【模板】线段树

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int val[MAXN];struct tree{  int maxx,sum;  int add;  tree()  {maxx=sum=add=0;}}use[MAXN*4];//[ lc(x); rc(x); ]inline int lc(int x){ return x << 1;}inline int rc(int x){ return x << 1 | 1;}//[ up(x); ]inline void up(int x){  int l=lc(x),r=rc(x);  use[x].maxx=max(use[l].maxx,use[r].maxx);  use[x].sum=use[l].sum+use[r].sum;    }//[ down(x); ]//inline max_down(int x){  if(use[x].add!=0)  {    int l=lc(x),r=rc(x);    use[l].maxx+=use[x].add;    use[r].maxx+=use[x].add;    use[l].add+=use[x].add;    use[r].add+=use[x].add;    use[x].add=0;     }       }//[ down(x,r-l+1); ]inline sum_down(int x,int len){ //len=r-l+1  if(use[x].add!=0)  {    int l=lc(x),r=rc(x);    use[l].sum+=use[x].add*(len-(len>>1));    use[r].sum+=use[x].add*(len>>1);    use[l].add+=use[x].add;    use[r].add+=use[x].add;    use[x].add=0;     }        }//[ build(1,1,n); ]</span>void build(int x,int l,int r){  if(l==r)  {    use[x].maxx=use[x].sum=val[l];    return ;  }       int mid=(l+r)>>1;  build(lc(x),l,mid);  build(rc(x),mid+1,r);  up(x);}//[ updata(1,1,n,l,r,v); ]void updata(int x,int l,int r,int op,int ed,int v){  if(op<=l&&r<=ed)  {    use[x].sum+=v*(r-l+1);    use[x].maxx+=v;    use[x].add+=v;    return ;                  }        down(x);     //根据情况选择函数和参数（max or sum)</span></span>  int mid=(l+r)>>1;  if(ed<=m)updata(lc(x),l,mid,op,ed,v);  else if(op>mid)updata(rc(x),mid+1,r,op,ed,v);  else   {    updata(lc(x),l,mid,op,ed,v);    updata(rc(x),mid+1,r,op,ed,v);  }  up(x);}<span style="font-size:18px;color:#FF0000;">[ query(1,1,n,l,r); ]<span style="color:#993399;">int query(int x,int l,int r,int op,int ed){  if(op<=l&&r<=ed)  {   //max:      return use[x].maxx;   //sum:      return use[x].sum;                  }      down(x);      //根据情况选择函数和参数（max or sum)  int mid=(l+r)>>1;  if(ed<=mid)return query(lc(x),l,mid,op,ed);  else if(op>mid)return query(rc(x),mid+1,r,op,ed);  else  {   //max:      return max( query(lc(x),l,mid,op,ed) , query(rc(x),mid+1,r,op,ed) );   //sum:      return query(lc(x),l,mid,op,ed) + query(rc(x),mid+1,r,op,ed);      } }
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