c/c++语言开发共享洛谷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 %dn", ans[i].fi, ans[i].se);     return 0; }