山海经

一颗比较麻烦的线段树，维护变量比较多

sum区间和

Max区间最大子序列和

Ls区间从左端点开始最大子序列和，Lto终点位置

Rs从右端点开始最大子序列和，Rto终点位置 

pushup()的时候，比较麻烦，细节比较多

更新Max的情况

更新Ls，Rs直接更新就可以了

#include<iostream>
#include<cstdio>
#include<cstring>
#define maxn 100005
using namespace std;
int n,m;

int a[maxn];
struct tree
{
     int l,r,Ls,Lto,Rs,Rto,Max,Ml,Mr,sum;  
}b[maxn*4];
void update(int z)
{
     if(b[2*z].Max>=b[2*z+1].Max){ b[z].Ml=b[2*z].Ml; b[z].Mr=b[2*z].Mr; }
     else { b[z].Ml=b[2*z+1].Ml; b[z].Mr=b[2*z+1].Mr;}
     b[z].Max=max(b[2*z].Max,b[2*z+1].Max);
     if(b[z].Max==b[2*z].Rs+b[2*z+1].Ls&&b[2*z].Rto<b[z].Ml)
     {  b[z].Ml=b[2*z].Rto;b[z].Mr=b[2*z+1].Lto; }
     if(b[z].Max==b[2*z].Rs+b[2*z+1].Ls&&b[2*z].Rto==b[z].Ml&&b[2*z+1].Lto<b[z].Mr)
        b[z].Mr=b[2*z+1].Lto;
     if(b[z].Max<b[2*z].Rs+b[2*z+1].Ls){
        b[z].Ml=b[2*z].Rto;b[z].Mr=b[2*z+1].Lto;
        b[z].Max=b[2*z].Rs+b[2*z+1].Ls;
     }
     b[z].Ls=b[2*z].Ls; b[z].Lto=b[2*z].Lto;
     if(b[2*z].Lto==b[2*z].r&&b[2*z+1].Ls>0){ 
       b[z].Ls=b[2*z].Ls+b[2*z+1].Ls;
       b[z].Lto=b[2*z+1].Lto;
     }
     if(b[2*z].sum+b[2*z+1].Ls>b[z].Ls){
        b[z].Ls=b[2*z].sum+b[2*z+1].Ls;
        b[z].Lto=b[2*z+1].Lto;  
     } 
     b[z].Rs=b[2*z+1].Rs;b[z].Rto=b[2*z+1].Rto;
     if(b[2*z+1].Rto==b[2*z+1].l&&b[2*z].Rs>0){ 
       b[z].Rs=b[2*z+1].Rs+b[2*z].Rs;
       b[z].Rto=b[2*z].Rto;
     }
     if(b[2*z+1].sum+b[2*z].Rs>b[z].Rs){
         b[z].Rs=b[2*z+1].sum+b[2*z].Rs;
         b[z].Rto=b[2*z].Rto;
     }
     b[z].sum=b[2*z].sum+b[2*z+1].sum;
}
void pushup(tree tmp1,tree tmp2,tree &tmp3)
{
     if(tmp1.Max>=tmp2.Max){ tmp3.Ml=tmp1.Ml; tmp3.Mr=tmp1.Mr; }
     else { tmp3.Ml=tmp2.Ml; tmp3.Mr=tmp2.Mr;}
     tmp3.Max=max(tmp1.Max,tmp2.Max);
     if(tmp3.Max==tmp1.Rs+tmp2.Ls&&tmp1.Rto<tmp3.Ml)
     {  tmp3.Ml=tmp1.Rto;tmp3.Mr=tmp2.Lto; }
     if(tmp3.Max==tmp1.Rs+tmp2.Ls&&tmp1.Rto==tmp3.Ml&&tmp2.Lto<tmp3.Mr)
        tmp3.Mr=tmp2.Lto;
     if(tmp3.Max<tmp1.Rs+tmp2.Ls){
       tmp3.Ml=tmp1.Rto;tmp3.Mr=tmp2.Lto;
       tmp3.Max=tmp1.Rs+tmp2.Ls;
     } 
     tmp3.Ls=tmp1.Ls; tmp3.Lto=tmp1.Lto;
     if(tmp1.Lto==tmp2.l-1&&tmp2.Ls>0){ 
       tmp3.Ls=tmp1.Ls+tmp2.Ls;
       tmp3.Lto=tmp2.Lto;
     }
     if(tmp1.sum+tmp2.Ls>tmp3.Ls){
        tmp3.Ls=tmp1.sum+tmp2.Ls;
        tmp3.Lto=tmp2.Lto;  
     }
     tmp3.Rs=tmp2.Rs;tmp3.Rto=tmp2.Rto;
     if(tmp2.Rto==tmp1.r+1&&tmp1.Rs>0){ 
       tmp3.Rs=tmp2.Rs+tmp1.Rs;
       tmp3.Rto=tmp1.Rto;
     }
     if(tmp2.sum+tmp1.Rs>tmp3.Rs){
         tmp3.Rs=tmp2.sum+tmp1.Rs;
         tmp3.Rto=tmp1.Rto;
     }
     tmp3.sum=tmp2.sum+tmp1.sum;
     //printf("%d %d %d %d %d\n",tmp3.l,tmp3.r,tmp3.Ml,tmp3.Mr,tmp3.Max);
}
tree quiry(int l,int r,int z)
{
    //printf("%d** %d\n",l,r);
    //if(i==23)  printf("%d %d \n",b[z].l,b[z].r);
     if(l<=b[z].l&&b[z].r<=r)
      return b[z];
     int mid=(b[z].l+b[z].r)/2;
     if(r<=mid) return quiry(l,r,2*z);
     else
      if(l>mid) return quiry(l,r,2*z+1);
      else
      {
          tree tmp1=quiry(l,mid,2*z);
          tree tmp2=quiry(mid+1,r,2*z+1);
          tree tmp3;tmp3.l=l;tmp3.r=r;
          pushup(tmp1,tmp2,tmp3); //printf("%d %d %d\n",tmp3.l,tmp3.r,tmp3.Max);
          return tmp3;
      }
}
void build(int l,int r,int z)
{
     b[z].l=l;b[z].r=r;
     if(l==r){
         b[z].Max=b[z].Ls=b[z].Rs=b[z].sum=a[l];
         b[z].Lto=b[z].Rto=b[z].Ml=b[z].Mr=l;
         return ;
     }
     int mid=(l+r)/2;
     build(l,mid,2*z);
     build(mid+1,r,2*z+1);
     update(z);
    // printf("l=%d, r=%d, max= %d, %d, %d\n",b[z].l,b[z].r,b[z].Max,b[z].Ls,b[z].Rs);
}
int main()
{
    freopen("hill.in","r",stdin);
   freopen("hill.out","w",stdout);
   // freopen("in.txt","r",stdin);
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++)  scanf("%d",&a[i]);
    build(1,n,1);
    int L,R;
    for(int i=1;i<=m;i++){
       scanf("%d%d",&L,&R);
      /// cout<<i<<" ";
        tree ans=quiry(L,R,1);   
       printf("%d %d %d\n",ans.Ml,ans.Mr,ans.Max);
    }
   // while(1);
    return 0;
}