UVa 3938 "Ray, Pass me the dishes!"

        题意：给一个长度为n的序列D，对m个询问做出回答。对于询问(a,b)，需要找到两个下标x，y，使得a<=x<=y<=b，并且Dx~Dy的和尽量大。如果有多解，x应该尽量小，如果还有多解，y应该尽量小。

        思路：线段树。每个节点维护这些内容：区间和，区间的最大子段和，最大前后缀和，最大子段和以及它们的下标。查询直接返回一个区间节点，建树时的节点合并其实和查询时的节点合并是一样的。

        照着大白书写了这题，感觉思路大为开阔~

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<string.h>
#include<math.h>
using namespace std;

#define maxn 500010
#define ll long long

struct node{
	ll sum;
	ll max_sub;
	ll max_suffix;
	ll max_prefix;
	int subl,subr;
	int sl,pr;
	int l;
	int r;
};

node tree[maxn*4];

int build_tree(int n,int l,int r){
	tree[n].l=l; tree[n].r=r;
	if(l==r){
		int t;
		scanf("%d",&t);
		tree[n].sum=tree[n].max_sub=tree[n].max_prefix=tree[n].max_suffix=t;
		tree[n].subl=tree[n].subr=tree[n].sl=tree[n].pr=l;
	}else{
		int mid=(l+r)/2;
		build_tree(n*2,l,mid);
		build_tree(n*2+1,mid+1,r);
		tree[n].sum=tree[n*2].sum+tree[n*2+1].sum;
		if(tree[n*2].sum+tree[n*2+1].max_prefix > tree[n*2].max_prefix){
			tree[n].max_prefix=tree[n*2].sum+tree[n*2+1].max_prefix;
			tree[n].pr=tree[n*2+1].pr;
		}else{
			tree[n].max_prefix=tree[n*2].max_prefix;
			tree[n].pr=tree[n*2].pr;
		}
		if(tree[n*2+1].max_suffix > tree[n*2+1].sum+tree[n*2].max_suffix ){
			tree[n].max_suffix = tree[n*2+1].max_suffix;
			tree[n].sl=tree[n*2+1].sl;
		}else{
			tree[n].max_suffix = tree[n*2+1].sum+tree[n*2].max_suffix;
			tree[n].sl=tree[n*2].sl;
		}
		ll tmp[3];
		tmp[0]=tree[n*2].max_sub; 
		tmp[1]=tree[n*2].max_suffix+tree[n*2+1].max_prefix;
		tmp[2]=tree[n*2+1].max_sub;
		sort(tmp,tmp+3);
		tree[n].max_sub=tmp[2];
		if(tmp[2]==tree[n*2].max_sub){
			tree[n].subl=tree[n*2].subl;
			tree[n].subr=tree[n*2].subr;
		}else if(tmp[2]==tree[n*2].max_suffix+tree[n*2+1].max_prefix){
			tree[n].subl=tree[n*2].sl;
			tree[n].subr=tree[n*2+1].pr;
		}else{
			tree[n].subl=tree[n*2+1].subl;
			tree[n].subr=tree[n*2+1].subr;
		}
	}
}

node query(int n,int l,int r){
	node re;
	if(tree[n].l==l&&tree[n].r==r){
		re=tree[n];
		return re;
	}
	int mid=(tree[n].l+tree[n].r)/2;
	if(r<=mid){
		return query(n*2,l,r);
	}else{
		if(l>mid){
			return query(n*2+1,l,r);
		}else{
			node nl=query(n*2,l,mid);
			node nr=query(n*2+1,mid+1,r);
			if(nl.sum+nr.max_prefix > nl.max_prefix){
				re.max_prefix=nl.sum+nr.max_prefix;
				re.pr=nr.pr;
			}else{
				re.max_prefix=nl.max_prefix;
				re.pr=nl.pr;
			}
			if(nr.max_suffix > nr.sum+nl.max_suffix ){
				re.max_suffix = nr.max_suffix;
				re.sl=nr.sl;
			}else{
				re.max_suffix = nr.sum+nl.max_suffix;
				re.sl=nl.sl;
			}
			ll tmp[3];
			tmp[0]=nl.max_sub; 
			tmp[1]=nl.max_suffix+nr.max_prefix;
			tmp[2]=nr.max_sub;
			sort(tmp,tmp+3);
			re.max_sub=tmp[2];
			if(tmp[2]==nl.max_sub){
				re.subl=nl.subl;
				re.subr=nl.subr;
			}else if(tmp[2]==nl.max_suffix+nr.max_prefix){
				re.subl=nl.sl;
				re.subr=nr.pr;
			}else{
				re.subl=nr.subl;
				re.subr=nr.subr;
			}
			return re;
		}
	}
}

int main(){
	int n,m;
	int cas=0;
	while(cin>>n>>m){
		cas++;
		build_tree(1,1,n);
		printf("Case %d:\n",cas);
		
		for(int i=1;i<=m;i++){
			int l,r;
			scanf("%d%d",&l,&r);
			node ans=query(1,l,r);
			printf("%d %d\n",ans.subl,ans.subr);
		}
	}
	return 0;
}