洛谷P4396 [AHOI2013]作业(树套树)

题意

题目链接

sol

为什么一堆分块呀。。三维数点不应该是套路离线/可持久化+树套树么。。

亲测树状数组套权值线段树可过

复杂度\(o(nlog^2n)\)，空间\(o(nlogn)\)(离线)

#include<bits/stdc++.h> 
#define pair pair<int, int>
#define mp(x, y) make_pair(x, y)
#define fi first
#define se second
#define fin(x) {freopen(#x".in","r",stdin);}
#define fout(x) {freopen(#x".out","w",stdout);}
using namespace std;
const int maxn = 4e5 + 10, ss = 1e7 + 10;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int n, m, a[maxn], pre[maxn], las[maxn], lim = 1e5, tot;
pair ans[maxn];
#define lb(x) (x & (-x))
struct bit {
    int t[maxn];
    void add(int x, int v) {
        x++;
        while(x <= lim) t[x] += v, x += lb(x); 
    }
    int sum(int x) {
        x++;
        int ans = 0;
        while(x) ans += t[x], x -= lb(x);
        return ans;
    }
    int query(int x, int y) {return sum(y) - sum(x - 1);}
}q1;
struct query {
    int k, a, b, id, p;
    bool operator < (const query &rhs) const {
        return k < rhs.k;   
    }
}q[maxn];
int root[ss], sum[ss], ls[ss], rs[ss], cnt;
void update(int k) {
    sum[k] = sum[ls[k]] + sum[rs[k]];
}
void insert(int &k, int l, int r, int p, int v) {
    if(!k) k = ++cnt;
    if(l == r) {sum[k]++; return ;}
    int mid = l + r >> 1;
    if(p <= mid) insert(ls[k], l, mid, p, v);
    else insert(rs[k], mid + 1, r, p, v);
    update(k);
}
int query(int k, int l, int r, int ql, int qr) {
    if(!k) return 0;
    if(ql <= l && r <= qr) return sum[k];
    int mid = l + r >> 1;   
    if(ql > mid) return query(rs[k], mid + 1, r, ql, qr);
    else if(qr <= mid) return query(ls[k], l, mid, ql, qr);
    else return query(ls[k], l, mid, ql, qr) + query(rs[k], mid + 1, r, ql, qr);
}
void add(int x, int v) {
    x++;
    while(x <= lim) insert(root[x], 0, lim, v, 1), x += lb(x); 
}
int query(int x, int a, int b) {
    x++;
    int ans = 0;
    while(x) ans += query(root[x], 0, lim, a, b), x -= lb(x);
    return ans;
}
void solve() {
    int x = 0;
    for(int i = 1; i <= tot; i++) {
        while(x < q[i].k) 
            q1.add(a[++x], 1), add(pre[x], a[x]);
        ans[abs(q[i].id)].fi += (q[i].id / (abs(q[i].id))) * q1.query(q[i].a, q[i].b);
        ans[abs(q[i].id)].se += (q[i].id / (abs(q[i].id))) * query(q[i].p, q[i].a, q[i].b);
    }
}
signed main() {
    n = read(); m = read();
    for(int i = 1; i <= n; i++) {
        a[i] = read();
        pre[i] = las[a[i]]; las[a[i]] = i;
    }
    for(int i = 1; i <= m; i++) {
        int l = read(), r = read(), a = read(), b = read();
        q[++tot].k = l - 1; q[tot].a = a; q[tot].b = b; q[tot].id = -i; q[tot].p = l - 1;
        q[++tot].k = r;     q[tot].a = a; q[tot].b = b; q[tot].id = i;  q[tot].p = l - 1;
    }
    sort(q + 1, q + tot + 1); 
    solve();
    for(int i = 1; i <= m; i++) printf("%d %d\n", ans[i].fi, ans[i].se);
    return 0;
}