BZOJ5312 冒险（线段树）

　　记录区间and/or，修改时如果对整个区间影响都相同就打标记，否则递归。复杂度不太会证。

#include<iostream> 
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 200010
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
    int x=0,f=1;char c=getchar();
    while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
    while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
    return x*f;
}
int n,m,a[N];
struct data{int l,r,max,AND,OR,lazy_and,lazy_or;
}tree[N<<2];
void up(int k)
{
    tree[k].max=max(tree[k<<1].max,tree[k<<1|1].max),
    tree[k].AND=tree[k<<1].AND&tree[k<<1|1].AND,
    tree[k].OR=tree[k<<1].OR|tree[k<<1|1].OR;
}
void update(int k,int x,int op)
{
    if (op==0)
    {
        tree[k].AND&=x,tree[k].OR&=x;
        if (tree[k].lazy_and==-1) tree[k].lazy_and=x;
        else tree[k].lazy_and&=x;
        if (tree[k].lazy_or!=-1) tree[k].lazy_or&=x;
        tree[k].max&=x;
    }
    else
    {
        tree[k].AND|=x,tree[k].OR|=x;
        if (tree[k].lazy_and!=-1) tree[k].lazy_and|=x;
        if (tree[k].lazy_or==-1) tree[k].lazy_or=x;
        else tree[k].lazy_or|=x;
        tree[k].max|=x;
    }
}
void down(int k)
{
    if (tree[k].lazy_and!=-1)
    {
        update(k<<1,tree[k].lazy_and,0),
        update(k<<1|1,tree[k].lazy_and,0),
        tree[k].lazy_and=-1;
    }
    if (tree[k].lazy_or!=-1)
    {
        update(k<<1,tree[k].lazy_or,1),
        update(k<<1|1,tree[k].lazy_or,1),
        tree[k].lazy_or=-1;
    }
}
void build(int k,int l,int r)
{
    tree[k].l=l,tree[k].r=r,tree[k].lazy_and=tree[k].lazy_or=-1;
    if (l==r) {update(k,a[l],1);return;}
    int mid=l+r>>1;
    build(k<<1,l,mid);
    build(k<<1|1,mid+1,r);
    up(k);
}
void modify(int k,int l,int r,int x,int op)
{
    if (tree[k].l==l&&tree[k].r==r&&(tree[k].OR&(~tree[k].AND))==0) {update(k,x,op);return;}
    down(k);
    int mid=tree[k].l+tree[k].r>>1;
    if (r<=mid) modify(k<<1,l,r,x,op);
    else if (l>mid) modify(k<<1|1,l,r,x,op);
    else modify(k<<1,l,mid,x,op),modify(k<<1|1,mid+1,r,x,op);
    up(k);
}
int query(int k,int l,int r)
{
    if (tree[k].l==l&&tree[k].r==r) return tree[k].max;
    down(k);
    int mid=tree[k].l+tree[k].r>>1;
    if (r<=mid) return query(k<<1,l,r);
    else if (l>mid) return query(k<<1|1,l,r);
    else return max(query(k<<1,l,mid),query(k<<1|1,mid+1,r));
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("bzoj5312.in","r",stdin);
    freopen("bzoj5312.out","w",stdout);
    const char LL[]="%I64d
";
#else
    const char LL[]="%lld
";
#endif
    n=read(),m=read();
    for (int i=1;i<=n;i++) a[i]=read();
    build(1,1,n);
    while (m--)
    {
        int op=read();
        if (op==1)
        {
            int l=read(),r=read(),x=read();
            modify(1,l,r,x,0);
        }
        else if (op==2)
        {
            int l=read(),r=read(),x=read();
            modify(1,l,r,x,1);
        }
        else
        {
            int l=read(),r=read();
            printf("%d
",query(1,l,r));
        }
    }
    return 0;
}