线段树
蒟蒻年轻的时候照着hzwer神犇的写的，勿喷= =
#include<cstdio>
#include<bits/stdc++.h>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#define R0(i,n) for(int i=0;i<n;++i)
#define R1(i,n) for(int i=1;i<=n;++i)
#define CLR(x,c) memset(x,c,sizeof x)
using namespace std;
typedef long long ll;
int read(){
int x=0,f=1;char ch=getchar();
while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
return x*f;
}
struct data{
int l,r;
ll sum;
int tag;
}tr[800001];
int aa[200001],n,q;
void build(int k,int a,int b){
tr[k].l=a,tr[k].r=b;
if(a==b){
tr[k].sum=aa[a];
return ;
}
int mid=(a+b)>>1;
build(k<<1,a,mid);
build(k<<1|1,mid+1,b);
tr[k].sum=tr[k<<1].sum+tr[k<<1|1].sum;
}
void pd(int k){
int x=tr[k].r-tr[k].l+1;
tr[k<<1].tag+=tr[k].tag,tr[k<<1|1].tag+=tr[k].tag;
tr[k<<1].sum+=(x- (x>>1) )*tr[k].tag;
tr[k<<1|1].sum+=(x>>1)*tr[k].tag;
tr[k].tag=0;
}
void update(int k,int a,int b,int x){
int l=tr[k].l,r=tr[k].r;
if(l==a&&r==b){
tr[k].tag+=x;
tr[k].sum+=(b-a+1)*x;
return;
}
if(tr[k].tag)pd(k);
int mid=(l+r)>>1;
if(b<=mid)update(k<<1,a,b,x);
else if(a>mid)update(k<<1|1,a,b,x);
else update(k<<1,a,mid,x),update(k<<1|1,mid+1,b,x);
tr[k].sum=tr[k<<1].sum+tr[k<<1|1].sum;
}
ll ask(int k,int a,int b){
int l=tr[k].l,r=tr[k].r;
if(a==l&&b==r)return tr[k].sum;
if(tr[k].tag)pd(k);
int mid=(l+r)>>1;
if(b<=mid)return ask(k<<1,a,b);
else if(a>mid)return ask(k<<1|1,a,b);
else return (ask(k<<1,a,mid)+ask(k<<1|1,mid+1,b));
}
int main(){
n=read();
R1(i,n)aa[i]=read();
build(1,1,n);
q=read();
R1(i,q){
int t,a,b,x;
t=read();
if(t==1){
a=read(),b=read(),x=read();
update(1,a,b,x);
}
else{
a=read(),b=read();
printf("%lld\n",ask(1,a,b));
}
}
return 0;
}zkw线段树
代码来自在被窝里写zkw线段树的神wxjlzbcd,太神了！！！
/*
Name:wxjlzbcd
Memory:8KB
Time:651ms
*/
#include<cstdio>
#include<cstdlib>
#include<cctype>
#include<cstring>
#define LL long long
LL x,a,b,c,n,m,pre=1;
LL num[524288],numm[1000000];
inline int getc()
{
static const int L = 1 << 15;
static char buf[L] , *S = buf , *T = buf;
if (S == T) {
T = (S = buf) + fread(buf , 1 , L , stdin);
if (S == T)
return EOF;
}
return *S++;
}
inline int getint() {
static char c;
while(!isdigit(c = getc()) && c != ‘-‘);
bool sign = (c == ‘-‘);
int tmp = sign ? 0 : c - ‘0‘;
while(isdigit(c = getc()))
tmp = (tmp << 1) + (tmp << 3) + c - ‘0‘;
return sign ? -tmp : tmp;
}
inline char getch() {
char c;
while((c = getc()) != ‘Q‘ && c != ‘C‘);
return c;
}
inline void add(int a,int b)
{
num[a+=pre]+=b;
numm[a]=(a-pre)*num[a];
a>>=1;
while(a)
{
num[a]=num[a<<1]+num[(a<<1)+1];
numm[a]=numm[a<<1]+numm[(a<<1)+1];
a>>=1;
}
}
inline LL query(int a,int b)
{
LL ans=0;
a+=pre-1;
b+=pre+1;
while(a^b^1)
{
if(~a&1)
ans+=num[a^1];
if(b&1)
ans+=num[b^1];
a>>=1;
b>>=1;
}
return ans;
}
inline LL nquery(int a,int b)
{
LL ans=0;
a+=pre-1;
b+=pre+1;
while(a^b^1)
{
if(~a&1)
ans+=numm[a^1];
if(b&1)
ans+=numm[b^1];
a>>=1;
b>>=1;
}
return ans;
}
int main()
{
int i;
n=getint();
while(pre<=n)
pre<<=1;
for(i=pre+1;i<=pre+n;++i)
num[i]=getint();
for(i=pre+n;i>pre;--i)
{
num[i]-=num[i-1];
numm[i]=(i-pre)*num[i];
}
for(i=pre-1;i>=1;--i)
{
num[i]=num[i<<1]+num[(i<<1)+1];
numm[i]=numm[i<<1]+numm[(i<<1)+1];
}
/* for(i=1;i<=pre+n;++i)
printf("%d ",num[i]);*/
m=getint();
for(i=1;i<=m;++i)
{
x=getint();
if(x==1)
{
a=getint(),b=getint(),c=getint();
add(a,c),add(b+1,-c);
}
else
{
a=getint(),b=getint();
printf("%lld\n",(query(1,b)*(b+1)-nquery(1,b))-query(1,a-1)*a+nquery(1,a-1));
}
}
return 0;
} 树状数组
参考了某神的花神游历各国的写法，然后就会bit的区间修改了
= =等等这个和zkw好像
由于本题更新的时候是区间更新
所以不能直接去一个个更新区间内的点，肯定会超时
对于每次更新C(a,b,d)表示区间[a,b]内的值增加d
用ans[a]表示a~n区间元素增加的值,所以对于C(a,b,d)有:ans[a]+=d,ans[b+1]-=d;
则每次询问的时候Q(a,b),求a~b的和SUM=sum(a,b)+ans[a](b-a+1)+ans[a+1](b-a)…+ans[b]//sum(a,b)表示a,b的和
Sum=sum(a,b)+sum(ans[a+t](b-a-t+1))=sum(a,b)+sum(ans[i](b-i+1));a<=i<=b;
Sum=sum(a,b)+(b+1)*sum(ans[i])-sum(ans[i]*i);//1~b所以(b+1)*sum(ans[i]),1~a-1则a*sum(ans[i])
所以可以用两个树状数组分别表示ans[i]的前缀和 和 ans[i]*i的前缀和
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<bits/stdc++.h>
#define R0(i,n) for(int i=0;i<n;++i)
#define R1(i,n) for(int i=1;i<=n;++i)
#define CLR(x,c) memset(x,c,sizeof x)
using namespace std;
typedef long long LL;
inline int getc() {
static const int L = 1 << 15;
static char buf[L] , *S = buf , *T = buf;
if (S == T) {
T = (S = buf) + fread(buf , 1 , L , stdin);
if (S == T)
return EOF;
}
return *S++;
}
inline int read() {
static char c;
while(!isdigit(c = getc()) && c != ‘-‘);
bool sign = (c == ‘-‘);
int tmp = sign ? 0 : c - ‘0‘;
while(isdigit(c = getc()))
tmp = (tmp << 1) + (tmp << 3) + c - ‘0‘;
return sign ? -tmp : tmp;
}
LL n,q;
LL sum[200005],c1[200005],c2[200005];
inline LL lowbit(LL x){
return x&(-x);
}
void update(LL x,LL d,LL *c){
while(x<=n){
c[x]+=d;
x+=lowbit(x);
}
}
LL query(LL x,LL *c){
LL sum=0;
while(x>0){
sum+=c[x];
x-=lowbit(x);
}
return sum;
}
int main(){
int op;
LL x,y,d;
n=read();
R1(i,n)sum[i]=read(),sum[i]+=sum[i-1];
q=read();
R0(i,q){
op=read();
if(op==1){//ans[x]+=d,ans[y+1]-=d
x=read(),y=read(),d=read();
update(x,d,c1),update(y+1,-d,c1);
update(x,x*d,c2),update(y+1,-(y+1)*d,c2);
}
else{
x=read(),y=read();
printf("%lld\n",sum[y]-sum[x-1]+query(y,c1)*(y+1)-query(x-1,c1)*x-query(y,c2)+query(x-1,c2));
}
}
return 0;
}时间对比
从上到下分别是zkw，bit，普通线段树
别问我素质为何这么低= =
时间： 2024-08-17 12:28:53