Y总的线段树板子，能运行但是只有10分，悬赏一个三连

ljx_gkx @ 2023-04-07 14:50:34

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;
const int N = 2e6 + 10;
int a[N];
int n, m;

struct Node{
    int l, r;
    int val;
}tr[N*4-1];

void pushup(int u)  //利用子节点的信息向上更新父节点的信息，本题中需要更新的是父节点的区间的最小值！
{
    tr[u].val = min(tr[u<<1].val, tr[u<<1|1].val);
    return ;
}

void build(int u, int l, int r)
{
    tr[u] = {l, r};
    if (l == r){
        tr[u].val = a[r];
        return ;
    }
    //给当前区间分配左右端点！
    int mid = (l+r) >> 1;
    build(u<<1, l, mid), build(u<<1|1, mid+1, r);
    pushup(u);
}

int query (int u, int l, int r)
{
    if (tr[u].l >= l && tr[u].r <= r)
    {
        return tr[u].val;
    }

    int mid = (tr[u].l + tr[u].r) >> 1;
    int res = 0x3f3f3f3f;

    if (l <= mid) res = min(res, query(u<<1, l, r));
    if (r > mid) res = min(res, query(u<<1|1, l, r));

    return res;
}

int main()
{
    cin >> n >> m;
    for (int i=1; i <= n; i ++)
        cin >> a[i];

    build(1, 1, n);

    for (int i=1; i <= n; i ++)
    {
        int l = i-m, r = i-1;
        if (l <= 0) cout << 0 << endl;
        else 
            cout << query(1, l, r) << endl;
    }

    return 0;
}

by jacklee10086 @ 2023-04-07 14:57:43

题目要求为若前面的数不足m项则从第1个数开始，若前面没有数则输出0。你的代码似乎是若前面的数不足m项则输出0了

by jacklee10086 @ 2023-04-07 15:03:30

给你简单改了一下

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;
const int N = 2e6 + 10;
int a[N];
int n, m;

struct Node{
    int l, r;
    int val;
}tr[N*4-1];

void pushup(int u)  //利用子节点的信息向上更新父节点的信息，本题中需要更新的是父节点的区间的最小值！
{
    tr[u].val = min(tr[u<<1].val, tr[u<<1|1].val);
    return ;
}

void build(int u, int l, int r)
{
    tr[u] = {l, r};
    if (l == r){
        tr[u].val = a[r];
        return ;
    }
    //给当前区间分配左右端点！
    int mid = (l+r) >> 1;
    build(u<<1, l, mid), build(u<<1|1, mid+1, r);
    pushup(u);
}

int query (int u, int l, int r)
{
    if (tr[u].l >= l && tr[u].r <= r)
    {
        return tr[u].val;
    }

    int mid = (tr[u].l + tr[u].r) >> 1;
    int res = 0x3f3f3f3f;

    if (l <= mid) res = min(res, query(u<<1, l, r));
    if (r > mid) res = min(res, query(u<<1|1, l, r));

    return res;
}

int main()
{
    cin >> n >> m;
    for (int i=1; i <= n; i ++)
        cin >> a[i];

    build(1, 1, n);

    for (int i=1; i <= n; i ++)
    {
        int l = max(i-m,1), r = i-1;
        if (r <= 0) cout << 0 << endl;
        else 
            cout << query(1, l, r) << endl;
    }

    return 0;
}

by jacklee10086 @ 2023-04-07 15:12:35

btw，这道题的正解是滑动窗口，虽然线段树可做但是要卡个常数。