hihoCoder1586 Minimum 线段树

题目：

http://hihocoder.com/problemset/problem/1586?sid=1197111

题意：

给定一个序列，有两种操作：

1 l r：从区间[l,r]内选出数字ai, aj，是的ai∗aj的值最小

2 x y：把第x个元素的值更新为y

思路：

求区间内乘积的最小值，无非以下情况：都是正数时，直接取最小值相乘即可，有正数有负数的时候，取最小值和最大值相乘，都是负数时，取最大值相乘。维护最大值最小值用线段树即可

#include <bits/stdc++.h>

using namespace std;

const int N = 1000000 + 10, INF = 0x3f3f3f3f;

struct node
{
int l, r, maxx, minn;
}tr[N*4];
int a
;
void push_up(int k)
{
tr[k].maxx = max(tr[k<<1].maxx, tr[k<<1|1].maxx);
tr[k].minn = min(tr[k<<1].minn, tr[k<<1|1].minn);
}
void build(int l, int r, int k)
{
tr[k].l = l, tr[k].r = r;
if(l == r)
{
tr[k].maxx = tr[k].minn = a[l];
return;
}
int mid = (l + r) >> 1;
build(l, mid, k<<1);
build(mid+1, r, k<<1|1);
push_up(k);
}
void update(int x, int val, int k)
{
if(tr[k].l == tr[k].r && x == tr[k].l)
{
tr[k].maxx = tr[k].minn = val; return;
}
int mid = (tr[k].l + tr[k].r) >> 1;
if(x <= mid) update(x, val, k<<1);
else update(x, val, k<<1|1);
push_up(k);
}
int query(int l, int r, int f, int k)
{
if(l <= tr[k].l && tr[k].r <= r)
{
return f ? tr[k].maxx : tr[k].minn;
}
int mid = (tr[k].l + tr[k].r) >> 1;
int ans = f ? -INF : INF;
if(l <= mid)
{
int t = query(l, r, f, k<<1);
ans = f ? max(ans, t) : min(ans, t);
}
if(r > mid)
{
int t = query(l, r, f, k<<1|1);
ans = f ? max(ans, t) : min(ans, t);
}
return ans;
}
int main()
{
int t, n, m;
scanf("%d", &t);
while(t--)
{
scanf("%d", &n);
n = 1 << n;
for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
build(1, n, 1);
scanf("%d", &m);
int opt, x, y;
for(int i = 1; i <= m; i++)
{
scanf("%d%d%d", &opt, &x, &y);
if(opt == 1)
{
++x, ++y;
int maxx = query(x, y, 1, 1), minn = query(x, y, 0, 1);
printf("%lld\n", min(1LL*minn*minn, min(1LL*minn*maxx, 1LL*maxx*maxx)));
}
else
{
++x;
update(x, y, 1);
}
}
}
return 0;
}