luoguP2781 传教

Posted 2021-01-07 ljc00118

https://www.luogu.org/problemnew/show/P2781

简化版题意：有 n 个数，初始值为 0，进行 m 次操作，每次操作支持将 [l, r] 加 v 和查询 [l, r] 中所有的数的和

n <= 1e9，m <= 1e3

博主 zz 的打了一个支持分裂节点的 splay，AC 后发现可以 m 方暴力过

方法和方伯伯的OJ这题类似，可以参考它的做法

#include <bits/stdc++.h>
using namespace std;

typedef unsigned long long ull;
typedef long long ll;

template <typename T>
inline void read(T &f) {
    f = 0; T fu = 1; char c = getchar();
    while(c < '0' || c > '9') {if(c == '-') fu = -1; c = getchar();}
    while(c >= '0' && c <= '9') {f = (f << 3) + (f << 1) + (c & 15); c = getchar();}
    f *= fu;
}

struct Node {
    ll val, tag, sum;
    int l, r, size;
    Node *ch[2];
    Node () {
        val = tag = l = r = size = 0;
        ch[0] = ch[1] = NULL;
    }
}*root;

int n, m;

void update(Node *u) {
    u -> size = u -> r - u -> l + 1;
    u -> val = u -> sum;
    if(u -> ch[0]) u -> size += u -> ch[0] -> size, u -> val += u -> ch[0] -> val;
    if(u -> ch[1]) u -> size += u -> ch[1] -> size, u -> val += u -> ch[1] -> val;
}

void pushdown(Node *u) {
    if(u -> tag) {
        if(u -> ch[0]) {
            u -> ch[0] -> tag += u -> tag;
            u -> ch[0] -> sum += (ll)(u -> ch[0] -> r - u -> ch[0] -> l + 1) * u -> tag;
            u -> ch[0] -> val += (ll)u -> ch[0] -> size * u -> tag; 
        }
        if(u -> ch[1]) {
            u -> ch[1] -> tag += u -> tag;
            u -> ch[1] -> sum += (ll)(u -> ch[1] -> r - u -> ch[1] -> l + 1) * u -> tag; 
            u -> ch[1] -> val += (ll)u -> ch[1] -> size * u -> tag;
        }
        u -> tag = 0;
    }
}

void rotate(Node *&u, int d) {
    Node *tmp = u -> ch[d];
    u -> ch[d] = tmp -> ch[d ^ 1];
    tmp -> ch[d ^ 1] = u;
    update(u); update(tmp);
    u = tmp;
}

void splay(Node *&u, int k) {
    if(u == NULL) return;
    pushdown(u);
    int ltree = u -> ch[0] ? u -> ch[0] -> size : 0;
    if(k > ltree && ltree + (u -> r - u -> l + 1) >= k) return;
    int d = k > ltree; 
    splay(u -> ch[d], d ? k - ltree - (u -> r - u -> l + 1) : k);
    rotate(u, d);
}

void split(Node *&u, int x) {
    splay(u, x); ll sum = u -> sum; int l = u -> l, r = u -> r;
    if(u -> l != x) {
        Node *tmp = new Node();
        tmp -> sum = sum / (ll)(r - l + 1) * (x - l);
        tmp -> l = l, tmp -> r = x - 1;
        tmp -> ch[0] = u -> ch[0]; update(tmp);
        u -> ch[0] = tmp; u -> l = x;
    }
    if(u -> r != x) {
        Node *tmp = new Node();
        tmp -> sum = sum / (ll)(r - l + 1) * (ll)(r - x);
        tmp -> l = x + 1, tmp -> r = r;
        tmp -> ch[1] = u -> ch[1]; update(tmp);
        u -> ch[1] = tmp; u -> r = x;
    }
    u -> sum /= (ll)(r - l + 1); update(u);
}

int main() {
    cin >> n >> m;
    root = new Node();
    root -> val = root -> tag = root -> sum = 0;
    root -> l = 1, root -> r = n; root -> size = n;
    root -> ch[0] = root -> ch[1] = NULL;
    for(int i = 1; i <= m; i++) {
//      printf("root -> size = %d
", root -> size);
        int t; read(t);
        if(t == 1) {
            int a, b; ll c;
            read(a); read(b); read(c);
            if(a == 1 && b == n) {
                root -> val += (ll)n * c;
                root -> tag += c;
                root -> sum += (ll)(root -> r - root -> l + 1) * c;
            } else if(a == 1) {
                split(root, b + 1);
                root -> ch[0] -> val += (ll)root -> ch[0] -> size * c;
                root -> ch[0] -> tag += c;
                root -> ch[0] -> sum += (ll)(root -> ch[0] -> r - root -> ch[0] -> l + 1) * c;
                update(root);
            } else if(b == n) {
                split(root, a - 1);
                root -> ch[1] -> val += (ll)root -> ch[1] -> size * c;
                root -> ch[1] -> tag += c;
                root -> ch[1] -> sum += (ll)(root -> ch[1] -> r - root -> ch[1] -> l + 1) * c;
                update(root);
            } else {
                split(root, b + 1);
                split(root -> ch[0], a - 1);
                root -> ch[0] -> ch[1] -> val += (ll)root -> ch[0] -> ch[1] -> size * c;
                root -> ch[0] -> ch[1] -> tag += c;
                root -> ch[0] -> ch[1] -> sum += (ll)(root -> ch[0] -> ch[1] -> r - root -> ch[0] -> ch[1] -> l + 1) * c;
                update(root -> ch[0]); update(root);
            }
        }
        if(t == 2) {
            int a, b; read(a); read(b);
            if(a == 1 && b == n) {
                printf("%lld
", root -> val);
            } else if(a == 1) {
                split(root, b + 1);
                printf("%lld
", root -> ch[0] -> val);
            } else if(b == n) {
                split(root, a - 1);
                printf("%lld
", root -> ch[1] -> val);
            } else {
                split(root, b + 1);
                split(root -> ch[0], a - 1);
                printf("%lld
", root -> ch[0] -> ch[1] -> val);
            }
        } 
    }
    return 0;
}