2569. 更新数组后处理求和查询（模板 + 普通线段树熟练掌握）

题目

• leetcode.cn/problems/ha…

思路

• 操作2和操作3都非常好实现，直接累加一个和就行
• 关键在于操作1（nums1数组所有元素大小在0~1之间）翻转
• 线段树相比树状数组更好理解
• 具体细节见代码

代码

class Solution {

typedef long long LL;

static const int N = 100010;

int n, m;
int w[N];
struct Node
{
    int l, r;
    LL sum, add;//懒标记：是否翻转过
}tr[N * 4];

void pushup(int u)
{
    tr[u].sum = tr[u << 1].sum + tr[u << 1 | 1].sum;
}

void pushdown(int u)
{
    auto &root = tr[u], &left = tr[u << 1], &right = tr[u << 1 | 1];
    if (root.add)
    {
        left.add ^= root.add;
        left.sum = (LL)(left.r - left.l + 1 - left.sum);
        right.add ^= root.add;
        right.sum = (LL)(right.r - right.l + 1 - right.sum);
        root.add = 0;
    }
}

void build(int u, int l, int r)
{
    if (l == r) tr[u] = {l, r, w[r], 0};
    else
    {
        tr[u] = {l, r};
        int mid = l + r >> 1;
        build(u << 1, l, mid), build(u << 1 | 1, mid + 1, r);
        pushup(u);
    }
}

void modify(int u, int l, int r, int d)
{
    if (tr[u].l >= l && tr[u].r <= r)
    {
        tr[u].add ^= d;
        tr[u].sum = (LL)(tr[u].r - tr[u].l + 1 - tr[u].sum);
        
    }
    else    // 一定要分裂
    {
        pushdown(u);
        int mid = tr[u].l + tr[u].r >> 1;
        if (l <= mid) modify(u << 1, l, r, d);
        if (r > mid) modify(u << 1 | 1, l, r, d);
        pushup(u);
    }
}

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

    pushdown(u);
    int mid = tr[u].l + tr[u].r >> 1;
    LL sum = 0;
    if (l <= mid) sum = query(u << 1, l, r);
    if (r > mid) sum += query(u << 1 | 1, l, r);
    return sum;
}


public:
    vector<long long> handleQuery(vector<int>& nums1, vector<int>& nums2, vector<vector<int>>& queries) {
        int n = nums1.size();
        LL add = 0;
        for(auto &x:nums2)add += x;

        for(int i = 1;i <= n;i ++)
        {
            w[i] = nums1[i-1];
        }
        build(1,1,n);
        
        vector<LL> ans;
        for(auto &t:queries)
        {
            if(t[0] == 1)modify(1,t[1]+1,t[2]+1,1);
            else if(t[0] == 2)add += tr[1].sum*t[1];
            else ans.push_back(add);
        }

        return ans;
    }
};