区间最大子区间和

规定：
ls:区间靠左部分子区间最大和
rs:区间靠右部分子区间最大和
ms:区间子区间最大和
s:区间和

方程与数量关系如图所示，可以用动态规划解决

山海经题解

这题是上述方法在线段树中的应用

solution

#include<bits/stdc++.h>
using namespace std;
const int maxn=2000010;
int lid(int id){
	return id<<1;
} 
int rid(int id){
	return id<<1|1;
}
int a[maxn],n,z,b,r,t,x,y,now,q,k,f,e,ans;
struct tree{
	int lid,rid;//子区间左边界、右边界
    int ms,ls,rs,s;//同上区间最大子区间和的定义
    int l,r,ml,mr;//区间左边界，右边界。最大子区间和左边界，右边界
}m[maxn*4+1];
void push(int id){//对上述一堆东西的更新维护
	m[id].s=m[lid(id)].s+m[rid(id)].s; 
	
	if(m[lid(id)].ls>=m[rid(id)].ls+m[lid(id)].s){m[id].ls=m[lid(id)].ls;m[id].rid=m[lid(id)].rid;}
	else{m[id].ls=m[rid(id)].ls+m[lid(id)].s;m[id].rid=m[rid(id)].rid;}
	
	if(m[rid(id)].rs>=m[rid(id)].s+m[lid(id)].rs){m[id].rs=m[rid(id)].rs;m[id].lid=m[rid(id)].lid;}
	else{m[id].rs=m[rid(id)].s+m[lid(id)].rs;m[id].lid=m[lid(id)].lid;}
	
	if(m[lid(id)].ms>=m[rid(id)].ms){m[id].ms=m[lid(id)].ms;m[id].ml=m[lid(id)].ml;m[id].mr=m[lid(id)].mr;}
	else{m[id].ms=m[rid(id)].ms;m[id].ml=m[rid(id)].ml;m[id].mr=m[rid(id)].mr;}
	
	if(m[id].ms>m[lid(id)].rs+m[rid(id)].ls) return;
	if(m[id].ms<m[lid(id)].rs+m[rid(id)].ls){m[id].ms=m[lid(id)].rs+m[rid(id)].ls;m[id].ml=m[lid(id)].lid;m[id].mr=m[rid(id)].rid;return;}
	
	if(m[id].ml<m[lid(id)].lid) return;
	if(m[id].ml>m[lid(id)].lid){m[id].ml=m[lid(id)].lid;m[id].mr=m[rid(id)].rid;return;}
	
	m[id].mr=min(m[id].mr,m[rid(id)].rid);
}
void build(int id,int l,int r){//建树
	m[id].l=l;
	m[id].r=r;
	if(l==r){
		m[id].ml=m[id].mr=m[id].lid=m[id].rid=l;
		m[id].ls=m[id].rs=m[id].s=m[id].ms=a[l];
		return; 
	}
	int mid=(l+r)>>1;
	build(lid(id),l,mid);
	build(rid(id),mid+1,r);
	push(id);
} 
tree find(int id,int s,int tt){
	int l=m[id].l,r=m[id].r;
	if(s<=l&&r<=tt) return m[id];
	int mid=(l+r)>>1;
	tree x, y, w;
    if(tt<=mid) return find(lid(id),s,tt);
    if(s>mid) return find(rid(id),s,tt);
   
    x=find(lid(id),s,tt);
    y=find(rid(id),s,tt);
    if(x.ls>=x.s+y.ls){w.ls=x.ls;w.rid=x.rid;w.l=x.l;}
    else{w.ls=x.s+y.ls;w.rid=y.rid;w.l=x.l;}
        
    if(y.rs>=x.rs+y.s){w.rs=y.rs;w.lid=y.lid;w.r=y.r;}
    else{w.rs=x.rs+y.s;w.lid=x.lid;w.r=y.r;}
        
    if(x.ms>=y.ms){w.ms=x.ms;w.ml=x.ml;w.mr=x.mr;}
    else{w.ms=y.ms;w.ml=y.ml;w.mr=y.mr;}
		
	if(w.ms>x.rs+y.ls)return w;
    if(w.ms<x.rs+y.ls){w.ms=x.rs+y.ls;w.ml=x.lid;w.mr=y.rid;return w;}
        
    if(w.ml<x.lid) return w;
    if(w.ml>x.lid){w.ml=x.lid;w.mr=y.rid;return w;}
        
	w.mr=min(w.mr,y.rid);
	
    return w;
}
int main(){
	scanf("%d%d",&n,&q);
	for(int i=1;i<=n;i++){
		scanf("%d",&a[i]);
//		modify (1,i,a[i]); 
	} 
	if(n!=0) build(1,1,n); 
	for(int i=1;i<=q;i++){ 
		scanf("%d%d",&f,&e);
		tree mm=find(1,f,e);
		printf("%d %d %d\n",mm.ml,mm.mr,mm.ms);
	}
    return 0;
}