线段树(区间合并) LA 3989 "Ray, Pass me the dishes!"

 

题目传送门

题意：动态最大连续子序列和，静态的题目

分析：nlogn的归并思想。线段树维护结点的三个信息，最大前缀和，最大后缀和，该区间的最大和的两个端点，然后答案是三个的better。书上用pair保存端点，用自带的<来得到最优。

#include <bits/stdc++.h>
using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
typedef pair<int, int> P;
const int N = 5e5 + 5;
ll sum[N];
struct	ST	{
	int pre[N<<2], suf[N<<2];
	P sub[N<<2];
	ll get_sum(P p)	{
		return sum[p.second] - sum[p.first-1];
	}
	P better(P a, P b)	{
		ll v1 = get_sum (a), v2 = get_sum (b);
		if (v1 != v2)	return v1 > v2 ? a : b;
		else	return a < b ? a : b;
	}
	void push_up(int l, int r, int rt)	{
		pre[rt] = better (make_pair (l, pre[rt<<1]), make_pair (l, pre[rt<<1|1])).second;	//该区间的最大前缀
		suf[rt] = better (make_pair (suf[rt<<1], r), make_pair (suf[rt<<1|1], r)).first;	//该区间的最大后缀
		sub[rt] = better (sub[rt<<1], sub[rt<<1|1]);		//该区间的最大连续和：max (左前缀，右后缀+左前缀，右后缀)
		sub[rt] = better (sub[rt], make_pair (suf[rt<<1], pre[rt<<1|1]));	//不一定就是最大前缀或最大后缀
	}
	void build(int l, int r, int rt)	{
		if (l == r)	{
			pre[rt] = suf[rt] = l;
			sub[rt] = make_pair (l, l);
			return ;
		}
		int mid = (l + r) >> 1;
		build (lson);	build (rson);
		push_up (l, r, rt);
	}
	P query_pre(int ql, int qr, int l, int r, int rt)	{
		if (pre[rt] <= qr)	return make_pair (l, pre[rt]);
		int mid = (l + r) >> 1;
		if (qr <= mid)	return query_pre (ql, qr, lson);
		P p = query_pre (ql, qr, rson);	p.first = l;
		return better (p, make_pair (l, pre[rt<<1]));
	}
	P query_suf(int ql, int qr, int l, int r, int rt)	{
		if (suf[rt] >= ql)	return make_pair (suf[rt], r);
		int mid = (l + r) >> 1;
		if (ql > mid)	return query_suf (ql, qr, rson);
		P p = query_suf (ql, qr, lson);	p.second = r;
		return better (p, make_pair (suf[rt<<1|1], r));
	}
	P query(int ql, int qr, int l, int r, int rt)	{
		if (ql <= l && r <= qr)	return sub[rt];
		int mid = (l + r) >> 1;
		if (qr <= mid)	return query (ql, qr, lson);
		if (ql > mid)	return query (ql, qr, rson);
		P p1 = query_suf (ql, qr, lson);				//ql <= mid < qr
		P p2 = query_pre (ql, qr, rson);				//和push_up一样
		P p3 = better (query (ql, qr, lson), query (ql, qr, rson));	//该区间的最大连续和：max (左前缀，右后缀+左前缀，右后缀)
		return better (p3, make_pair (p1.first, p2.second));
	}
}st;

int main(void)	{
	int n, q, cas = 0;
	while (scanf ("%d%d", &n, &q) == 2)	{
		for (int i=1; i<=n; ++i)	{
			scanf ("%lld", &sum[i]);	sum[i] += sum[i-1];
		}
		st.build (1, n, 1);
		printf ("Case %d:\n", ++cas);
		int ql, qr;
		while (q--)	{
			scanf ("%d%d", &ql, &qr);
			P ans = st.query (ql, qr, 1, n, 1);
			printf ("%d %d\n", ans.first, ans.second);
		}
	}

	return 0;
}