hdu 3308(线段树区间合并)

线段树的区间并，这道题是比较简单的一个，，重点是细节的处理

#include<bits/stdc++.h>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;#define sf scanf#define pf printf#define mem(a,b) memset(a,b,sizeof(a));#define rep(i,a,b) for(int i=(a);i<=(b);++i)#define MP make_pair#define ULL unsigned long long#define LL   long long#define inf 0x3f3f3f3f#define md ((ll[i]+rr[i])>>1)#define ls (i<<1)#define rs (ls|1)#define eps 1e-5#define N 200050#define ree freopen("in.txt","r",stdin);#define bug pf("----------------");//2017年09月25日21:29:24int ll[N<<2],rr[N<<2],a[N],lx[N<<2],rx[N<<2],mx[N<<2];int n,m;int len(int i){    return rr[i]-ll[i]+1;}void up(int i){    mx[i]=max(mx[ls],mx[rs]);    if(a[md]<a[md+1]){        mx[i]=max(mx[i],rx[ls]+lx[rs]);        if(lx[ls]==len(ls))            lx[i]=len(ls)+lx[rs];        else            lx[i]=lx[ls];        if(rx[rs]==len(rs))            rx[i]=len(rs)+rx[ls];        else            rx[i]=rx[rs];    }    else{        lx[i]=lx[ls],rx[i]=rx[rs];    }}void build(int l,int r,int i){    ll[i]=l,rr[i]=r;    if(l==r){        mx[i]=lx[i]=rx[i]=1;        return ;    }    build(l,md,ls);build(md+1,r,rs);    up(i);}void update(int p,int val,int i){    if(ll[i]==rr[i])return ;    if(p<=md)        update(p,val,ls);    else        update(p,val,rs);    up(i);}int query(int l,int r,int i){    if(ll[i]==l&&rr[i]==r)        return mx[i];    if(r<=md)return query(l,r,ls);    if(l>md)return query(l,r,rs);    if(a[md]>=a[md+1])        return max(query(l,md,ls),query(md+1,r,rs));    int ret=max(query(l,md,ls),query(md+1,r,rs));    int head=max(l,md-rx[ls]+1);    int tail=min(r,md+lx[rs]);    ret=max(ret,tail-head+1);    return ret;}int main(){    //ree    int cas;    sf("%d",&cas);    while(cas--){        sf("%d%d",&n,&m);        for(int i=1;i<=n;++i)sf("%d",&a[i]);        build(1,n,1);        int k,v,l,r;        while(m--){            char op[2];sf("%s",op);            if(op[0]=='Q'){                sf("%d%d",&l,&r);                l++;r++;                pf("%d\n",query(l,r,1));                continue;            }            sf("%d%d",&k,&v);            k++; a[k]=v;            update(k,v,1);        }    }}