SPOJ 1043 Can you answer these queries I 求任意区间最大连续子段和 线段树

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题目链接：点击打开链接

维护区间左起连续的最大和，右起连续的和。。

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <string.h>
#include <math.h>
#include <vector>
#include <map>
using namespace std;
#define N 50050
#define Lson(x) tree[x].l
#define Rson(x) tree[x].r
#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define Sum(x) tree[x].sum
#define Max(x) tree[x].max
#define Lmax(x) tree[x].lmax
#define Rmax(x) tree[x].rmax
struct node{
	int l, r;
	int mid(){return (l+r)>>1;}
	int lmax, rmax, max, sum;
}tree[N<<4];
int n, a[N], Q;
void push_down(int id){}
void push_up(int id){
	Lmax(id) = max(Lmax(L(id)), Sum(L(id)) + Lmax(R(id)));
	Rmax(id) = max(Rmax(R(id)), Sum(R(id)) + Rmax(L(id)));
	Sum(id) = Sum(L(id)) + Sum(R(id));
	Max(id) = max(max(Max(L(id)), Max(R(id))), Rmax(L(id)) + Lmax(R(id)));
}
void updata_point(int val, int id){Lmax(id) = Rmax(id) = Max(id) = Sum(id) = val;}
void build(int l, int r, int id){
	Lson(id) = l; Rson(id) = r;
	if(l == r)
	{ 
		updata_point(a[l], id);
		return;
	}
	int mid = tree[id].mid();
	build(l, mid, L(id));
	build(mid+1, r, R(id));
	push_up(id);
}
void updata(int pos, int val, int id){
	push_down(id);
	if(Lson(id) == Rson(id))
	{
		updata_point(val, id);
		return ;
	}
	int mid = tree[id].mid();
	if(mid < pos)
		updata(pos, val, R(id));
	else 
		updata(pos, val, L(id));
}
int query_l(int l, int r, int id){
	push_down(id);
	if(l == Lson(id) && Rson(id) == r) return Lmax(id);
	int mid = tree[id].mid();
	if(mid < l)
		return query_l(l, r, R(id));
	else if(r <= mid)
		return query_l(l, r, L(id));
	int lans = query_l(l, mid, L(id)), rans = query_l(mid+1, r, R(id));
	return max(lans, Sum(L(id)) + rans);
}
int query_r(int l, int r, int id){
	push_down(id);
	if(l == Lson(id) && Rson(id) == r) return Rmax(id);
	int mid = tree[id].mid();
	if(mid < l)
		return query_r(l, r, R(id));
	else if(r <= mid)
		return query_r(l, r, L(id));
	int lans = query_r(l, mid, L(id)), rans = query_r(mid+1, r, R(id));
	return max(rans, Sum(R(id)) + lans);
}
int query(int l, int r, int id){
	push_down(id);
	if(l == Lson(id) && Rson(id) == r)return Max(id);
	int mid = tree[id].mid();
	if(mid < l)
		return query(l, r, R(id));
	else if(r<=mid)
		return query(l, r, L(id));
	int lans = query(l, mid, L(id)), rans = query(mid+1, r, R(id));
	int ans = max(lans, rans);
	return max(ans, query_r(l, mid, L(id)) + query_l(mid+1, r, R(id)));
}
int main(){
	while(~scanf("%d",&n)){
		for(int i = 1; i <= n; i++)scanf("%d",&a[i]);
		build(1, n, 1);
		scanf("%d",&Q);
		while(Q--){
			int l, r;
			scanf("%d %d",&l,&r);
			printf("%d\n", query(l, r, 1));
		}
	}
	return 0;
}
/*
3 
-1 2 3
1
1 2

*/

SPOJ 1043 Can you answer these queries I 求任意区间最大连续子段和 线段树,布布扣,bubuko.com

SPOJ 1043 Can you answer these queries I 求任意区间最大连续子段和 线段树

标签：http   os   io   for   ar   amp   ef   on   

原文地址：http://blog.csdn.net/qq574857122/article/details/38669803