【cogs 775】山海经 ——Segment Tree

题目链接：

　     TP

题解：

　　  我数据结构真心是弱啊= =。

　　线段树好厉害啊，一直不会区间最大连续和，今天刚学习了一下233。

　　维护前缀最大和后缀最大，越界最大（？），再维护一个区间最大，瞎搞搞就好了，RE了一遍233。

代码： 

　　

 #define Troy 

 #include <bits/stdc++.h>

 using namespace std;

 inline int read(){
     int s=,k=;char ch=getchar();
     while(ch<''|ch>'')    ch=='-'?k=-:,ch=getchar();
     while(ch>&ch<='')    s=s*+(ch^),ch=getchar();
     return s*k;
 }

 const int N=;

 int n,m,a[N];

 struct node {
     int l,r,val;
     friend bool operator <(node x,node y){
         return x.val!=y.val?x.val<y.val:(x.l!=y.l?x.l>y.l:x.r>y.r);
     }
     friend node operator +(node x,node y){
         node z;
         z.l=min(x.l,y.l);z.r=max(x.r,y.r);
         z.val=x.val+y.val;return z;
     }
     inline void out(){printf("%d %d %d\n",l,r,val);}
 };

 struct Tree{
     node prefix,suffix,middle,section;
     Tree *lc,*rc;
 }*root,tree[N*],*ans;int cnt;

 inline void update( Tree *u){
     u->middle=u->lc->suffix+u->rc->prefix;
     u->prefix=max(u->lc->prefix,u->lc->section+u->rc->prefix);
     u->suffix=max(u->rc->suffix,u->rc->section+u->lc->suffix);
     u->section=u->lc->section+u->rc->section;
     u->middle=max(u->middle,max(u->lc->middle,max(u->rc->middle,max(u->prefix,u->suffix))));
 }

 inline void build(Tree *&u,int l,int r){
     u=tree+cnt;++cnt;
     if(l==r){
         u->prefix=u->suffix=u->middle=u->section=(node){l,r,a[l]};
         return ;
     }
     int mid=l+r>>;
     build(u->lc,l,mid);
     build(u->rc,mid+,r);
     update(u);
 }

 inline void query(Tree *u,int l,int r,int x,int y){
     if(x<=l&&r<=y){
         if(ans==NULL)   ans=u;
         else{
             Tree *now=ans;ans=tree+cnt,++cnt;
             ans->lc=now,ans->rc=u;
             update(ans);
         }
         return ;
     }
     int mid=l+r>>;
     if(x<=mid)  query(u->lc,l,mid,x,y);
     if(y>mid)   query(u->rc,mid+,r,x,y);
 }

 int main(){
     freopen("hill.in","r",stdin);
     freopen("hill.out","w",stdout);
     n=read(),m=read();
     for(int i=;i<=n;++i)   a[i]=read();
     build(root,,n);
     for(int i=;i<=m;++i){
         int l=read(),r=read();
         ans=NULL;query(root,,n,l,r);
         ans->middle.out();
     }
 }