D 巅峰对决
线段树:单点更新+区间查询。
既然任何时候这个数字均互不相同,那么就维护这个区间的最大值或者最小值即可。
#include <bits/stdc++.h>
#define rep(i,x,y) for (int i=(x);i<=(y);i++)
using namespace std;
const int maxn=1e6+10;
const int inf=0x3f3f3f3f;
typedef long long ll;
ll a[maxn<<2];
ll n,m,q,op1,x,y,l,r;
struct sa{
ll mn;
ll mx;
}tr[maxn<<2];
inline void pushup(ll i)
{
tr[i].mn = min(tr[i<<1].mn,tr[i<<1|1].mn);
tr[i].mx = max(tr[i<<1].mx,tr[i<<1|1].mx);
}
void bulid(ll i,ll l,ll r)
{
if(l==r){
tr[i].mn=a[l];
tr[i].mx=a[l];
return;
}
ll mid=(l+r)>>1;
bulid(i<<1,l,mid);
bulid(i<<1|1,mid+1,r);
pushup(i);
}
inline void update (ll i,ll l,ll r,ll x,ll y)
{
if(l>x||r<x) return;
if(l==x&&l==r) {
tr[i].mn=y;
tr[i].mx=y;
return;
}
ll mid=(l+r)>>1;
update(i<<1,l,mid,x,y);
update(i<<1|1,mid+1,r,x,y);
pushup(i);
}
ll query_min(ll i,ll l,ll r,ll x,ll y)
{
ll res=inf;
if(l>y||r<x) return 0;
if(l>=x&&r<=y) return tr[i].mn;
ll mid=(l+r)>>1;
if(x<=mid) res=min(res,query_min(i<<1,l,mid,x,y));
if(y>mid) res=min(res,query_min(i<<1|1,mid+1,r,x,y));
return res;
}
ll query_max(ll i,ll l,ll r,ll x,ll y)
{
ll res=-inf;
if(l>y||r<x) return 0;
if(l>=x&&r<=y) return tr[i].mx;
ll mid=(l+r)>>1;
if(x<=mid) res=max(res,query_max(i<<1,l,mid,x,y));
if(y>mid) res=max(res,query_max(i<<1|1,mid+1,r,x,y));
return res;
}
namespace IO{
char ibuf[1<<21],*ip=ibuf,*ip_=ibuf;
char obuf[1<<21],*op=obuf,*op_=obuf+(1<<21);
inline char gc(){
if(ip!=ip_)return *ip++;
ip=ibuf;ip_=ip+fread(ibuf,1,1<<21,stdin);
return ip==ip_?EOF:*ip++;
}
inline void pc(char c){
if(op==op_)fwrite(obuf,1,1<<21,stdout),op=obuf;
*op++=c;
}
inline ll read(){
register ll x=0,ch=gc(),w=1;
for(;ch<'0'||ch>'9';ch=gc())if(ch=='-')w=-1;
for(;ch>='0'&&ch<='9';ch=gc())x=x*10+ch-48;
return w*x;
}
template<class I>
inline void write(I x){
if(x<0)pc('-'),x=-x;
if(x>9)write(x/10);pc(x%10+'0');
}
class flusher_{
public:
~flusher_(){if(op!=obuf)fwrite(obuf,1,op-obuf,stdout);}
}IO_flusher;
}
using namespace IO;
int main()
{
n=read();q=read();
rep(i,1,n) a[i]=read();
bulid(1,1,n);
rep(i,1,q){
op1=read();x=read();y=read();
if(op1==1){
update(1,1,n,x,y);
}
if(op1==2){
l=query_min(1, 1, n, x, y);
r=query_max(1, 1, n, x, y);
if(y-x==r-l) puts("YES");
else puts("NO");
}
}
return 0;
} 
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