1. 程式人生 > >【trie樹】POJ2001Shortest Prefixes

【trie樹】POJ2001Shortest Prefixes

Shortest Prefixes
Time Limit: 1000MS        Memory Limit: 30000K
Total Submissions: 22161        Accepted: 9470
Description

A prefix of a string is a substring starting at the beginning of the given string. The prefixes of "carbon" are: "c", "ca", "car", "carb", "carbo", and "carbon". Note that the empty string
is not considered a prefix in this problem, but every non-empty string is considered to be a prefix of itself. In everyday language, we tend to abbreviate words by prefixes. For example, "carbohydrate" is commonly abbreviated by "carb". In this problem, given a set of words, you will find for each word the shortest prefix that uniquely identifies the word it represents. In the sample input below,
"carbohydrate" can be abbreviated to "carboh", but it cannot be abbreviated to "carbo" (or anything shorter) because there are other words in the list that begin with "carbo". An exact match will override a prefix match. For example, the prefix "car" matches the given word "car" exactly. Therefore, it is
understood without ambiguity that "car" is an abbreviation for "car" , not for "carriage" or any of the other words in the list that begins with "car". Input The input contains at least two, but no more than 1000 lines. Each line contains one word consisting of 1 to 20 lower case letters. Output The output contains the same number of lines as the input. Each line of the output contains the word from the corresponding line of the input, followed by one blank space, and the shortest prefix that uniquely (without ambiguity) identifies this word. Sample Input carbohydrate cart carburetor caramel caribou carbonic cartilage carbon carriage carton car carbonate Sample Output carbohydrate carboh cart cart carburetor carbu caramel cara caribou cari carbonic carboni cartilage carti carbon carbon carriage carr carton carto car car carbonate carbona Source Rocky Mountain 2004
T

這道題就是一個trie樹(別問我怎麼知道的)

這道題就是要求每個串的最短辨識串

思路也非常簡單

就是在trie樹剛剛分叉的地方輸出就好了

我的方式就是在每個節點記個cnt,在加的時候走過一次就把cnt++,

然後遇到cnt等於1的時候就return就好了

 1 //#include<bits/stdc++.h>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 #define idx(i) (i-'a')
 6 #define N 1011
 7 using namespace std;
 8 char s[N][30];
 9 int inp;
10 struct TRIE{int to[26],val,cnt;}tree[N*30];
11 int cnt=1;
12 int regist()
13 {
14     memset(&tree[cnt],0,sizeof(TRIE));
15     return cnt++;
16 }
17 void insert(char *now)
18 {
19     int len=strlen(now),rt=0,c;
20     for(int i=0;i<len;i++)
21     {
22         c=idx(now[i]);
23         if(!tree[rt].to[c])
24         {
25             tree[rt].to[c]=regist();
26         }
27         rt=tree[rt].to[c];
28         tree[rt].cnt++;
29     }
30     tree[rt].val++;
31 }
32 void find(char *now)
33 {
34     int len=strlen(now),rt=0,c;
35     for(int i=0;i<len;i++)
36     {
37         c=idx(now[i]);
38         if(tree[rt].cnt==1)return;
39         printf("%c",now[i]);
40         rt=tree[rt].to[c];
41     }
42 }
43 int main()
44 {
45     while(scanf("%s",s[++inp]+1)!=EOF)
46         insert(s[inp]+1);    
47     for(int i=1;i<=inp;i++)
48     {
49         printf("%s ",s[i]+1);
50         find(s[i]+1);
51         printf(" \n");
52     }
53     return 0;
54 }