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 %dn", ans[i].fi, ans[i].se); return 0; }
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