POJ 2001 Shortest Prefixes

Language: Default Shortest Prefixes Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 16792   Accepted: 7288 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

【题目分析】

    一看题目，是很简单的字典树的题目。只要建立一棵字典树，然后记录一下节点的经过此数，如果唯一就找到了最短的前缀。如果没有最小前缀，就需要输出整个字符串了。

#include <cstdio>
#include <iostream>
#include <string>
#include <cstring>
#include <algorithm>
using namespace std;
struct node{
	int now;
	int v;
};
int trie[50001][26];
node pos[50001];//每一个节点的信息 别问我为什么要记录当前节点的内容 其实可以删 但是我有空间任性 
string sl[5001],s;
int t=0;
int root=1;//这个变量就是搞笑的 
int sum=1;//总共的节点数 
void insert(string s)//插入字符串 字典树构建过程 
{
	int pp=1;
	pos[1].v++;
	for (int i=0;i<s.size();++i){
		if (trie[pp][s[i]-'a']!=0){
			pp=trie[pp][s[i]-'a'];
			pos[pp].v++;
		}
		else if (trie[pp][s[i]-'a']==0){
			trie[pp][s[i]-'a']=++sum;
			pos[sum].now=s[i]-'a';
			pos[sum].v++;
			pp=sum;
		}
	}
}
void find(string s)
{
	for (int i=0;i<s.size();++i) printf("%c",s[i]);
	printf(" ");
	int pp=1;
	for (int i=0;i<s.size();++i){
		printf("%c",s[i]);
		if (pos[trie[pp][s[i]-'a']].v==1){
			break;
		}
		pp=trie[pp][s[i]-'a'];
	}
	printf("\n");
}
int main()
{
	while(cin>>s){
		sl[++t]=s;
		insert(s); 
	}
	for (int i=1;i<=t;++i){
		find(sl[i]);
	}
}