题意
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;
}