DZY Loves Colors

题面翻译

• 有一个 \(n\) 个元素组成的序列，每个元素有两个属性：颜色 \(c_i\) 和权值 \(w_i\)。\(c_i\) 初始为 \(i\)，\(w_i\) 初始为 \(0\)。\(m\) 次操作，操作有两种：
  1. 1 l r x：对 \(i\in [l,r]\) 的所有 \(i\) 进行如下操作：设第 \(i\) 个元素 原来 的颜色为 \(y\)，您要把第 \(i\) 个元素的颜色改为 \(x\)，权值 增加 \(|y-x|\)。
  2. 2 l r：求 \(\displaystyle\sum_{i=l}^r w_i\)。
• \(1\le n,m\le 10^5\)，\(1\le x\le 10^8\)。

输入格式

The first line contains two space-separated integers $ n,m (1<=n,m<=10^{5}) $ .

Each of the next $ m $ lines begins with a integer $ type (1<=type<=2) $ , which represents the type of this operation.

If $ type=1 $ , there will be $ 3 $ more integers $ l,r,x (1<=l<=r<=n; 1<=x<=10^{8}) $ in this line, describing an operation $ 1 $ .

If $ type=2 $ , there will be $ 2 $ more integers $ l,r (1<=l<=r<=n) $ in this line, describing an operation $ 2 $ .

输出格式

For each operation $ 2 $ , print a line containing the answer — sum of colorfulness.

样例 #1

样例输入 #1

3 3
1 1 2 4
1 2 3 5
2 1 3

样例输出 #1

8

样例 #2

样例输入 #2

3 4
1 1 3 4
2 1 1
2 2 2
2 3 3

样例输出 #2

3
2
1

样例 #3

样例输入 #3

10 6
1 1 5 3
1 2 7 9
1 10 10 11
1 3 8 12
1 1 10 3
2 1 10

样例输出 #3

129

分析：

修改涉及到绝对值，可以通过判断区间内的数是否全部相等再进行区间操作；
通过懒标记 same 判断该区间的所有颜色是否都相等，
· 相等时就可以将懒标记下传，更新add;

实现：

#include <bits/stdc++.h>
using namespace std;
#define mst(x, y) memset(x, y, sizeof x)
#define endl '\n'
#define INF LONG_LONG_MAX
#define int long long
#define Lson u << 1, l, mid
#define Rson u << 1 | 1, mid + 1, r
#define FAST ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
const int N = 500010, MOD = 1e9 + 7;
const double EPS = 1e-6;
typedef pair<int, int> PII;
typedef unordered_map<int, int> Ump;
int T;
int n, m;
int op;
struct Node
{
    int l, r;
    int sum, same, add;
} tr[N << 2];
void pushup(Node &u, Node &l, Node &r)
{
    u.sum = l.sum + r.sum;
    if (l.same == r.same)
        u.same = l.same;
    else
        u.same = 0;
}
void pushup(int u)
{
    pushup(tr[u], tr[u << 1], tr[u << 1 | 1]);
}
void pushdown(int u)
{
    if (tr[u].same)
    {
        tr[u << 1].sum += (tr[u << 1].r - tr[u << 1].l + 1) * tr[u].add;
        tr[u << 1 | 1].sum += (tr[u << 1 | 1].r - tr[u << 1 | 1].l + 1) * tr[u].add;
        tr[u << 1].add += tr[u].add;
        tr[u << 1 | 1].add += tr[u].add;
        tr[u << 1].same = tr[u].same;
        tr[u << 1 | 1].same = tr[u].same;
        tr[u].add = 0;
    }
}
void build(int u, int l, int r)
{
    if (l == r)
        tr[u] = {r, r, 0, l, 0};
    else
    {
        tr[u] = {l, r};
        int mid = l + r >> 1;
        build(Lson), build(Rson);
        pushup(u);
    }
}
void modify(int u, int l, int r, int k)
{
    if (tr[u].l >= l && tr[u].r <= r && tr[u].same)
    {
        tr[u].sum += (tr[u].r - tr[u].l + 1) * abs(k - tr[u].same);
        tr[u].add += abs(k - tr[u].same);
        tr[u].same = k;
    }
    else
    {
        pushdown(u);
        int mid = tr[u].l + tr[u].r >> 1;
        if (l <= mid)
            modify(u << 1, l, r, k);
        if (r > mid)
            modify(u << 1 | 1, l, r, k);
        pushup(u);
    }
}
int query(int u, int l, int r)
{
    if (tr[u].l >= l && tr[u].r <= r)
        return tr[u].sum;
    else
    {
        pushdown(u);
        int mid = tr[u].l + tr[u].r >> 1;
        int res = 0;
        if (l <= mid)
            res = query(u << 1, l, r);
        if (r > mid)
            res += query(u << 1 | 1, l, r);
        return res;
    }
}
void solve()
{
    cin >> n >> m;
    build(1, 1, n);
    while (m--)
    {
        int op;
        cin >> op;
        if (op == 1)
        {
            int l, r, x;
            cin >> l >> r >> x;
            modify(1, l, r, x);
        }
        else if (op == 2)
        {
            int l, r;
            cin >> l >> r;
            cout << query(1, l, r) << endl;
        }
    }
}
signed main()
{
    FAST;
    T = 1;
    // cin >> T;
    while (T--)
        solve();
    return 0;
}