JZOJ 1073. 【GDOI2005】山海经

\(\text{Solution}\)

非常经典的求区间最大字段和

不难想到线段树，考虑处理区间答案的合并

维护前缀后缀最大和与区间答案，合并考虑跨中点贡献即可

代码打得非常恶心。。。

\(\text{Code}\)

#include <cstdio>
#define RE register
#define IN inline
#define ls (p << 1)
#define rs (ls | 1)
using namespace std;

const int N = 1e5 + 5, INF = 0x3f3f3f3f;
int n, m, a[N], s[N];
struct node{int pl, pr, sl, sr, l, r;}seg[N * 4];
IN int Get(int l, int r){if (!l || !r) return -INF; return s[r] - s[l - 1];}
IN node operator + (const node &a, const node &b)
{
	node c = node{0, 0, 0, 0, 0, 0};
	c.pl = a.pl, c.pr = a.pr; if (Get(a.pl, b.pr) > Get(c.pl, c.pr)) c.pr = b.pr;
	c.sl = b.sl, c.sr = b.sr; if ((Get(a.sl, b.sr) > Get(c.sl, c.sr) || (Get(a.sl, b.sr) == Get(c.sl, c.sr) && a.sl < c.sl))) c.sl = a.sl;
	if (Get(a.l, a.r) >= Get(b.l, b.r)) c.l = a.l, c.r = a.r; else c.l = b.l, c.r = b.r;
	if (Get(c.pl, c.pr) > Get(c.l, c.r) || (Get(c.pl, c.pr) == Get(c.l, c.r) && (c.pl < c.l || (c.pl == c.l && c.pr < c.r)))) c.l = c.pl, c.r = c.pr;
	if (Get(c.sl, c.sr) > Get(c.l, c.r) || (Get(c.sl, c.sr) == Get(c.l, c.r) && (c.sl < c.l || (c.sl == c.l && c.sr < c.r)))) c.l = c.sl, c.r = c.sr;
	if (Get(a.sl, b.pr) > Get(c.l, c.r) || (Get(a.sl, b.pr) == Get(c.l, c.r) && (a.sl < c.l || (a.sl == c.l && b.pr < c.r)))) c.l = a.sl, c.r = b.pr;
	return c;
}
void build(int p, int l, int r)
{
	if (l == r) return seg[p] = node{l, l, l, l, l, l}, void();
	int mid = l + r >> 1; build(ls, l, mid), build(rs, mid + 1, r), seg[p] = seg[ls] + seg[rs];
}
node Query(int p, int l, int r, int x, int y)
{
	if (x <= l && r <= y) return seg[p]; int mid = l + r >> 1; node ret = node{0, 0, 0, 0, 0, 0};
	if (x <= mid) ret = Query(ls, l, mid, x, y); if (y > mid) ret = ret + Query(rs, mid + 1, r, x, y);
	return ret;
}

int main()
{
	scanf("%d%d", &n, &m); for(RE int i = 1; i <= n; i++) scanf("%d", &a[i]), s[i] = s[i - 1] + a[i]; build(1, 1, n); node ret;
	for(int l, r; m; --m) scanf("%d%d", &l, &r), ret = Query(1, 1, n, l, r), printf("%d %d %d\n", ret.l, ret.r, Get(ret.l, ret.r));
}