POJ 2104 K-th Number
Description
You are working for Macrohard company in data structures department. After failing your previous task about key insertion you were asked to write a new data structure that would be able to return quickly k-th order statistics in the array segment.
That is, given an array a[1...n] of different integer numbers, your program must answer a series of questions Q(i, j, k) in the form: "What would be the k-th number in a[i...j] segment, if this segment was sorted?"
For example, consider the array a = (1, 5, 2, 6, 3, 7, 4). Let the question be Q(2, 5, 3). The segment a[2...5] is (5, 2, 6, 3). If we sort this segment, we get (2, 3, 5, 6), the third number is 5, and therefore the answer to the question is 5.
That is, given an array a[1...n] of different integer numbers, your program must answer a series of questions Q(i, j, k) in the form: "What would be the k-th number in a[i...j] segment, if this segment was sorted?"
For example, consider the array a = (1, 5, 2, 6, 3, 7, 4). Let the question be Q(2, 5, 3). The segment a[2...5] is (5, 2, 6, 3). If we sort this segment, we get (2, 3, 5, 6), the third number is 5, and therefore the answer to the question is 5.
Input
The first line of the input file contains n --- the size of the array, and m --- the number of questions to answer (1 <= n <= 100 000, 1 <= m <= 5 000).
The second line contains n different integer numbers not exceeding 10 9 by their absolute values --- the array for which the answers should be given.
The following m lines contain question descriptions, each description consists of three numbers: i, j, and k (1 <= i <= j <= n, 1 <= k <= j - i + 1) and represents the question Q(i, j, k).
The second line contains n different integer numbers not exceeding 10 9 by their absolute values --- the array for which the answers should be given.
The following m lines contain question descriptions, each description consists of three numbers: i, j, and k (1 <= i <= j <= n, 1 <= k <= j - i + 1) and represents the question Q(i, j, k).
Output
For each question output the answer to it --- the k-th number in sorted a[i...j] segment.
Sample Input
7 3 1 5 2 6 3 7 4 2 5 3 4 4 1 1 7 3
Sample Output
5 63
静态区间第k大,这题时限比较长，暴力大法什么的就不说了，主要用来熟悉各种log的算法
第一发，二分答案+线段树
#include<cstdio> #include<cstring> #include<cmath> #include<queue> #include<vector> #include<iostream> #include<algorithm> #include<bitset> #include<functional> using namespace std; typedef long long LL; const int maxn = 1e5 + 10; int n, m, l, r, k; struct tree { int f[20][maxn]; void insert(int x, int l, int r) { if (l == r) { scanf("%d", &f[x][l]); return; } int mid = l + r >> 1; insert(x + 1, l, mid); insert(x + 1, mid + 1, r); for (int i = l, j = l, k = mid + 1; i <= r; i++) { if ((f[x + 1][j]<f[x + 1][k] || k>r) && j <= mid) f[x][i] = f[x + 1][j++]; else f[x][i] = f[x + 1][k++]; } } int find(int x, int l, int r, int ll, int rr, int v) { if (ll <= l&&r <= rr) { return lower_bound(f[x] + l, f[x] + r + 1, v) - (f[x] + l); } else { int mid = l + r >> 1, ans = 0; if (ll <= mid) ans += find(x + 1, l, mid, ll, rr, v); if (rr > mid) ans += find(x + 1, mid + 1, r, ll, rr, v); return ans; } } }st; int main() { while (scanf("%d%d", &n, &m) != EOF) { st.insert(0, 1, n); while (m--) { scanf("%d%d%d", &l, &r, &k); int q = 1, h = n; while (q < h) { int mid = q + h >> 1; int u = st.find(0, 1, n, l, r, st.f[0][mid]); if (u > k - 1) h = mid - 1; if (u == k - 1) q = mid; if (u < k - 1) q = mid + 1; if (st.find(0, 1, n, l, r, st.f[0][h]) == k - 1) q = h; else h--; } printf("%d\n", st.f[0][q]); } } return 0; }第二发，可持久化线段树（静态主席树）（非离散化）（线段树动态建点）
#include<cstdio> #include<cstring> #include<cmath> #include<queue> #include<vector> #include<iostream> #include<algorithm> #include<bitset> #include<functional> using namespace std; typedef long long LL; const int maxn = 4e6 + 10; const int mod = 2e9; int n, m, l, r, k; struct tree { int f[maxn], L[maxn], R[maxn], tot, first[maxn]; int newnode() { f[tot] = L[tot] = R[tot] = 0; return tot++; } void insert(int bef, int now, int l, int r, int v) { f[now] = f[bef] + 1; if (l == r) return; int mid = l + (r - l >> 1); L[now] = L[bef]; R[now] = R[bef]; if (v <= mid) { L[now] = newnode(); insert(L[bef], L[now], l, mid, v); } else { R[now] = newnode(); insert(R[bef], R[now], mid + 1, r, v); } } void add(int n) { tot = 0; for (int i = 0; i <= n; i++) first[i] = newnode(); for (int i = 1; i <= n; i++) { int x; scanf("%d", &x); insert(first[i - 1], first[i], 0, mod, x + mod / 2); } } int find(int bef, int now, int l, int r, int v) { if (l == r) return l - mod / 2; int mid = l + (r - l >> 1); if (v <= f[L[now]] - f[L[bef]]) return find(L[bef], L[now], l, mid, v); else return find(R[bef], R[now], mid + 1, r, v - f[L[now]] + f[L[bef]]); } }st; int main() { while (scanf("%d%d", &n, &m) != EOF) { st.add(n); while (m--) { scanf("%d%d%d", &l, &r, &k); printf("%d\n", st.find(st.first[l - 1], st.first[r], 0, mod, k)); } } return 0; }
第三发，可持久化线段树+离散化#include<cstdio> #include<cstring> #include<cmath> #include<queue> #include<vector> #include<iostream> #include<algorithm> #include<bitset> #include<functional> using namespace std; typedef long long LL; const int maxn = 1e5 + 10; int n, m, l, r, k, a[maxn], b[maxn], c[maxn]; struct tree { int f[maxn * 20], L[maxn * 20], R[maxn * 20], tot, first[maxn]; int newnode() { f[tot] = L[tot] = R[tot] = 0; return tot++; } void insert(int bef, int now, int l, int r, int v) { f[now] = f[bef] + 1; if (l == r) return; int mid = l + (r - l >> 1); L[now] = L[bef]; R[now] = R[bef]; if (v <= mid) { L[now] = newnode(); insert(L[bef], L[now], l, mid, v); } else { R[now] = newnode(); insert(R[bef], R[now], mid + 1, r, v); } } void add() { tot = 0; for (int i = 0; i <= n; i++) first[i] = newnode(); for (int i = 1; i <= n; i++) { insert(first[i - 1], first[i], 0, n - 1, c[i - 1]); } } int find(int bef, int now, int l, int r, int v) { if (l == r) return a[l]; int mid = l + (r - l >> 1); if (v <= f[L[now]] - f[L[bef]]) return find(L[bef], L[now], l, mid, v); else return find(R[bef], R[now], mid + 1, r, v - f[L[now]] + f[L[bef]]); } }st; int main() { while (scanf("%d%d", &n, &m) != EOF) { for (int i = 0; i < n; i++) scanf("%d", &a[i]), b[i] = a[i]; sort(a, a + n); for (int i = 0; i < n; i++) c[i] = lower_bound(a, a + n, b[i]) - a; st.add(); while (m--) { scanf("%d%d%d", &l, &r, &k); printf("%d\n", st.find(st.first[l - 1], st.first[r], 0, n - 1, k)); } } return 0; }莫队分块+treap 超时。。。#include<cstdio> #include<cstring> #include<cmath> #include<queue> #include<vector> #include<iostream> #include<algorithm> #include<bitset> #include<functional> using namespace std; typedef long long LL; const int maxn = 1e5 + 10; const int mod = 1e9 + 7; int n, m, a[maxn], ans[maxn]; struct query { int l, r, k, b, id; void read(int x){ scanf("%d%d%d", &l, &r, &k); id = x; } }c[maxn]; bool cmpl(const query&a, const query&b) { return a.b == b.b ? a.l < b.l : a.b < b.b; } bool cmpr(const query&a, const query&b) { return a.r < b.r; } struct Node { int v, p, cnt, inc; Node *l, *r; Node(int v = 0, int p = 0) :v(v), p(p){ l = r = NULL; cnt = inc = 1; } void clear(){ if (l) l->clear(); if (r)r->clear(); delete[]this; } }; struct Treap { #define node Node* #define now (*root) Node *root; void Count(node root) { root->cnt = root->inc; if (root->l) root->cnt += root->l->cnt; if (root->r) root->cnt += root->r->cnt; } void rotate_left(node *root) { node right = now->r; now->r = right->l; right->l = now; now = right; Count(now->l); Count(now); } void rotate_right(node *root) { node left = now->l; now->l = left->r; left->r = now; now = left; Count(now->r); Count(now); } void Insert(node *root, int v, int p) { if (!now) { now = new Node(v, p); return; } if (now->v == v) now->inc++; else if (v < now->v) { Insert(&(now->l), v, p); if (now->l->p < now->p) rotate_right(root); } else { Insert(&(now->r), v, p); if (now->r->p < now->p) rotate_left(root); } Count(now); } void add(int x){ Insert(&root, x, rand() % mod); } void Delete(node *root, int v) { if (!now) return; if (v < now->v) Delete(&(now->l), v); if (v > now->v) Delete(&(now->r), v); if (v == now->v) { if (now->inc > 1) now->inc--; else if (!now->l || !now->r) { node temp = now; if (now->l) now = now->l; else now = now->r; delete[] temp; return; } else { if (now->l->p < now->r->p) { rotate_right(root); Delete(&(now->r), v); } else { rotate_left(root); Delete(&(now->l), v); } } } Count(now); } void dec(int x){ Delete(&root, x); } int find(int k) { for (node temp = root; temp;) { int left = temp->l ? temp->l->cnt : 0; if (k <= left) temp = temp->l; else if (k <= left + temp->inc) return temp->v; else { temp = temp->r; k -= left + temp->inc; } } } void clear(){ if (root) root->clear(); root = NULL; } }t; int main() { while (scanf("%d%d", &n, &m) != EOF) { t.clear(); for (int i = 1; i <= n; i++) scanf("%d", &a[i]); for (int i = 0; i < m; i++) c[i].read(i); sort(c, c + m, cmpr); for (int i = 0; i < m; i++) c[i].b = i / sqrt(m); sort(c, c + m, cmpl); for (int i = 0, l = 1, r = 0; i < m; i++) { while (l > c[i].l) t.add(a[--l]); while (r < c[i].r) t.add(a[++r]); while (l < c[i].l) t.dec(a[l++]); while (r > c[i].r) t.dec(a[r--]); ans[c[i].id] = t.find(c[i].k); } for (int i = 0; i < m; i++) printf("%d\n", ans[i]); } return 0; }第五发，二分+线段树套treap，超时。。#include<cstdio> #include<cstring> #include<cmath> #include<queue> #include<vector> #include<iostream> #include<algorithm> #include<bitset> #include<functional> using namespace std; typedef long long LL; const int maxn = 1e5 + 10; const int mod = 1e9 + 7; int n, m, l, r, k, a[maxn]; struct Node { int v, p, cnt, inc; Node *l, *r; Node(int v = 0, int p = 0) :v(v), p(p){ l = r = NULL; cnt = inc = 1; } void clear(){ if (l) l->clear(); if (r)r->clear(); delete[]this; } }; struct Treap { #define node Node* #define now (*root) Node *root; void Count(node root) { root->cnt = root->inc; if (root->l) root->cnt += root->l->cnt; if (root->r) root->cnt += root->r->cnt; } void rotate_left(node *root) { node right = now->r; now->r = right->l; right->l = now; now = right; Count(now->l); Count(now); } void rotate_right(node *root) { node left = now->l; now->l = left->r; left->r = now; now = left; Count(now->r); Count(now); } void Insert(node *root, int v, int p) { if (!now) { now = new Node(v, p); return; } if (now->v == v) now->inc++; else if (v < now->v) { Insert(&(now->l), v, p); if (now->l->p < now->p) rotate_right(root); } else { Insert(&(now->r), v, p); if (now->r->p < now->p) rotate_left(root); } Count(now); } void add(int x){ Insert(&root, x, rand() % mod); } void Delete(node *root, int v) { if (!now) return; if (v < now->v) Delete(&(now->l), v); if (v > now->v) Delete(&(now->r), v); if (v == now->v) { if (now->inc > 1) now->inc--; else if (!now->l || !now->r) { node temp = now; if (now->l) now = now->l; else now = now->r; delete[] temp; return; } else { if (now->l->p < now->r->p) { rotate_right(root); Delete(&(now->r), v); } else { rotate_left(root); Delete(&(now->l), v); } } } Count(now); } void dec(int x){ Delete(&root, x); } int find(int x) { int ans = 0; for (node temp = root; temp;) { int left = temp->l ? temp->l->cnt : 0; if (x < temp->v) temp = temp->l; else { if (x == temp->v) return ans + left; ans += left + temp->inc; temp = temp->r; } } return ans; } void clear(){ if (root) root->clear(); root = NULL; } }; struct Tree { Treap f[maxn * 4]; void insert(int x, int l, int r) { f[x].clear(); if (l == r) { scanf("%d", &a[l]); f[x].add(a[l]); return; } int mid = l + r >> 1; insert(x << 1, l, mid); insert(x << 1 | 1, mid + 1, r); for (int i = l; i <= r; i++) f[x].add(a[i]); } int find(int x, int l, int r, int ll, int rr, int v) { if (ll <= l&&r <= rr) return f[x].find(v); int mid = l + r >> 1, ans = 0; if (ll <= mid) ans += find(x << 1, l, mid, ll, rr, v); if (rr > mid) ans += find(x << 1 | 1, mid + 1, r, ll, rr, v); return ans; } }st; int main() { while (scanf("%d%d", &n, &m) != EOF) { st.insert(1, 1, n); sort(a + 1, a + n + 1); while (m--) { scanf("%d%d%d", &l, &r, &k); int q = 1, h = n; while (q < h) { int mid = q + h >> 1; int u = st.find(1, 1, n, l, r, a[mid]); if (u > k - 1) h = mid - 1; if (u == k - 1) q = mid; if (u < k - 1) q = mid + 1; if (st.find(1, 1, n, l, r, a[h]) == k - 1) q = h; else h--; } printf("%d\n", a[q]); } } return 0; }果然指针是在这种地方的应用实在是费时，现在改用数组版的treap就过了
#include<cstdio> #include<cstring> #include<cmath> #include<queue> #include<vector> #include<iostream> #include<algorithm> #include<bitset> #include<functional> using namespace std; typedef long long LL; const int maxn = 1e5 + 10; const int mod = 1e9 + 7; int n, m, l, r, k, a[maxn]; struct Treaps { const static int maxn = 3e6 + 3e5; int L[maxn], R[maxn], v[maxn], p[maxn], A[maxn], C[maxn], tot; void clear(){ A[0] = L[0] = R[0] = C[0] = 0; tot = 1; } int Node(int V, int P){ L[tot] = R[tot] = 0; v[tot] = V; p[tot] = P; A[tot] = C[tot] = 1; return tot++; } void Count(int x){ C[x] = A[x] + C[L[x]] + C[R[x]]; } void rotate_right(int &x) { int y = L[x]; L[x] = R[y]; R[y] = x; C[y] = C[x]; Count(x); x = y; } void rotate_left(int &x) { int y = R[x]; R[x] = L[y]; L[y] = x; C[y] = C[x]; Count(x); x = y; } void Insert(int& x, int V, int P) { if (!x) { x = Node(V, P); return; } if (v[x] == V) ++A[x]; else if (V < v[x]) { Insert(L[x], V, P); if (p[x]>p[L[x]]) rotate_right(x); } else { Insert(R[x], V, P); if (p[x] > p[R[x]]) rotate_left(x); } Count(x); } void add(int &x, int V){ Insert(x, V, rand()); }//外部直接调用，x是树根，v是值 void Delete(int &x, int V) { if (!x) return; if (V < v[x]) Delete(L[x], V); else if (V > v[x]) Delete(R[x], V); else if (A[x] > 1) --A[x]; else if (!L[x] || !R[x]) x = L[x] + R[x]; else if (p[L[x]] < p[R[x]]){ rotate_right(x); Delete(R[x], V); } else { rotate_left(x); Delete(L[x], V); } Count(x); } void dec(int &x, int V) { Delete(x, V); }//外部直接调用，x是树根，v是值 int find(int x, int V)//返回树x中小于等于V的数字个数 { int ans = 0; for (int i = x; i; i = V < v[i] ? L[i] : R[i]) { if (v[i] <= V) ans += C[L[i]] + A[i]; } return ans; } }; struct Tree { Treaps solve; int f[maxn * 4]; void insert(int x, int l, int r) { f[x] = 0; if (l == r) { scanf("%d", &a[l]); solve.add(f[x], a[l]); return; } int mid = l + r >> 1; insert(x << 1, l, mid); insert(x << 1 | 1, mid + 1, r); for (int i = l; i <= r; i++) solve.add(f[x], a[i]); } int find(int x, int l, int r, int ll, int rr, int v) { if (ll <= l&&r <= rr) return solve.find(f[x], v); int mid = l + r >> 1, ans = 0; if (ll <= mid) ans += find(x << 1, l, mid, ll, rr, v); if (rr > mid) ans += find(x << 1 | 1, mid + 1, r, ll, rr, v); return ans; } }st; int main() { while (scanf("%d%d", &n, &m) != EOF) { st.solve.clear(); st.insert(1, 1, n); sort(a + 1, a + n + 1); while (m--) { scanf("%d%d%d", &l, &r, &k); int q = 1, h = n; while (q <= h) { int mid = q + h >> 1; int u = st.find(1, 1, n, l, r, a[mid]); if (u <= k - 1) q = mid + 1; else h = mid - 1; } printf("%d\n", a[q]); } } return 0; }