SPOJ GSS2 Can you answer these queries II (线段树&离线) - xgtao -

Can you answer these queries II

 

这是一道线段树的题目，维护历史版本，给出N(<=100000)个数字(-100000<=x<=100000),要求求出在[l,r]区间里面的连续序列的最大值，并且重复的数字可以加入序列但是值不能再计算。

 

本题明显使用线段树,它只存在询问而没有修改操作,离线相对于在线更好维护。

定义s[i] = ai + ai+1 + ai+2 + ... an，以ai开头的数列的和,那么每次加入更新ai 那么s1,s2,...si都会相应的加一个ai,s[1~i]中出现过a[i]是不能重复加值的,那么为了避免重复加值,用pre[a[i]]表示a[i]上一次出现的位置,那么也就是s[pre[ai]+1]~s[i]这一个区间加上a[i]，每一次更新a[i]都要记录历史版本的最大值和懒惰标记的最大值。

 

#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;

const int N = 100010;
const int M = 100010;
const int C = 100001;
long long res[N];
int a[N],pre[N<<1],n,m;

struct Que{
	int l,r,ID;
	bool operator < (const Que &rhs)const{return r < rhs.r;}
}question[M];

struct Tree{long long s,ms,d,md;}tree[N<<2];

#define lson k<<1,l,mid
#define rson k<<1|1,mid+1,r

void processup(int k){
	tree[k].s = max(tree[k<<1].s,tree[k<<1|1].s);
	tree[k].ms = max(tree[k<<1].ms,tree[k<<1|1].ms);
}

void processdown(int k){
	if(!tree[k].d && !tree[k].md)return;
	tree[k<<1].ms = max(tree[k<<1].ms,tree[k<<1].s+tree[k].md);
	tree[k<<1].md = max(tree[k<<1].md,tree[k<<1].d+tree[k].md);
	tree[k<<1].s += tree[k].d,tree[k<<1].d += tree[k].d;
	
	tree[k<<1|1].ms = max(tree[k<<1|1].ms,tree[k<<1|1].s+tree[k].md);
	tree[k<<1|1].md = max(tree[k<<1|1].md,tree[k<<1|1].d+tree[k].md);
	tree[k<<1|1].s += tree[k].d,tree[k<<1|1].d += tree[k].d;
	
	tree[k].d = tree[k].md = 0;
}

long long query(int k,int l,int r,int xl,int xr){
	if(l == xl && r == xr)return tree[k].ms;
	processdown(k);
	int mid = (l+r)>>1;
	if(xr <= mid)return query(lson,xl,xr);
	else if(xl > mid)return query(rson,xl,xr);
	else return max(query(lson,xl,mid),query(rson,mid+1,xr));
	processup(k);
}

void update(int k,int l,int r,int xl,int xr,long long x){
	if(l == xl && r == xr){
		tree[k].s += x;
		tree[k].d += x;
		tree[k].ms = max(tree[k].ms,tree[k].s);
		tree[k].md = max(tree[k].md,tree[k].d);
		return;
	}
	processdown(k);
	int mid = (l+r)>>1;
	if(xr <= mid)update(lson,xl,xr,x);
	else if(xl > mid)update(rson,xl,xr,x);
	else update(lson,xl,mid,x),update(rson,mid+1,xr,x);
	processup(k);
}

#define clr(a,b) memset(a,b,sizeof(a))

int main(){
	while(scanf("%d",&n) == 1){
		clr(tree,0),clr(pre,0);
		for(int i = 1;i <= n;++i)scanf("%d",&a[i]);
		scanf("%d",&m);
		for(int i = 1;i <= m;++i){
			scanf("%d%d",&question[i].l,&question[i].r);
			question[i].ID = i;
		}
		sort(question+1,question+m+1);
		int ID = 1;
		for(int i = 1;i <= n;++i){
			update(1,1,n,pre[a[i]+C]+1,i,a[i]);
			pre[a[i]+C] = i;
			while(ID <= m && question[ID].r == i){
				res[question[ID].ID] = query(1,1,n,question[ID].l,question[ID].r);
				ID++;
			}
		}
		for(int i = 1;i <= m;++i)printf("%lld\n",res[i]);
	}
	return 0;
}