51NOD 1287 加农炮（不水的线段树）

》》点击进入原题测试《《

Input示例
9 11
1
2
0
4
3
2
1
5
7
2
8
0
7
6
5
3
4
5
6
5
Output示例
2
2
2
4
3
3
5
6
7

思路：刚开始以为结点存最大值就行了，然后大于左子树的最大值就能进入右子树；然后发现样例都过不了；后面发现，并不是这个样子，假如这个数小于等于右孩子最左边那个数的话，也不能进入有孩子，所以结点还得保存右孩子最左边的那个值；同时更新一个最大值，当输入值咸鱼等于a[0]或者大于最大值时跳过。

#include<climits>
#include<iostream>
#include<algorithm>
using namespace std;

#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define ll long long
const int maxn = 5e4 + 5;
ll tree[maxn << 2], mtree[maxn << 2], a[maxn];
ll ma = INT_MIN;

void build(int l, int r, int rt)
{
    if (l == r){
        tree[rt] = a[l];
        mtree[rt] = a[l];
        return;
    }
    int m = (l + r) >> 1;
    build(lson);
    build(rson);
    tree[rt] = max(tree[rt << 1], tree[rt << 1 | 1]);
    mtree[rt] = mtree[rt << 1];
}
void update(int x, int l, int r, int rt)
{
    if (l == r){
        a[l] += 1;
        tree[rt] += 1;
        mtree[rt] += 1;
        if (a[l] > ma)ma = a[l];
        return;
    }
    int m = (l + r) >> 1;
    if (tree[rt << 1] < x&&x>mtree[rt << 1 | 1])
    
        update(x, rson);
    else update(x, lson);
    tree[rt] = max(tree[rt << 1], tree[rt << 1 | 1]);
    mtree[rt] = mtree[rt << 1];
}
void Print(int l, int r, int rt)
{
    if (l == r){
        cout << rt << " = " << tree[rt] << endl;
        return;
    }
    cout << rt << " = " << tree[rt] << endl;
    int m = (r + l) >> 1;
    if (l <= m)Print(lson);
    if (r > m)Print(rson);
}
int main()
{
    std::ios::sync_with_stdio(false);
    int m, n; cin >> m >> n;

    
    for (int i = 1; i <= m; i++){
        cin >> a[i];
        if (ma < a[i])ma = a[i];
    }
    build(1, m, 1);
    
    int temp;
    while (n--){
        cin >> temp;
        if (temp <= a[1] || temp>ma)
            continue;
        update(temp, 1, m, 1);
        //Print(1, m, 1);    
    }
    for (int i = 1; i <= m; i++){
        if (i != 1)cout << endl;
        cout << a[i];
    }
    

}