线段树整理
区间最大值、单点修改
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#define re register
#define lson o<<1
#define rson o<<1|1
using namespace std ;
const int maxn = 200005 ;
inline int read(){
int f = 1 , x = 0 ;
char ch = getchar() ;
while(ch > '9' || ch < '0') {if(ch == '-') f = -1 ; ch = getchar() ;}
while(ch >= '0' && ch <= '9') {x = (x << 1) + (x << 3) + ch - '0' ; ch = getchar () ;}
return x * f ;
}
int n , m , a , b ;
int tree[maxn * 4] ;
inline void pushup(int o) {
tree[o] = max(tree[lson] , tree[rson]) ;
}
inline void build(int o , int l , int r) {
if(l == r) {
tree[o] = read() ;
return ;
}
int mid = (l + r) / 2 ;
build(lson , l , mid) ;
build(rson , mid + 1 , r) ;
pushup(o) ;
}
inline int query(int o , int l , int r , int x , int y) {
if(x <= l && r <= y)
return tree[o] ;
int maxx = 0 ;
int mid = (l + r) / 2 ;
if(x <= mid) maxx = max(maxx , query(lson , l , mid , x , y)) ;
if(y > mid) maxx = max(maxx , query(rson , mid + 1 , r , x , y)) ;
return maxx ;
}
inline void update(int o , int l , int r , int x , int y) {
if(l == r) {
tree[o] = y ;
return ;
}
int mid = (l + r) / 2 ;
if(x <= mid) update(lson , l , mid , x , y) ;
else update(rson , mid + 1 , r , x , y) ;
pushup(o) ;
}
int main () {
freopen("hate it.in" , "r" , stdin) ;
n = read () ; m = read () ;
build(1 , 1 , n) ;
char flag ;
while(m--) {
cin >> flag ;
if(flag == 'Q') {
a = read (); b = read () ;
printf("%d\n" , query(1 , 1 , n , a , b)) ;
}
else {
a = read () ; b = read () ;
update(1 , 1 , n , a , b) ;
}
}
return 0 ;
}
区间最小值、单点修改
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <cmath>
#define re register
using namespace std ;
const int maxm = 300005 ;
#define lson o<<1
#define rson o<<1|1
inline int read() {
int f = 1 , x = 0 ;
char ch = getchar () ;
while(ch > '9' || ch < '0') {if(ch == '-') f = -1 ; ch = getchar () ;}
while(ch >= '0' && ch <= '9') {x = (x << 1) + (x << 3) + ch - '0' ; ch = getchar () ;}
return x * f;
}
int m , n , a , b ;
int tree[maxm] ;
inline void pushup(int o) {
tree[o] = min(tree[lson] , tree[rson]) ;
}
inline void build(int o , int l , int r) {
if(l == r) {
tree[o] = read() ;
return ;
}
int mid = (l + r) / 2 ;
build(lson , l , mid) ;
build(rson , mid + 1 , r) ;
pushup(o) ;
}
inline int query(int o , int l , int r , int x , int y) {
if(x <= l && r <= y)
return tree[o] ;
int minn = 1e9 ;
int mid = (l + r) / 2;
if(x <= mid) minn = min(minn, query(lson , l , mid , x , y)) ;
if(y > mid) minn = min(minn, query(rson , mid + 1 , r , x , y)) ;
return minn;
}
inline void update(int o , int l , int r , int x , int y) {
if(l == r) {
tree[o] = max(tree[o] , y) ;
return ;
}
int mid = (l + r) / 2 ;
if(x <= mid) update(lson , l , mid , x , y) ;
else update(rson , mid + 1 , r , x , y) ;
pushup(o) ;
}
int main () {
m = read () ; n = read () ;
build(1 , 1 , m) ;
char flag ;
while(n --) {
cin >> flag ;
if(flag == 'Q') {
a = read (); b = read () ;
printf("%d\n" , query(1 , 1 , m , a , b)) ;
}
else {
a = read () ; b = read () ;
update(1 , 1 , m , a , b) ;
}
}
return 0 ;
}
例题：洛古P1816 忠诚
区间求和、区间修改
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#define re register
#define lson o<<1
#define rson o<<1|1
#define ll long long
using namespace std ;
const int maxn = 100005 ;
inline long long read(){
long long f = 1 , x = 0 ;
char ch = getchar() ;
while(ch > '9' || ch < '0') {if(ch == '-') f = -1 ; ch = getchar() ;}
while(ch >= '0' && ch <= '9') {x = (x << 1) + (x << 3) + ch - '0' ; ch = getchar () ;}
return x * f ;
}
long long n , m , x , y , k ;
long long tree[maxn * 4] , add[maxn * 4] ;
long long flag ;
inline void pushup(ll o) {
tree[o] = tree[lson] + tree[rson] ;
}
inline void pushdown(ll o , ll l , ll r) {
if(add[o]) {
ll mid = (l + r) / 2 ;
add[lson] += add[o] ;
add[rson] += add[o] ;
tree[lson] += (mid - l + 1) * add[o] ;
tree[rson] += (r - mid) * add[o] ;
add[o] = 0 ;
}
}
inline void build(ll o , ll l , ll r) {
if(l == r) {
tree[o] = read() ;
return ;
}
int mid = (l + r) >> 1 ;
build(lson , l , mid) ;
build(rson , mid + 1 , r) ;
pushup(o) ;
}
inline void update(ll o , ll l , ll r , ll x , ll y , ll val) {
if(x <= l && r <= y) {
tree[o] +=(r - l + 1) * val ;
add[o] += val ;
return ;
}
pushdown(o , l , r) ;
ll mid = (l + r) / 2 ;
if(x <= mid) update(lson , l , mid , x , y , val) ;
if(y > mid) update(rson , mid + 1 , r , x , y , val) ;
pushup(o) ;
}
inline ll query(ll o , ll l , ll r , ll x , ll y) {
if(x <= l && r <= y) {
return tree[o] ;
}
pushdown(o , l , r) ;
int mid = (l + r) / 2 ;
ll ans = 0 ;
if(x <= mid) ans += query(lson , l , mid , x , y) ;
if(y > mid) ans += query(rson , mid + 1 , r , x , y) ;
return ans ;
}
int main() {
n = read() ; m = read() ;
build(1 , 1 , n) ;
while(m--) {
flag = read() ;
if(flag == 1) {
x = read() ; y = read() ; k = read () ;
update(1 , 1 , n , x , y , k) ;
}
else {
x = read () ; y = read () ;
printf("%lld\n" , query(1 , 1 , n , x , y)) ;
}
}
return 0 ;
}
例题：洛古P3372 线段树1
单点乘、单点加、区间求和
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#define re register
#define lson o<<1
#define rson o<<1|1
#define ll long long
using namespace std ;
const int maxn = 100005 ;
inline long long read(){
long long f = 1 , x = 0 ;
char ch = getchar() ;
while(ch > '9' || ch < '0') {if(ch == '-') f = -1 ; ch = getchar() ;}
while(ch >= '0' && ch <= '9') {x = (x << 1) + (x << 3) + ch - '0' ; ch = getchar () ;}
return x * f ;
}
ll n , m , mod , flag , x , y , k ;
ll tree[maxn * 4] , add[maxn * 4] , mul[maxn * 4] ;
inline void pushup(ll o) {
tree[o] = tree[lson] + tree[rson] ;
mul[o] = 1 ;
}
inline void pushdown(ll o , ll l , ll r) {
ll mid = (l + r) / 2 ;
if(mul[o] != 1) {
mul[lson] = (mul[lson] * mul[o]) % mod ;
mul[rson] = (mul[rson] * mul[o]) % mod ;
add[lson] = (add[lson] * mul[o]) % mod ;
add[rson] = (add[rson] * mul[o]) % mod ;
tree[lson] = (tree[lson] * mul[o]) % mod ;
tree[rson] = (tree[rson] * mul[o]) % mod ;
mul[o] = 1 ;
}
if(add[o]) {
tree[lson] = (tree[lson] + (add[o] * (mid - l + 1)) % mod) % mod ;
tree[rson] = (tree[rson] + (add[o] * (r - mid)) % mod) % mod ;
add[lson] = (add[lson] + add[o]) % mod ;
add[rson] = (add[rson] + add[o]) % mod ;
add[o] = 0 ;
}
}
inline void build(ll o , ll l , ll r) {
if(l == r) {
tree[o] = read() ;
return ;
}
ll mid = (l + r) / 2 ;
build(lson , l , mid) ;
build(rson , mid + 1 , r) ;
pushup(o) ;
}
inline void Mul(ll o , ll l , ll r , ll x , ll y , ll k) {
if(x <= l && r <= y) {
tree[o] = (tree[o] * k) % mod ;
mul[o] = (mul[o] * k) % mod ;
add[o] = (add[o] * k) % mod ;
return ;
}
pushdown(o , l , r) ;
ll mid = (l + r) / 2 ;
if(x <= mid) Mul(lson , l , mid , x , y , k) ;
if(y > mid) Mul(rson , mid + 1 , r , x , y , k) ;
pushup(o) ;
}
inline void Add(ll o , ll l , ll r , ll x , ll y , ll k) {
if(x <= l && r <= y) {
tree[o] = (tree[o] + (r - l + 1) * k % mod) % mod ;
add[o] = (add[o] + k) % mod ;
return ;
}
pushdown(o , l , r) ;
ll mid = (l + r) / 2 ;
if(x <= mid) Add(lson , l , mid , x , y , k) ;
if(y > mid) Add(rson , mid + 1 , r , x , y , k) ;
pushup(o) ;
}
inline ll query(ll o , ll l , ll r , ll x , ll y) {
if(x <= l && r <= y) {
return tree[o] ;
}
ll mid = (l + r) / 2 ;
ll ans = 0 ;
pushdown(o , l , r) ;
if(x <= mid) ans = (ans + query(lson , l , mid , x , y)) % mod ;
if(y > mid) ans = (ans + query(rson , mid + 1 , r , x , y)) % mod ;
return ans % mod ;
}
int main () {
n = read () ; m = read () ; mod = read() ;
build(1 , 1 , n) ;
while(m --) {
flag = read ();
if(flag == 1) {
x = read () ; y = read () ; k = read() ;
Mul(1 , 1 , n , x , y , k) ;
}
else if(flag == 2) {
x = read() ; y = read() ; k = read() ;
Add(1 , 1 , n , x , y , k) ;
}
else {
x = read () ; y = read () ;
printf("%lld\n" , query(1 , 1 , n , x , y)) ;
}
}
return 0 ;
}
例题：洛古P3373 线段树2
顺风不浪，逆风不怂。
