Light OJ 1114 Easily Readable 字典树

Posted 2020-07-02 phlsheji

题目来源：Light OJ 1114 Easily Readable

题意：求一个句子有多少种组成方案 仅仅要满足每一个单词的首尾字符一样 中间顺序能够变化

思路：每一个单词除了首尾 中间的字符排序 然后插入字典树 记录每一个单词的数量

输入一个句子 每一个单词也排序之后查找 依据乘法原理 答案就是每一个单词的数量之积

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;

const int maxnode = 111111;
const int sigma_size = 52;
int ch[maxnode][sigma_size];
int val[maxnode];
int sz;
void init()
{
	sz = 1;
	memset(ch[0], -1, sizeof(ch[0]));
}
int idx(char c)
{
	if(c >= 'a' && c <= 'z')
		return c - 'a';
	else
		return c - 'A' + 26;
}

void insert(char *s)
{
	int u = 0, n = strlen(s);
	for(int i = 0; i < n; i++)
	{
		int c = idx(s[i]);
		if(ch[u][c] == -1)
		{
			memset(ch[sz], -1, sizeof(ch[sz]));
			val[sz] = 0;
			ch[u][c] = sz++;
		}
		u = ch[u][c];
	}
	val[u]++;
}
int find(char *s)
{
	int u = 0, n = strlen(s);
	for(int i = 0; i < n; i++)
	{
		int c = idx(s[i]);
		u = ch[u][c];
		if(val[u] == -1)
			return 0;	
	}
	return val[u];
}

int main()
{
	int cas = 1;
	int T;
	scanf("%d", &T);
	while(T--)
	{
		init();
		int n;
		scanf("%d", &n);
		for(int i = 0; i < n; i++)
		{
			char s[11111];
			scanf("%s", s);
			int len = strlen(s);
			if(len > 2)
			sort(s+1, s+len-1);
			insert(s);
		}
		scanf("%d", &n);
		getchar();
		printf("Case %d:\n", cas++);
		while(n--)
		{
			int sum = 1;
			char s[11111];
			gets(s);
			char *p = strtok(s, " ");
			while(p)
			{
				char str[11111];
				strcpy(str, p);
				int len = strlen(str);
				if(len > 2)
					sort(str+1, str+len-1);
				sum *= find(str);
				p = strtok(NULL, " ");
			}
			printf("%d\n", sum);
		}
	}
	return 0;
}