1.如果知道两个子范围上的最值，通过比较就可以知道整个范围上的最值
2.如果知道两个子范围上分别出现的次数最多的数，但无法知道整个范围上出现最多的数
范围修改logn的前提：如果维护的是区域和，要把区域上的每个数字加上a，只需要知道区域中数字的个数乘以a加到原数字和上就可以得到新的和，时间复杂度为O(1)，那么这样的修改复杂度就是logn；如果是将区域上的每个数数位上的数字都倒置得到新的数字，那么这样是无法直接得到新的累加和的，这样的修改操作就不是logn
P3372 【模板】线段树 1 - 洛谷 | 计算机科学教育新生态 (luogu.com.cn)
范围查询，范围修改
java
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.io.StreamTokenizer;
public class Main {
public static int MAXN = 100001;
public static long[] arr = new long[MAXN];
public static long[] sum = new long[MAXN << 2];
public static long[] add = new long[MAXN << 2];
// 累加和信息的汇总
public static void up(int i) {
// 父范围的累加和 = 左范围累加和 + 右范围累加和
sum[i] = sum[i << 1] + sum[i << 1 | 1];
}
// 懒信息的下发
public static void down(int i, int ln, int rn) {
if (add[i] != 0) {
// 发左
lazy(i << 1, add[i], ln);
// 发右
lazy(i << 1 | 1, add[i], rn);
// 父范围懒信息清空
add[i] = 0;
}
}
// 当前来到l~r范围，对应的信息下标是i，范围上数字的个数是n = r-l+1
// 现在收到一个懒更新任务 : l~r范围上每个数字增加v
// 这个懒更新任务有可能是任务范围把当前线段树范围全覆盖导致的
// 也有可能是父范围的懒信息下发下来的
// 总之把线段树当前范围的sum数组和add数组调整好
// 就不再继续往下下发了，懒住了
public static void lazy(int i, long v, int n) {
sum[i] += v * n;
add[i] += v;
}
// 建树
public static void build(int l, int r, int i) {
if (l == r) {
sum[i] = arr[l];
} else {
int mid = (l + r) >> 1;
build(l, mid, i << 1);
build(mid + 1, r, i << 1 | 1);
up(i);
}
add[i] = 0;
}
// 范围修改
// jobl ~ jobr范围上每个数字增加jobv
public static void add(int jobl, int jobr, long jobv, int l, int r, int i) {
if (jobl <= l && r <= jobr) {
lazy(i, jobv, r - l + 1);
} else {
int mid = (l + r) >> 1;
down(i, mid - l + 1, r - mid);
if (jobl <= mid) {
add(jobl, jobr, jobv, l, mid, i << 1);
}
if (jobr > mid) {
add(jobl, jobr, jobv, mid + 1, r, i << 1 | 1);
}
up(i);
}
}
// 查询累加和
public static long query(int jobl, int jobr, int l, int r, int i) {
if (jobl <= l && r <= jobr) {
return sum[i];
}
int mid = (l + r) >> 1;
down(i, mid - l + 1, r - mid);
long ans = 0;
if (jobl <= mid) {
ans += query(jobl, jobr, l, mid, i << 1);
}
if (jobr > mid) {
ans += query(jobl, jobr, mid + 1, r, i << 1 | 1);
}
return ans;
}
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StreamTokenizer in = new StreamTokenizer(br);
PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
in.nextToken(); int n = (int) in.nval;
in.nextToken(); int m = (int) in.nval;
for (int i = 1; i <= n; i++) {
in.nextToken();
arr[i] = (long) in.nval;
}
build(1, n, 1);
long jobv;
for (int i = 1, op, jobl, jobr; i <= m; i++) {
in.nextToken();
op = (int) in.nval;
if (op == 1) {
in.nextToken(); jobl = (int) in.nval;
in.nextToken(); jobr = (int) in.nval;
in.nextToken(); jobv = (long) in.nval;
add(jobl, jobr, jobv, 1, n, 1);
} else {
in.nextToken(); jobl = (int) in.nval;
in.nextToken(); jobr = (int) in.nval;
out.println(query(jobl, jobr, 1, n, 1));
}
}
out.flush();
out.close();
br.close();
}
}范围查询，单点修改
java
public class Code02_SegmentTreeUpdateQuerySum {
public static int MAXN = 100001;
public static long[] arr = new long[MAXN];
public static long[] sum = new long[MAXN << 2];
public static long[] change = new long[MAXN << 2];
public static boolean[] update = new boolean[MAXN << 2];
public static void up(int i) {
sum[i] = sum[i << 1] + sum[i << 1 | 1];
}
public static void down(int i, int ln, int rn) {
if (update[i]) {
lazy(i << 1, change[i], ln);
lazy(i << 1 | 1, change[i], rn);
update[i] = false;
}
}
public static void lazy(int i, long v, int n) {
sum[i] = v * n;
change[i] = v;
update[i] = true;
}
public static void build(int l, int r, int i) {
if (l == r) {
sum[i] = arr[l];
} else {
int mid = (l + r) >> 1;
build(l, mid, i << 1);
build(mid + 1, r, i << 1 | 1);
up(i);
}
change[i] = 0;
update[i] = false;
}
public static void update(int jobl, int jobr, long jobv, int l, int r, int i) {
if (jobl <= l && r <= jobr) {
lazy(i, jobv, r - l + 1);
} else {
int mid = (l + r) >> 1;
down(i, mid - l + 1, r - mid);
if (jobl <= mid) {
update(jobl, jobr, jobv, l, mid, i << 1);
}
if (jobr > mid) {
update(jobl, jobr, jobv, mid + 1, r, i << 1 | 1);
}
up(i);
}
}
public static long query(int jobl, int jobr, int l, int r, int i) {
if (jobl <= l && r <= jobr) {
return sum[i];
}
int mid = (l + r) >> 1;
down(i, mid - l + 1, r - mid);
long ans = 0;
if (jobl <= mid) {
ans += query(jobl, jobr, l, mid, i << 1);
}
if (jobr > mid) {
ans += query(jobl, jobr, mid + 1, r, i << 1 | 1);
}
return ans;
}
// 对数器逻辑
// 展示了线段树的建立和使用
// 使用验证结构来检查线段树是否正常工作
public static void main(String[] args) {
System.out.println("测试开始");
int n = 1000;
int v = 2000;
int t = 5000000;
// 生成随机值填入arr数组
randomArray(n, v);
// 建立线段树
build(1, n, 1);
// 生成验证的结构
long[] check = new long[n + 1];
for (int i = 1; i <= n; i++) {
check[i] = arr[i];
}
for (int i = 1; i <= t; i++) {
// 生成操作类型
// op = 0 更新操作
// op = 1 查询操作
int op = (int) (Math.random() * 2);
// 下标从1开始，不从0开始，生成两个随机下标
int a = (int) (Math.random() * n) + 1;
int b = (int) (Math.random() * n) + 1;
// 确保jobl <= jobr
int jobl = Math.min(a, b);
int jobr = Math.max(a, b);
if (op == 0) {
// 更新操作
// 线段树、验证结构同步更新
int jobv = (int) (Math.random() * v * 2) - v;
update(jobl, jobr, jobv, 1, n, 1);
checkUpdate(check, jobl, jobr, jobv);
} else {
// 查询操作
// 线段树、验证结构同步查询
// 比对答案
long ans1 = query(jobl, jobr, 1, n, 1);
long ans2 = checkQuery(check, jobl, jobr);
if (ans1 != ans2) {
System.out.println("出错了!");
}
}
}
System.out.println("测试结束");
}
java
public class Code03_SegmentTreeAddQueryMax {
public static int MAXN = 100001;
public static long[] arr = new long[MAXN];
public static long[] max = new long[MAXN << 2];
public static long[] add = new long[MAXN << 2];
public static void up(int i) {
max[i] = Math.max(max[i << 1], max[i << 1 | 1]);
}
public static void down(int i) {
if (add[i] != 0) {
lazy(i << 1, add[i]);
lazy(i << 1 | 1, add[i]);
add[i] = 0;
}
}
public static void lazy(int i, long v) {
max[i] += v;
add[i] += v;
}
public static void build(int l, int r, int i) {
if (l == r) {
max[i] = arr[l];
} else {
int mid = (l + r) >> 1;
build(l, mid, i << 1);
build(mid + 1, r, i << 1 | 1);
up(i);
}
add[i] = 0;
}
public static void add(int jobl, int jobr, long jobv, int l, int r, int i) {
if (jobl <= l && r <= jobr) {
lazy(i, jobv);
} else {
down(i);
int mid = (l + r) >> 1;
if (jobl <= mid) {
add(jobl, jobr, jobv, l, mid, i << 1);
}
if (jobr > mid) {
add(jobl, jobr, jobv, mid + 1, r, i << 1 | 1);
}
up(i);
}
}
public static long query(int jobl, int jobr, int l, int r, int i) {
if (jobl <= l && r <= jobr) {
return max[i];
}
down(i);
int mid = (l + r) >> 1;
long ans = Long.MIN_VALUE;
if (jobl <= mid) {
ans = Math.max(ans, query(jobl, jobr, l, mid, i << 1));
}
if (jobr > mid) {
ans = Math.max(ans, query(jobl, jobr, mid + 1, r, i << 1 | 1));
}
return ans;
}
// 对数器逻辑
// 展示了线段树的建立和使用
// 使用验证结构来检查线段树是否正常工作
public static void main(String[] args) {
System.out.println("测试开始");
int n = 1000;
int v = 2000;
int t = 5000000;
// 生成随机值填入arr数组
randomArray(n, v);
// 建立线段树
build(1, n, 1);
// 生成验证的结构
long[] check = new long[n + 1];
for (int i = 1; i <= n; i++) {
check[i] = arr[i];
}
for (int i = 1; i <= t; i++) {
// 生成操作类型
// op = 0 增加操作
// op = 1 查询操作
int op = (int) (Math.random() * 2);
// 下标从1开始，不从0开始，生成两个随机下标
int a = (int) (Math.random() * n) + 1;
int b = (int) (Math.random() * n) + 1;
// 确保jobl <= jobr
int jobl = Math.min(a, b);
int jobr = Math.max(a, b);
if (op == 0) {
// 增加操作
// 线段树、验证结构同步增加
int jobv = (int) (Math.random() * v * 2) - v;
add(jobl, jobr, jobv, 1, n, 1);
checkAdd(check, jobl, jobr, jobv);
} else {
// 查询操作
// 线段树、验证结构同步查询
// 比对答案
long ans1 = query(jobl, jobr, 1, n, 1);
long ans2 = checkQuery(check, jobl, jobr);
if (ans1 != ans2) {
System.out.println("出错了!");
}
}
}
System.out.println("测试结束");
}
java
public class Code04_SegmentTreeUpdateQueryMax {
public static int MAXN = 100001;
public static long[] arr = new long[MAXN];
public static long[] max = new long[MAXN << 2];
public static long[] change = new long[MAXN << 2];
public static boolean[] update = new boolean[MAXN << 2];
public static void up(int i) {
max[i] = Math.max(max[i << 1], max[i << 1 | 1]);
}
public static void down(int i) {
if (update[i]) {
lazy(i << 1, change[i]);
lazy(i << 1 | 1, change[i]);
update[i] = false;
}
}
public static void lazy(int i, long v) {
max[i] = v;
change[i] = v;
update[i] = true;
}
public static void build(int l, int r, int i) {
if (l == r) {
max[i] = arr[l];
} else {
int mid = (l + r) >> 1;
build(l, mid, i << 1);
build(mid + 1, r, i << 1 | 1);
up(i);
}
change[i] = 0;
update[i] = false;
}
public static void update(int jobl, int jobr, long jobv, int l, int r, int i) {
if (jobl <= l && r <= jobr) {
lazy(i, jobv);
} else {
down(i);
int mid = (l + r) >> 1;
if (jobl <= mid) {
update(jobl, jobr, jobv, l, mid, i << 1);
}
if (jobr > mid) {
update(jobl, jobr, jobv, mid + 1, r, i << 1 | 1);
}
up(i);
}
}
public static long query(int jobl, int jobr, int l, int r, int i) {
if (jobl <= l && r <= jobr) {
return max[i];
}
down(i);
int mid = (l + r) >> 1;
long ans = Long.MIN_VALUE;
if (jobl <= mid) {
ans = Math.max(ans, query(jobl, jobr, l, mid, i << 1));
}
if (jobr > mid) {
ans = Math.max(ans, query(jobl, jobr, mid + 1, r, i << 1 | 1));
}
return ans;
}
// 对数器逻辑
// 展示了线段树的建立和使用
// 使用验证结构来检查线段树是否正常工作
public static void main(String[] args) {
System.out.println("测试开始");
int n = 1000;
int v = 2000;
int t = 5000000;
// 生成随机值填入arr数组
randomArray(n, v);
// 建立线段树
build(1, n, 1);
// 生成验证的结构
long[] check = new long[n + 1];
for (int i = 1; i <= n; i++) {
check[i] = arr[i];
}
for (int i = 1; i <= t; i++) {
// 生成操作类型
// op = 0 更新操作
// op = 1 查询操作
int op = (int) (Math.random() * 2);
// 下标从1开始，不从0开始，生成两个随机下标
int a = (int) (Math.random() * n) + 1;
int b = (int) (Math.random() * n) + 1;
// 确保jobl <= jobr
int jobl = Math.min(a, b);
int jobr = Math.max(a, b);
if (op == 0) {
// 更新操作
// 线段树、验证结构同步更新
int jobv = (int) (Math.random() * v * 2) - v;
update(jobl, jobr, jobv, 1, n, 1);
checkUpdate(check, jobl, jobr, jobv);
} else {
// 查询操作
// 线段树、验证结构同步查询
// 比对答案
long ans1 = query(jobl, jobr, 1, n, 1);
long ans2 = checkQuery(check, jobl, jobr);
if (ans1 != ans2) {
System.out.println("出错了!");
}
}
}
System.out.println("测试结束");
}
1.lazy：如果来的是add操作，那么就在之前update操作上继续加即可，update操作不需要改变；如果是update操作，那么之前的add操作就相当于没发生过，所以add数组置为0；
2.如果父节点既有add信息又有update信息，我们知道，如果有update操作，就会把add信息取消，所以update信息一定早于add信息，所以先传递update信息，在传递add信息
java
public class Code05_SegmentTreeUpdateAddQuerySum {
public static int MAXN = 1000001;
public static long[] arr = new long[MAXN];
public static long[] sum = new long[MAXN << 2];
public static long[] add = new long[MAXN << 2];
public static long[] change = new long[MAXN << 2];
public static boolean[] update = new boolean[MAXN << 2];
public static void up(int i) {
sum[i] = sum[i << 1] + sum[i << 1 | 1];
}
public static void down(int i, int ln, int rn) {
if (update[i]) {
updateLazy(i << 1, change[i], ln);
updateLazy(i << 1 | 1, change[i], rn);
update[i] = false;
}
if (add[i] != 0) {
addLazy(i << 1, add[i], ln);
addLazy(i << 1 | 1, add[i], rn);
add[i] = 0;
}
}
public static void updateLazy(int i, long v, int n) {
sum[i] = v * n;
add[i] = 0;
change[i] = v;
update[i] = true;
}
public static void addLazy(int i, long v, int n) {
sum[i] += v * n;
add[i] += v;
}
public static void build(int l, int r, int i) {
if (l == r) {
sum[i] = arr[l];
} else {
int mid = (l + r) >> 1;
build(l, mid, i << 1);
build(mid + 1, r, i << 1 | 1);
up(i);
}
add[i] = 0;
change[i] = 0;
update[i] = false;
}
public static void update(int jobl, int jobr, long jobv, int l, int r, int i) {
if (jobl <= l && r <= jobr) {
updateLazy(i, jobv, r - l + 1);
} else {
int mid = (l + r) >> 1;
down(i, mid - l + 1, r - mid);
if (jobl <= mid) {
update(jobl, jobr, jobv, l, mid, i << 1);
}
if (jobr > mid) {
update(jobl, jobr, jobv, mid + 1, r, i << 1 | 1);
}
up(i);
}
}
public static void add(int jobl, int jobr, long jobv, int l, int r, int i) {
if (jobl <= l && r <= jobr) {
addLazy(i, jobv, r - l + 1);
} else {
int mid = (l + r) >> 1;
down(i, mid - l + 1, r - mid);
if (jobl <= mid) {
add(jobl, jobr, jobv, l, mid, i << 1);
}
if (jobr > mid) {
add(jobl, jobr, jobv, mid + 1, r, i << 1 | 1);
}
up(i);
}
}
public static long query(int jobl, int jobr, int l, int r, int i) {
if (jobl <= l && r <= jobr) {
return sum[i];
}
int mid = (l + r) >> 1;
down(i, mid - l + 1, r - mid);
long ans = 0;
if (jobl <= mid) {
ans += query(jobl, jobr, l, mid, i << 1);
}
if (jobr > mid) {
ans += query(jobl, jobr, mid + 1, r, i << 1 | 1);
}
return ans;
}
public static void main(String[] args) {
System.out.println("测试开始");
int n = 1000;
int v = 2000;
int t = 5000000;
// 生成随机值填入arr数组
randomArray(n, v);
// 建立线段树
build(1, n, 1);
// 生成验证的结构
long[] check = new long[n + 1];
for (int i = 1; i <= n; i++) {
check[i] = arr[i];
}
for (int i = 1; i <= t; i++) {
// 生成操作类型
// op = 0 增加操作
// op = 1 更新操作
// op = 2 查询操作
int op = (int) (Math.random() * 3);
// 下标从1开始，不从0开始，生成两个随机下标
int a = (int) (Math.random() * n) + 1;
int b = (int) (Math.random() * n) + 1;
// 确保jobl <= jobr
int jobl = Math.min(a, b);
int jobr = Math.max(a, b);
if (op == 0) {
// 增加操作
// 线段树、验证结构同步增加
int jobv = (int) (Math.random() * v * 2) - v;
add(jobl, jobr, jobv, 1, n, 1);
checkAdd(check, jobl, jobr, jobv);
} else if (op == 1) {
// 更新操作
// 线段树、验证结构同步更新
int jobv = (int) (Math.random() * v * 2) - v;
update(jobl, jobr, jobv, 1, n, 1);
checkUpdate(check, jobl, jobr, jobv);
} else {
// 查询操作
// 线段树、验证结构同步查询
// 比对答案
long ans1 = query(jobl, jobr, 1, n, 1);
long ans2 = checkQuery(check, jobl, jobr);
if (ans1 != ans2) {
System.out.println("出错了!");
}
}
}
System.out.println("测试结束");
}P1253 扶苏的问题 - 洛谷 | 计算机科学教育新生态 (luogu.com.cn)
java
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.io.StreamTokenizer;
public class Main {
public static int MAXN = 1000001;
public static long[] arr = new long[MAXN];
public static long[] max = new long[MAXN << 2];
public static long[] add = new long[MAXN << 2];
public static long[] change = new long[MAXN << 2];
public static boolean[] update = new boolean[MAXN << 2];
public static void up(int i) {
max[i] = Math.max(max[i << 1], max[i << 1 | 1]);
}
public static void down(int i) {
if (update[i]) {
updateLazy(i << 1, change[i]);
updateLazy(i << 1 | 1, change[i]);
update[i] = false;
}
if (add[i] != 0) {
addLazy(i << 1, add[i]);
addLazy(i << 1 | 1, add[i]);
add[i] = 0;
}
}
public static void updateLazy(int i, long v) {
max[i] = v;
add[i] = 0;
change[i] = v;
update[i] = true;
}
public static void addLazy(int i, long v) {
max[i] += v;
add[i] += v;
}
public static void build(int l, int r, int i) {
if (l == r) {
max[i] = arr[l];
} else {
int mid = (l + r) >> 1;
build(l, mid, i << 1);
build(mid + 1, r, i << 1 | 1);
up(i);
}
add[i] = 0;
change[i] = 0;
update[i] = false;
}
public static void update(int jobl, int jobr, long jobv, int l, int r, int i) {
if (jobl <= l && r <= jobr) {
updateLazy(i, jobv);
} else {
int mid = (l + r) >> 1;
down(i);
if (jobl <= mid) {
update(jobl, jobr, jobv, l, mid, i << 1);
}
if (jobr > mid) {
update(jobl, jobr, jobv, mid + 1, r, i << 1 | 1);
}
up(i);
}
}
public static void add(int jobl, int jobr, long jobv, int l, int r, int i) {
if (jobl <= l && r <= jobr) {
addLazy(i, jobv);
} else {
int mid = (l + r) >> 1;
down(i);
if (jobl <= mid) {
add(jobl, jobr, jobv, l, mid, i << 1);
}
if (jobr > mid) {
add(jobl, jobr, jobv, mid + 1, r, i << 1 | 1);
}
up(i);
}
}
public static long query(int jobl, int jobr, int l, int r, int i) {
if (jobl <= l && r <= jobr) {
return max[i];
}
int mid = (l + r) >> 1;
down(i);
long ans = Long.MIN_VALUE;
if (jobl <= mid) {
ans = Math.max(ans, query(jobl, jobr, l, mid, i << 1));
}
if (jobr > mid) {
ans = Math.max(ans, query(jobl, jobr, mid + 1, r, i << 1 | 1));
}
return ans;
}
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StreamTokenizer in = new StreamTokenizer(br);
PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
in.nextToken(); int n = (int) in.nval;
in.nextToken(); int m = (int) in.nval;
for (int i = 1; i <= n; i++) {
in.nextToken();
arr[i] = (long) in.nval;
}
build(1, n, 1);
long jobv;
for (int i = 1, op, jobl, jobr; i <= m; i++) {
in.nextToken();
op = (int) in.nval;
if (op == 1) {
in.nextToken(); jobl = (int) in.nval;
in.nextToken(); jobr = (int) in.nval;
in.nextToken(); jobv = (long) in.nval;
update(jobl, jobr, jobv, 1, n, 1);
} else if (op == 2) {
in.nextToken(); jobl = (int) in.nval;
in.nextToken(); jobr = (int) in.nval;
in.nextToken(); jobv = (long) in.nval;
add(jobl, jobr, jobv, 1, n, 1);
} else {
in.nextToken(); jobl = (int) in.nval;
in.nextToken(); jobr = (int) in.nval;
out.println(query(jobl, jobr, 1, n, 1));
}
}
out.flush();
out.close();
br.close();
}
}