@ZZHow(ZZHow1024)
字符串
KMP字符串
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致敬三位前辈：D.E.Knuth、J.H.Morris 和 V.R.Pratt！
字符串模式匹配算法，大大避免重复遍历的情况，克努特-莫里斯-普拉特算法。
java
import java.io.*;
import java.util.*;
public class Main {
static final int N = 100010;
static final int M = 1000010;
static char[] p = new char[N];
static char[] s = new char[M];
static int[] ne = new int[N];
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
int n = Integer.parseInt(br.readLine());
String pStr = " " + br.readLine();
int m = Integer.parseInt(br.readLine());
String sStr = " " + br.readLine();
p = pStr.toCharArray();
s = sStr.toCharArray();
// 计算 next 数组
for (int i = 2, j = 0; i <= n; i++) {
while (j != 0 && p[i] != p[j + 1])
j = ne[j];
if (p[i] == p[j + 1])
j++;
ne[i] = j;
}
// KMP 匹配
for (int i = 1, j = 0; i <= m; i++) {
while (j != 0 && s[i] != p[j + 1])
j = ne[j];
if (s[i] == p[j + 1])
j++;
if (j == n) {
bw.write((i - n) + " ");
j = ne[j];
}
}
br.close();
bw.close();
}
}Trie树
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用于高效地存储和查找字符串集合的数据结构。
java
import java.io.*;
import java.util.*;
public class Main {
static final int N = 100010;
static int[][] son = new int[N][26]; // 存储所有节点的儿子是什么
static int[] cnt = new int[N]; // 字符串结束标记且进行计数
static int idx; // 下标为 0 的点，既是跟节点，又是空节点
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
int n = Integer.parseInt(br.readLine());
while (n-- != 0) {
String[] lines = br.readLine().split(" ");
char op = lines[0].charAt(0);
String str = lines[1];
switch (op) {
case 'I': {
insert(str.toCharArray());
break;
}
case 'Q': {
bw.write(query(str.toCharArray()) + "");
bw.newLine();
}
}
}
br.close();
bw.close();
}
// 插入字符串
public static void insert(char[] str) {
int p = 0;
for (int i = 0; i < str.length; i++) {
int u = str[i] - 'a';
if (son[p][u] == 0)
son[p][u] = ++idx;
p = son[p][u];
}
cnt[p]++;
}
// 查询字符串出现的次数
public static int query(char[] str) {
int p = 0;
for (int i = 0; i < str.length; i++) {
int u = str[i] - 'a';
if (son[p][u] == 0)
return 0;
p = son[p][u];
}
return cnt[p];
}
}并查集
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两个操作：
- 将两个集合合并。
- 询问两个元素是否在一个集合当中。
基本原理：每个集合用一棵树来表示。树根的编号就是整个集合的编号。每个节点存储
它的父节点，Px 表示 x 的父节点。
问题 1：如何判断树根 if (p[x] == ×) {return true;}
问题 2：如何求 x 的集合编号 while (p[x] != x) x = p[x];
问题 3：如何合并两个集合，px 是 x 的集合编号，py 是 y 的集合编号 p[x] = y;
java
import java.util.Scanner;
public class Main {
static int[] p;
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int n = scanner.nextInt();
p = new int[n + 10];
for (int i = 1; i <= n; i++)
p[i] = i;
int m = scanner.nextInt();
while (m-- != 0) {
char op = scanner.next().charAt(0);
int a = scanner.nextInt();
int b = scanner.nextInt();
switch (op) {
case 'M': {
p[find(a)] = find(b);
break;
}
case 'Q': {
System.out.println((find(a) == find(b)) ? "Yes" : "No");
break;
}
}
}
}
// 返回祖宗节点 + 路径压缩
public static int find(int x) {
if (x != p[x])
p[x] = find(p[x]);
return p[x];
}
}堆
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手写堆，功能：
- 插入一个数：
heap[++size] = x; up(size); - 求集合中的最小值：
heap[1]; - 删除最小值：
heap[1] = heap[size]; size--; down(1); - 删除任意一个元素：
heap[1] = heap[k]; size--; down(1); up(1); - 修改任意一个元素：
heap[k] = x; down(k); up(k);
使用一维数组存储：
- x 左儿子：2x
- x 右儿子：2x + 1
java
import java.util.Scanner;
public class Main {
static final int N = 100010;
static int[] h;
static int[] ph;
static int[] hp;
static int size;
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int n = scanner.nextInt();
h = new int[n + 10];
ph = new int[n + 10];
hp = new int[n + 10];
int m = 0;
while (n-- != 0) {
String op = scanner.next();
switch (op) {
case "I": {
int x = scanner.nextInt();
size++;
m++;
ph[m] = size;
hp[size] = m;
h[size] = x;
up(size);
break;
}
case "PM": {
System.out.println(h[1]);
break;
}
case "DM": {
heapSwap(1, size);
size--;
down(1);
break;
}
case "D": {
int k = scanner.nextInt();
k = ph[k];
heapSwap(k, size);
size--;
down(k);
up(k);
break;
}
case "C": {
int k = scanner.nextInt();
int x = scanner.nextInt();
k = ph[k];
h[k] = x;
down(k);
up(k);
break;
}
}
}
}
public static void heapSwap(int a, int b) {
int temp = ph[hp[a]];
ph[hp[a]] = ph[hp[b]];
ph[hp[b]] = temp;
temp = hp[a];
hp[a] = hp[b];
hp[b] = temp;
temp = h[a];
h[a] = h[b];
h[b] = temp;
}
public static void up(int u) {
while (u / 2 != 0 && h[u / 2] > h[u]) {
heapSwap(u / 2, u);
u /= 2;
}
}
public static void down(int u) {
int t = u; // t 表示三个节点中最小节点的编号
if (u * 2 <= size && h[u * 2] < h[t])
t = u * 2;
if (u * 2 + 1 <= size && h[u * 2 + 1] < h[t])
t = u * 2 + 1;
if (u != t) {
heapSwap(u, t);
down(t);
}
}
}