UVA10887--Trie--哈希

A language is a set of strings. And the concatenation of two languages is the set of all strings that are formed by concatenating the strings of the second language at the end of the strings of the first language.

For example, if we have two language A and B such that:

A = {cat, dog, mouse}

B = {rat, bat}

The concatenation of A and B would be:

C = {catrat, catbat, dograt, dogbat, mouserat, mousebat}

Given two languages your task is only to count the number of strings in the concatenation of the two languages.

Input

There can be multiple test cases. The first line of the input file contains the number of test cases,T (1≤T≤25). ThenT test cases follow. The first line of each test case contains two integers,M andN (M,N<1500), the number of strings in each of the languages. Then the nextMlines contain the strings of the first language. TheN following lines give you the strings of the second language. You can assume that the strings are formed by lower case letters (‘a’ to‘z’) only, that they are less than10 characters long and that each string is presented in one line without any leading or trailing spaces. The strings in the input languages may not be sorted and there will be no duplicate string.

Output

For each of the test cases you need to print one line of output. The output for each test case starts with the serial number of the test case, followed by the number of strings in the concatenation of the second language after the first language.

Sample Input Output for Sample Input

2 3 2 cat dog mouse rat bat 1 1 abc cab

Case 1: 6 Case 2: 1

思路：简单的用trie树标记。。关键这题坑啊！！！输入要用gets。。而且初始化也要注意下。。有空串的。。所以val[0] = 0;

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
using namespace std;
#define maxn 52600000
int ans;
char str1[1600][12],str2[1600][12],str[28];
struct Trie
{
	int first[maxn],nxt[maxn],ch[maxn];
	bool val[maxn];
	int cnt;
	void init()
	{	
		val[0] = 0;
		first[0] = -1;
		cnt = 1;
	}
	
	int idx(char c)
	{
		return c;
	}

	void insert(char * s)
	{
		int len = strlen(s);
		int u = 0;
		for(int i = 0;i < len;i++)
		{
			int c = idx(s[i]);
			bool flag = false;
			int j;
			for(j = first[u];j != -1;j = nxt[j])
			{
				if(ch[j] == c)
				{
					flag = true;
					break;
				}
			}
			if(!flag)
			{
				first[cnt] = -1;
				ch[cnt] = c;
				val[cnt] = 0;
				nxt[cnt] = first[u];
				first[u] = cnt;
				cnt++;
			}
			if(flag)	u = j;
			else u = cnt-1;
		}
		if(!val[u])	ans++;
		val[u] = 1;
	}
}trie;
		
int main()
{
	//freopen("in.txt","r",stdin);
	int t;
	scanf("%d",&t);
	for(int cas = 1;cas <= t;cas++)
	{
		trie.init();
		int n,m;
		scanf("%d%d",&n,&m);
		getchar();
		ans = 0;
		for(int i = 1;i <= n;i++)
			gets(str1[i]);
		for(int i = 1;i <= m;i++)
			gets(str2[i]);
		for(int i = 1;i <= n;i++)
			for(int j = 1;j <= m;j++)
			{
				strcpy(str,str1[i]);
				strcpy(&str[strlen(str1[i])],str2[j]);
				//strcat(str,str2[j]);
				trie.insert(str);
			}
		printf("Case %d: %d\n",cas,ans);
	}
	return 0;
}

用哈希也写了下：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
//1500 1500 20
#define maxn 2260007
int first[maxn],nxt[maxn];
int ans;
char str[maxn][24];
char s[1588][14];
char ss[14];
int Hash(int x)
{
	int len = strlen(str[x]);
	int ans = 0;
	for(int i = 0;i < len;i++)
		ans = ((ans * 17) + str[x][i]) % maxn;
	return ans;
}

void Try_to_insert(int x)
{
	int h = Hash(x);
	for(int i = first[h];i != -1;i = nxt[i])
	{
		if(!strcmp(str[x],str[i]))	return;
	}
	nxt[x] = first[h];
	first[h] = x;
	ans++;
}

void init()
{
	ans = 0;
	memset(first,-1,sizeof(first));
}

int main()
{
	//freopen("in.txt","r",stdin);
	int t;	scanf("%d",&t);
	for(int cas = 1;cas <= t;cas++)
	{
		init();
		int n,m;
		scanf("%d%d",&n,&m);
		getchar();
		for(int i = 1;i <= n;i++)	
			gets(s[i]);
		for(int i = 1;i <= m;i++)
		{
			gets(ss);
			for(int j = 1;j <= n;j++)
			{
				strcpy(str[(i-1)*n+j],s[j]);
				strcpy(&str[(i-1)*n+j][strlen(s[j])],ss);
				Try_to_insert((i-1)*n+j);
			}
		}
		printf("Case %d: %d\n",cas,ans);
	}
	return 0;
}