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SP1716GSS3 - Can you answer these queries III参与者 3已保存回复 3

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@mi86bihn
此快照首次捕获于
2025/11/21 09:20
4 个月前
此快照最后确认于
2025/11/21 09:20
4 个月前
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一个多小时了,还没找到错误,样例未过……
CPP
#include <cstdio>
#include <iostream>
#include <cstring>

using namespace std;

#define MAXN 50010
#define INF 0x3f3f3f3f

struct node
{
    int a[4][4];
}tr[MAXN << 2];
int n, m, l, r, opt;
int a[MAXN];

inline node mul(node a, node b)
{
    node c;
    memset(c.a, -INF, sizeof(c.a));
    for (int i = 1; i <= 3; ++ i)
        for (int j = 1; j <= 3; ++ j)
            for (int k = 1; k <= 3; ++ k)
                c.a[i][j] = max(a.a[i][k] * b.a[k][j], c.a[i][j]);
    return c;
}

void upt(int rt)
{
    tr[rt] = mul(tr[rt << 1], tr[rt << 1 | 1]);
}

void build(int rt, int l, int r)
{
    if (l == r)
    {
        tr[rt].a[1][1] = tr[rt].a[1][3] = tr[rt].a[2][1] = tr[rt].a[2][3] = a[l];
        tr[rt].a[1][2] = tr[rt].a[3][1] = tr[rt].a[3][2] = -INF;
        tr[rt].a[2][2] = tr[rt].a[3][3] = 0;
        return ;
    }
    int mid = (l + r) >> 1;
    build(rt << 1, l, mid);
    build(rt << 1 | 1, mid + 1, r);
    upt(rt);
}

node query(int rt, int x, int y, int l, int r)
{
    if (l <= x && y <= r)
        return tr[rt];
    int mid = (x + y) >> 1;
    if (l <= mid && mid < r)
        return mul(query(rt << 1, x, mid, l, r), query(rt << 1 | 1, mid + 1, y, l, r));
    else 
        if (l <= mid)
            return query(rt << 1, x, mid, l, r);
        else 
            return query(rt << 1 | 1, mid + 1, y, l, r);
}

void modify(int rt, int l, int r, int pos, int v)
{
    if (l == r)
    {
        tr[rt].a[1][1] = tr[rt].a[1][3] = tr[rt].a[2][1] = tr[rt].a[2][3] = v;
        return ;
    }
    int mid = (l + r) >> 1;
    if (pos <= mid)
        modify(rt << 1, l, mid, pos, v);
    else 
        modify(rt << 1 | 1, mid + 1, r, pos, v);
    upt(rt);
}

int main()
{
    scanf("%d", &n);
    for (int i = 1; i <= n; ++ i)
        scanf("%d", &a[i]);
    build(1, 1, n);
    scanf("%d", &m);
    while (m --)
    {
        scanf("%d%d%d", &opt, &l, &r);
        if (opt == 1)
        {
            if (l > r)
                swap(l, r);
            node ans = query(1, 1, n, l, r);
            printf("%d\n", max(ans.a[2][1], ans.a[2][3]));
        }
        else 
        {
            a[l] = r;
            modify(1, 1, n, l, r);
        }
    }
    return 0;
}

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