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最新推荐文章于 2025-05-10 18:09:08 发布

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最新推荐文章于 2025-05-10 18:09:08 发布

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文章标签：算法

本文链接：https://blog.youkuaiyun.com/qq_52898168/article/details/120579403

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本文解析了四个编程题目，涉及数据结构（字典树、二进制操作）、字符串处理（立即可解编码）和算法应用（最大 XOR 对组合并）。通过实例演示了如何使用C++实现插入、查询和解码操作，适合算法竞赛和编程入门学习。

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题目链接：http://#10050. 「一本通 2.3 例 2」The XOR Largest Pair

#include<bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10, M = 31 * N;
int a[N];
int son[M][2],idx;//idx建字典树时给每一节点做标号
void insert(int x)
{
	int p = 0;
	for(int i = 30; i >= 0; i --)
	{
		int u = x >> i & 1;
		if(!son[p][u]) son[p][u] = ++ idx;
		p = son[p][u];
	}
}
int query(int x)
{
	int p = 0, res = 0;
	for(int i = 30; i >= 0; i --)
	{
		int u = x >> i & 1;
		if(son[p][!u])//二进制第i位不同
		{
			p = son[p][!u];
			res = res * 2 + !u;
		}
		else//相同
		{
			p = son[p][u];
			res = res * 2 + u;
		}
	}
	return res;
}
int main()
{
	int res=0,n;
	cin>>n;
	for(int i = 0; i < n; i ++)
	{
		scanf("%d", &a[i]);
		insert(a[i]);
		int t = query(a[i]);
		res = max(res, a[i] ^ t);
	}
	printf("%d\n", res);
	return 0;
}

题目链接：https://loj.ac/p/10052

#include <stdio.h>
#include <string.h>
int a[123224][3];
int book[1234];
int cnt;
int insert(char x[])
{
	int i, lx = strlen(x);
	int f = 0, id = 1;
	for (i = 0; i < lx; i++)
	{
		int xx = x[i] - '0';
		if (a[id][xx] == 0) a[id][xx] = ++cnt;
		else if (i==lx-1)	f = 1;//x[]为另一字符串的子串
		if (book[id]) f=1;//另一字符串为x[]的字串
		id = a[id][xx];
	}
	book[id] = 1;
	return f;
}
int main()
{
	char x[12];
	int f = 0;
	int q = 0;
	cnt = 1;
	memset(book, 0, sizeof(book));
	while (~scanf("%s", &x))
	{
		int l = strlen(x);
		if (l == 1 && x[0] == '9')
		{
			q++;
			if(!f) printf("Set %d is immediately decodable\n", q);
			else   printf("Set %d is not immediately decodable\n", q);
			f = 0;
			cnt = 1;
			memset(book, 0, sizeof(book));
			memset(a, 0, sizeof(a));
		}
		else if (insert(x))
				f = 1;
	}
	return 0;
}

题目链接：https://loj.ac/p/10054

#include<bits/stdc++.h>
using namespace std;
const int N=5e5+7;
int tree[N][2],a[N];
int sum[N],size[N],cnt;
void insert(int k)
{
	int p=0;
	for(int i=1; i<=k; p=tree[p][a[i]],++i)
	{
		if(!tree[p][a[i]]) tree[p][a[i]]=++cnt;
		size[p]++;
	}
	sum[p]++;
	size[p]++;
}
void quiry(int k)
{
	int ans=0,p=0;
	for(int i=1; i<=k; p=tree[p][a[i]],++i)
	{
		if(!tree[p][a[i]]) break;
		if(i!=k) ans+=sum[tree[p][a[i]]];//当前字符串为另一字符串的母串
		else ans+=size[tree[p][a[i]]];//当前字符串为另一字符串的子串
	}
	printf("%d\n",ans);
}
int main()
{
	int i,j,k;
	int n,m;
	cin>>n>>m;
	for(i=1; i<=n; ++i)
	{
		scanf("%d",&k);
		for(j=1; j<=k; ++j) scanf("%d",&a[j]);
		insert(k);
	}
	for(i=1; i<=m; ++i)
	{
		scanf("%d",&k);
		for(j=1; j<=k; ++j) scanf("%d",&a[j]);
		quiry(k);
	}
	return 0;
}

题目链接：https://loj.ac/p/10056

:https://blog.youkuaiyun.com/luyehao1/article/details/86495109

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int M = 1e5 + 10;
const int N = 1e6 + 10;
struct dd
{
	int to, w;
};
int n;
vector<dd>tree[M];
int trie[N][2];
int ans,cnt;
void Insert(int x)
{
	int root = 0;
	for (int i = 30; i >= 0; i--)
	{
		int v = (x >> i) & 1;
		if (!trie[root][v])	trie[root][v] = ++cnt;
		root = trie[root][v];
	}
}
int query(int x)
{
	int root = 0;
	int ans = 0;
	for (int i = 30; i >= 0; i--)
	{
		int v = (x >> i) & 1;
		if (trie[root][v ^ 1])
		{
			ans += (1 << i);
			root = trie[root][v ^ 1];
		}
		else root=trie[root][v];
	}
	return ans;
}
void dfs(int root, int last, int sum)
{
	ans = max(ans,query(sum));
	for (int i = 0; i < tree[root].size(); i++)
	{
		int now= tree[root][i].to;
		if (now == last)	continue;
		int w = tree[root][i].w^sum;
		Insert(w);
		dfs(now, root, w);
	}
}
int main()
{
	scanf("%d", &n);
	for (int i = 1; i < n; i++)
	{
		int u, v, w;
		scanf("%d%d%d", &u, &v, &w);
		tree[u].push_back(dd{ v,w });
		tree[v].push_back(dd{ u,w });
	}
	Insert(0);
	dfs(1, -1, 0);
	printf("%d\n", ans);
	return 0;
}