codevs1082 线段树练习3

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= =