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最新推荐文章于 2021-02-18 11:16:36 发布

four_knife

最新推荐文章于 2021-02-18 11:16:36 发布

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分类专栏：练习题

本文链接：https://blog.youkuaiyun.com/fourknife/article/details/23614109

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本文介绍了一种利用字典树解决字符串组合问题的方法。该方法通过构建字典树来存储字符串的各种可能组合，并以此来计算不同字符串组合的数量。文章详细介绍了字典树的构建过程及其在字符串匹配中的应用。

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题意：给出一个字符串s，可以任意分割成两个字符串s1和是s3，s1和s3可以得出顺序相反的反串s2和s4.由四个串可以两两组合，求一共能组合多少个不同的字符串。

思路：hash表或者字典树均可。字典树简单一点，而且叶子都在同一层，处理起来方便一些。

#include<iostream>
#include<cstring>

using namespace std;
char a[80];
int size;
struct node
{
	int val;
	node* ptr[26];
}mem[100000];

node* new_node(){
    return mem+(size++);
}
node* root;
int sum;

void build_str(char* str){
	node* p = root;
	while(*str){
		int t = *str - 'a';
		if (p->ptr[t] == NULL){
			node* q = new_node();
			q->val = 0;
			memset(q->ptr, 0, sizeof(q->ptr));
			p->ptr[t] = q;
			p = q;
		}
		else{
			p = p->ptr[t];
		}
		str++;
	}
	p->val = 1;
}

int check(char* str){
	node* p = root;
	char* s = str;
	while(*str){
		int t = *str - 'a';
		if (p->ptr[t] == NULL){
			build_str(s);
			return 1;
		}else{
			p = p->ptr[t];
			str++;
		}
	}
	
	return 0;
}

void match_up(int len, int in){
	char b[500];
	//F is order, L is inor
	for (int i = 0; i < len; i++){
		if (i <= in) b[i] = a[i];
		else b[i] = a[len-(i-in)];
	}
	b[len]='\0';
	//cout << b << endl;
	sum += check(b);
	//F is inor, L is order
	for (int i = 0; i < len; i++){
		if (i <= in) b[i] = a[in - i];
		else b[i] = a[i];
	}
	b[len] = '\0';
	sum += check(b);
	//cout << b << endl;
	//F is inor, L is inor
	for (int i = 0; i < len; i++){
		if (i <= in) b[i] = a[in-i];
		else b[i] = a[len-i+in];
	}
	b[len] = '\0';
	sum += check(b);
	//cout << b << endl;
	//L is order, F is order
	for (int i = 0; i < len; i++){
		if (len - i -1 > in) b[i] = a[in+i+1];
		else b[i] = a[in-len+i+1];
	}
	b[len] = '\0';
	sum += check(b);
	//cout << b << endl;
	//L is order, F is inor
	for (int i = 0; i < len; i++){
		if (len - i -1 > in) b[i] = a[in+i+1];
		else b[i] = a[len-i-1];
	}
	b[len] = '\0';
	sum += check(b);
	//cout << b << endl;
	//L is inor, F is order
	for(int i = 0; i < len; i++){
		if (len-i-1 > in) b[i] = a[len-i-1];
		else b[i] = a[in-len+i+1];
	}
	b[len] = '\0';
	sum += check(b);
	//cout << b << endl;
	//L is inor, F is inor
	for (int i = 0; i < len; i++){
		if (len-i-1 > in) b[i] = a[len-i-1];
		else b[i] = a[len-i-1];
	}
	b[len] = '\0';
	//cout << b << endl;
	sum += check(b);
}

int main(){
	int test;
	scanf("%d", &test);
	while(test--){
        size = 0;
        sum = 0;
        root = new_node();
        for (int i = 0; i < 26; i++){
            root->ptr[i] = NULL;
        }
		scanf("%s", a);
		build_str(a);

		int len = strlen(a);
		for (int i = 0; i < len; i++)
			match_up(len, i);
		printf("%d\n", sum+1);
	}
}