcpy(deal[n++],s); 122 dir[0]=deal[0]; 123 while(fgets(s,110,stdin)!=NULL && strcmp(s,"0 ")!=0) 124 { 125 s[strlen(s)-1]=' '; 126 strcpy(deal[n],s); 127 dir[n]=deal[n]; 128 n++; 129 } 130 sort(dir+0,dir+n,cmp); 131 for(int i=0;i<n;i++) 132 { 133 tried.insert(0,tried.head,dir[i]); 134 } 135 printf("Case %d: ",++kase); 136 tried.output(tried.head,0); 137 } 138 return 0; 139 }