zoj 3228 trie tree(字典树)

本文介绍了一种高效的字符串搜索算法,该算法能够快速查找一个长字符串中多个短子串出现的次数,支持子串出现的重叠与非重叠两种情况。通过使用前缀树(Trie)的数据结构来实现高效搜索,特别适用于子串长度较短的情况。

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Searching the String

Time Limit: 7 Seconds      Memory Limit: 129872 KB

Little jay really hates to deal with string. But moondy likes it very much, and she's so mischievous that she often gives jay some dull problems related to string. And one day, moondy gave jay another problem, poor jay finally broke out and cried, " Who can help me? I'll bg him! "

So what is the problem this time?

First, moondy gave jay a very long string A. Then she gave him a sequence of very short substrings, and asked him to find how many times each substring appeared in string A. What's more, she would denote whether or not founded appearances of this substring are allowed to overlap.

At first, jay just read string A from begin to end to search all appearances of each given substring. But he soon felt exhausted and couldn't go on any more, so he gave up and broke out this time.

I know you're a good guy and will help with jay even without bg, won't you?

Input

Input consists of multiple cases( <= 20 ) and terminates with end of file.

For each case, the first line contains string A ( length <= 10^5 ). The second line contains an integer N ( N <= 10^5 ), which denotes the number of queries. The next N lines, each with an integer type and a string a ( length <= 6 ), type = 0 denotes substring a is allowed to overlap and type = 1 denotes not. Note that all input characters are lowercase.

There is a blank line between two consecutive cases.

Output

For each case, output the case number first ( based on 1 , see Samples ).

Then for each query, output an integer in a single line denoting the maximum times you can find the substring under certain rules.

Output an empty line after each case.

Sample Input

ab
2
0 ab
1 ab

abababac
2
0 aba
1 aba

abcdefghijklmnopqrstuvwxyz
3
0 abc
1 def
1 jmn

Sample Output

Case 1
1
1

Case 2
3
2

Case 3
1
1
0

Hint

In Case 2,you can find the first substring starting in position (indexed from 0) 0,2,4, since they're allowed to overlap. The second substring starts in position 0 and 4, since they're not allowed to overlap.

For C++ users, kindly use scanf to avoid TLE for huge inputs.


Author: LI, Jie
Source: ZOJ Monthly, July 2009

字符串可重复就直接是模板题

对于不可以重复的串,因为串长度<=6 所以可以用标记 和枚举1到6 的串

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
const int N = 1e5 + 10;
int n,flag;
int ans[N],
      f[N],
      pre[N],
      cnt = 0;

struct trie
{
    trie *next[26];
    int id;
    trie (){
    for(int i = 0; i < 26 ; ++i) next[i] = NULL;
    id = 0;
    }
}e[N*6];
void init(){
    memset(pre,0,sizeof(pre));
    memset(ans,0,sizeof(ans));

    for(int i = 0; i <= cnt; i++){
            e[i].id = 0 ;
            for(int j = 0 ; j < 26 ; j++) e[i].next[j] = NULL;
    }
      cnt = 0;
}
class Trie
{
    public :
    trie * root ;
    void init() {
    root = &e[cnt++];
    }
    void insert(char *s, int id){
        int len = strlen(s);
        trie *p = root;
        for(int i = 0 ; i < len ; ++i){
            int idx = s[i] - 'a';
            if(!p->next[idx]) p->next[idx] = &e[cnt++];
            p = p->next[idx];
        }
            p->id = id;
        }
    int query(char *s, int len){
        trie * p = root;
        for(int i = 0; i < len ; ++i){
            int id = s[i] - 'a';
            if(!p->next[id]) return 0;
            p = p->next[id];
        }
        return p->id;
    }


}T[2];
char s[N],str[N][10];
void run()
{   int cas = 1;
    while(~scanf("%s",s)){
        init();
        T[1].init();
        T[0].init();
        scanf("%d",&n);
        for(int i = 1; i <= n; ++i){
            scanf("%d%s",&flag,str[i]);
            T[flag].insert(str[i],i);
            f[i] = flag;
        }
        int len = strlen(s);
        for(int i = 0 ; i < len ; ++i){
            for(int x = 1; i + x <= len && x <= 6; ++x){
                flag = T[0].query(s + i, x);
                ans[flag]++;
                flag = T[1].query(s + i, x);
                if(flag && pre[flag] <= i){
                    ans[flag]++;
                    pre[flag] = x + i;
            }

            }
        }
        //
        printf("Case %d\n",cas++);
        for(int i = 1 ; i <= n; i ++){
            flag = f[i];
            int len = strlen(str[i]);
            int id =  T[flag].query(str[i],len);
            printf("%d\n",ans[id]);

        }
        puts("");
    }
}
int main()
{
   // freopen("in]","r",stdin);
    run();
    return 0;
}


### ZOJ 1088 线段树 解题思路 #### 题目概述 ZOJ 1088 是一道涉及动态维护区间的经典问题。通常情况下,这类问题可以通过线段树来高效解决。题目可能涉及到对数组的区间修改以及单点查询或者区间查询。 --- #### 线段树的核心概念 线段树是一种基于分治思想的数据结构,能够快速处理区间上的各种操作,比如求和、最大值/最小值等。其基本原理如下: - **构建阶段**:通过递归方式将原数组划分为多个小区间,并存储在二叉树形式的节点中。 - **更新阶段**:当某一段区间被修改时,仅需沿着对应路径向下更新部分节点即可完成全局调整。 - **查询阶段**:利用懒惰标记(Lazy Propagation),可以在 $O(\log n)$ 时间复杂度内完成任意范围内的计算。 具体到本题,假设我们需要支持以下两种主要功能: 1. 对指定区间 `[L, R]` 执行某种操作(如增加固定数值 `val`); 2. 查询某一位置或特定区间的属性(如总和或其他统计量)。 以下是针对此场景设计的一种通用实现方案: --- #### 实现代码 (Python) ```python class SegmentTree: def __init__(self, size): self.size = size self.tree_sum = [0] * (4 * size) # 存储区间和 self.lazy_add = [0] * (4 * size) # 延迟更新标志 def push_up(self, node): """ 更新父节点 """ self.tree_sum[node] = self.tree_sum[2*node+1] + self.tree_sum[2*node+2] def build_tree(self, node, start, end, array): """ 构建线段树 """ if start == end: # 到达叶节点 self.tree_sum[node] = array[start] return mid = (start + end) // 2 self.build_tree(2*node+1, start, mid, array) self.build_tree(2*node+2, mid+1, end, array) self.push_up(node) def update_range(self, node, start, end, l, r, val): """ 区间更新 [l,r], 加上 val """ if l <= start and end <= r: # 当前区间完全覆盖目标区间 self.tree_sum[node] += (end - start + 1) * val self.lazy_add[node] += val return mid = (start + end) // 2 if self.lazy_add[node]: # 下传延迟标记 self.lazy_add[2*node+1] += self.lazy_add[node] self.lazy_add[2*node+2] += self.lazy_add[node] self.tree_sum[2*node+1] += (mid - start + 1) * self.lazy_add[node] self.tree_sum[2*node+2] += (end - mid) * self.lazy_add[node] self.lazy_add[node] = 0 if l <= mid: self.update_range(2*node+1, start, mid, l, r, val) if r > mid: self.update_range(2*node+2, mid+1, end, l, r, val) self.push_up(node) def query_sum(self, node, start, end, l, r): """ 查询区间[l,r]的和 """ if l <= start and end <= r: # 完全匹配 return self.tree_sum[node] mid = (start + end) // 2 res = 0 if self.lazy_add[node]: self.lazy_add[2*node+1] += self.lazy_add[node] self.lazy_add[2*node+2] += self.lazy_add[node] self.tree_sum[2*node+1] += (mid - start + 1) * self.lazy_add[node] self.tree_sum[2*node+2] += (end - mid) * self.lazy_add[node] self.lazy_add[node] = 0 if l <= mid: res += self.query_sum(2*node+1, start, mid, l, r) if r > mid: res += self.query_sum(2*node+2, mid+1, end, l, r) return res def solve(): import sys input = sys.stdin.read data = input().split() N, Q = int(data[0]), int(data[1]) # 数组大小 和 操作数量 A = list(map(int, data[2:N+2])) # 初始化数组 st = SegmentTree(N) st.build_tree(0, 0, N-1, A) idx = N + 2 results = [] for _ in range(Q): op_type = data[idx]; idx += 1 L, R = map(int, data[idx:idx+2]); idx += 2 if op_type == 'Q': # 查询[L,R]的和 result = st.query_sum(0, 0, N-1, L-1, R-1) results.append(result) elif op_type == 'U': # 修改[L,R]+X X = int(data[idx]); idx += 1 st.update_range(0, 0, N-1, L-1, R-1, X) print("\n".join(map(str, results))) solve() ``` --- #### 关键点解析 1. **初始化与构建**:在线段树创建过程中,需要遍历输入数据并将其映射至对应的叶子节点[^1]。 2. **延迟传播机制**:为了优化性能,在执行批量更新时不立即作用于所有受影响区域,而是记录更改意图并通过后续访问逐步生效[^2]。 3. **时间复杂度分析**:由于每层最多只访问两个子树分支,因此无论是更新还是查询都维持在 $O(\log n)$ 范围内[^3]。 ---
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