HDU 5618 Jam's problem again
Problem Description
Jam like to solve the problem which on the 3D-axis,given
N(1≤N≤100000) points
(x,y,z)(1≤x,y,z≤100000)
If two point such as (xi,yi,zi) and (xj,yj,zj) xi≥xj yi≥yj zi≥zj , the bigger one level add 1
Ask for the each level of the point.
If two point such as (xi,yi,zi) and (xj,yj,zj) xi≥xj yi≥yj zi≥zj , the bigger one level add 1
Ask for the each level of the point.
Input
The first line is
T(1≤T≤15) means
T Case
For each case
The first line is N means the number of Point and next there are N line, each line has (x,y,z)
For each case
The first line is N means the number of Point and next there are N line, each line has (x,y,z)
Output
Output with N line,each line has one number means the lever of point
Sample Input
1 4 10 4 7 10 6 6 8 2 5 7 3 10
Sample Output
1 1 0 0
三维偏序，研究了挺久的，bc的时候因为没考虑相等的问题最后wa了，真的是。。。
BC的时候写的树状数组套线段树，缺点是容易爆内存，好像数据比较水？
#include<cstdio> #include<cstring> #include<cmath> #include<vector> #include<algorithm> #include<queue> #include<iostream> using namespace std; typedef long long LL; const int maxn = 2e5 + 10; const int low(int x){ return x&-x; } int T, n; struct point { int x, y, z, id, cnt; void read(){ scanf("%d%d%d", &x, &y, &z); cnt = 0; } bool operator<(const point& a) const { if (x == a.x&&y == a.y) return z < a.z; if (x == a.x) return y < a.y; return x < a.x; } }a[maxn]; bool cmp(const point&a, const point&b) { return a.id < b.id; } struct Tree { int tot, first[maxn], f[maxn * 50], L[maxn * 50], R[maxn * 50], lit, lim; void clear(int x, int y) { lit = x; lim = y; for (int i = 1; i <= lit; i++) { first[i] = i; f[i] = R[i] = L[i] = 0; } tot = lit + 1; } int newnode() { f[tot] = R[tot] = L[tot] = 0; return tot++; } void ins(int x, int l, int r, int v) { f[x] += 1; if (l == r) return; int mid = l + r >> 1; if (v <= mid) { if (!L[x]) L[x] = newnode(); ins(L[x], l, mid, v); } else { if (!R[x]) R[x] = newnode(); ins(R[x], mid + 1, r, v); } } void insert(int x, int y) { for (int i = x; i <= lit; i += low(i)) ins(first[i], 1, lim, y); } int find(int x, int l, int r, int v) { if (r <= v) return f[x]; int mid = l + r >> 1, ans = 0; if (v <= mid) ans = find(L[x], l, mid, v); else ans = f[L[x]] + find(R[x], mid + 1, r, v); return ans; } int get(int x, int y) { int res = 0; for (int i = x; i; i -= low(i)) res += find(first[i], 1, lim, y); return res; } }tree; int main() { scanf("%d", &T); while (scanf("%d", &n) != EOF, T--) { for (int i = 0; i < n; i++) a[i].read(), a[i].id = i; sort(a, a + n); for (int i = n - 2; i >= 0; i--) if (a[i].x == a[i + 1].x&&a[i].y == a[i + 1].y&&a[i].z == a[i + 1].z) a[i].cnt = a[i + 1].cnt + 1; int ans1 = 0, ans2 = 0; for (int i = 0; i < n; i++) ans1 = max(ans1, a[i].y), ans2 = max(ans2, a[i].z); tree.clear(ans1, ans2); for (int i = 0; i < n; i++) { a[i].cnt += tree.get(a[i].y, a[i].z); tree.insert(a[i].y, a[i].z); } sort(a, a + n, cmp); for (int i = 0; i < n; i++) printf("%d\n", a[i].cnt); } return 0; }
后来搞定的cdq分治，一维排序，二维分治，三维树状数组（四维的话就两次分治）#include<iostream> #include<algorithm> #include<math.h> #include<cstdio> #include<cstring> #include<string> using namespace std; const int maxn = 1e5 + 5; const int low(int x){ return (x&-x); } int T, n, f[maxn]; struct point { int x, y, z, cnt, id; void read(){ scanf("%d%d%d", &x, &y, &z); cnt = 0; } bool operator<(const point&a)const { if (x == a.x&&y == a.y) return z < a.z; if (x == a.x) return y < a.y; return x < a.x; } bool operator==(const point&a)const { return x == a.x&&y == a.y&&z == a.z; } }a[maxn], b[maxn]; bool cmp(const point&a, const point&b) { return a.id < b.id; } void add(int x, int y) { for (int i = x; i < maxn; i += low(i)) f[i] += y; } int get(int x) { int ans = 0; for (int i = x; i; i -= low(i)) ans += f[i]; return ans; } void merge(int l, int r) { if (l == r) return; int mid = l + r >> 1; merge(l, mid); merge(mid + 1, r); for (int i = l, j = l, k = mid + 1; i <= r; i++) { if ((a[j].y <= a[k].y || r < k) && j <= mid) { b[i] = a[j++]; add(b[i].z, 1); } else { b[i] = a[k++]; b[i].cnt += get(b[i].z); } } for (int i = l; i <= mid; i++) add(a[i].z, -1), a[i] = b[i]; for (int i = mid + 1; i <= r; i++) a[i] = b[i]; } int main(){ scanf("%d", &T); while (~scanf("%d", &n), T--) { for (int i = 0; i < n; i++) a[i].read(), a[i].id = i; sort(a, a + n); for (int i = n - 2; i >= 0; i--) if (a[i] == a[i + 1]) a[i].cnt = a[i + 1].cnt + 1; merge(0, n - 1); sort(a, a + n, cmp); for (int i = 0; i < n; i++) printf("%d\n", a[i].cnt); } return 0; }
后来自己鼓捣了一个分块，也能过去，应该是正确的，时间效率4nlog(n)+2n(sqrt(n)+log^2(sqrt(n)))比cdq稍微慢一点#include<iostream> #include<algorithm> #include<math.h> #include<cstdio> #include<vector> #include<cstring> #include<string> using namespace std; const int maxn = 5e2 + 5; const int low(int x){ return x&-x; } int T, n, f[maxn][maxn], sqn; struct point { int x, y, z, cnt, id, by, bz; void read(){ scanf("%d%d%d", &x, &y, &z); cnt = 0; } bool operator==(const point&a) { return x == a.x&&y == a.y&&z == a.z; } bool operator<(const point&a) { return x <= a.x&&y <= a.y&&z <= a.z; } }a[maxn * 200]; vector<point> p[maxn], q[maxn]; bool cmpx(const point&a, const point&b) { if (a.x == b.x&&a.y == b.y) return a.z < b.z; if (a.x == b.x) return a.y < b.y; return a.x < b.x; } bool cmpy(const point&a, const point&b) { if (a.y == b.y) return a.z < b.z; return a.y < b.y; } bool cmpz(const point&a, const point&b) { if (a.z == b.z) return a.y < b.y; return a.z < b.z; } bool cmpid(const point&a, const point&b) { return a.id < b.id; } void add(int x, int y) { for (int i = x; i <= sqn; i += low(i)) for (int j = y; j <= sqn; j += low(j)) f[i][j]++; } int get(int x, int y) { int ans = 0; for (int i = x; i ; i -= low(i)) for (int j = y; j ; j -= low(j)) ans += f[i][j]; return ans; } int main(){ scanf("%d", &T); while (~scanf("%d", &n), T--) { memset(f, 0, sizeof(f)); for (int i = 0; i < n; i++) a[i].read(), a[i].id = i; sqn = sqrt(n) + 1; for (int i = 1; i <= sqn; i++) p[i].clear(), q[i].clear(); sort(a, a + n, cmpy); for (int i = 0; i < n; i++) a[i].by = i / sqn + 1; sort(a, a + n, cmpz); for (int i = 0; i < n; i++) a[i].bz = i / sqn + 1; sort(a, a + n, cmpx); for (int i = n - 2; i >= 0; i--) if (a[i] == a[i + 1]) a[i].cnt = a[i + 1].cnt + 1; for (int i = 0; i < n; i++) { int x = a[i].by, y = a[i].bz; a[i].cnt += get(x - 1, y - 1); for (int j = 0; j < p[x].size(); j++) if (p[x][j] < a[i]) a[i].cnt++; for (int j = 0; j < q[y].size(); j++) if (q[y][j] < a[i] && q[y][j].by != x) a[i].cnt++; p[x].push_back(a[i]); q[y].push_back(a[i]); add(x, y); } sort(a, a + n, cmpid); for (int i = 0; i < n; i++) printf("%d\n", a[i].cnt); } return 0; }完全理解了cdq分治的思想以后再写的不用树状数组，直接排序+两次cdq分治的方法
#include<cstdio> #include<iostream> #include<algorithm> using namespace std; const int maxn = 1e5 + 10; int T, n, ans[maxn]; struct point { int x, y, z, fx; int *ans; void read(){ scanf("%d%d%d", &x, &y, &z); } bool operator==(const point&a)const { return x == a.x&&y == a.y&&z == a.z; } }a[maxn], b[maxn], c[maxn]; bool cmpx(const point&a, const point&b) { return a.x == b.x ? a.y == b.y ? a.z < b.z : a.y < b.y : a.x < b.x; } void solve(int l, int r) { if (l == r) return; int mid = l + r >> 1; solve(l, mid); solve(mid + 1, r); for (int i = l, j = l, k = mid + 1, cnt = 0; i <= r; i++) { if ((b[j].z <= b[k].z || k>r) && j <= mid) c[i] = b[j++], cnt += c[i].fx ^ 1; else { c[i] = b[k++]; if (c[i].fx) *c[i].ans += cnt; } } for (int i = l; i <= r; i++) b[i] = c[i]; } void merge(int l, int r) { if (l == r) return; int mid = l + r >> 1; merge(l, mid); merge(mid + 1, r); for (int i = l, j = l, k = mid + 1; i <= r; i++) { if ((a[j].y <= a[k].y || k>r) && j <= mid) b[i] = a[j++], b[i].fx = 0; else b[i] = a[k++], b[i].fx = 1; } for (int i = l; i <= r; i++) a[i] = b[i]; solve(l, r); } int main(){ scanf("%d", &T); while (T--) { scanf("%d", &n); for (int i = 0; i < n; i++) a[i].read(), a[i].ans = &ans[i], ans[i] = 0; sort(a, a + n, cmpx); for (int i = n - 2; i >= 0; i--) if (a[i] == a[i + 1]) *a[i].ans = *a[i + 1].ans + 1; merge(0, n - 1); for (int i = 0; i < n; i++) printf("%d\n", ans[i]); } return 0; }
分块+bitset的方法，但是会超时，这个在解决更高维的时候会有特效
#include<cstdio> #include<iostream> #include<algorithm> #include<bitset> #include<map> using namespace std; const int maxn = 1e5 + 10; int T, n, sqn, p[maxn][3]; bitset<maxn> ans[3][320], res[3]; map<int, int> M[3]; struct point { int x, id; void read(int y){ id = y; scanf("%d", &x); } bool operator<(const point&a) { return x == a.x ? id < a.id : x < a.x; } }a[3][maxn]; int main(){ scanf("%d", &T); while (T--) { scanf("%d", &n); sqn = sqrt(n); for (int i = 0; i < n; i++) for (int j = 0; j < 3; j++) a[j][i].read(i); for (int i = 0; i < n; i++) for (int j = 0; j < 3; j++) p[i][j] = a[j][i].x; for (int i = 0; i < 3; i++) sort(a[i], a[i] + n), M[i].clear(); for (int i = 0; i < 3; i++) { int block = 0; for (int j = 0; j <= n / sqn; j++) ans[i][j].reset(); for (int j = 0; j < n; j++) { if (j / sqn == block + 1) ans[i][j / sqn]|= ans[i][block++]; ans[i][j / sqn][a[i][j].id] = 1; M[i][a[i][j].x] = j; } } for (int i = 0; i < n; i++) { for (int j = 0, block, now; j < 3; j++) { now = M[j][p[i][j]]; res[j].reset(); if (block = now / sqn) res[j] |= ans[j][block - 1]; for (int k = now; k / sqn == block&&k >= 0; k--) res[j][a[j][k].id] = 1; } printf("%d\n", (res[0] & res[1] & res[2]).count() - 1); } } return 0; }排序+线段树套treap，常数比较大
#include<cstdio> #include<cstring> #include<cmath> #include<vector> #include<algorithm> #include<queue> #include<iostream> using namespace std; typedef long long LL; const int maxn = 2e5 + 10; const int mod = 1e9 + 7; int T, n, ans[maxn]; struct point { int x, y, z, id; void read(int d){ scanf("%d%d%d", &x, &y, &z); id = d; } bool operator<(const point& a) const { return x == a.x ? y == a.y ? z < a.z : y < a.y : x < a.x; } bool operator==(const point& a) const { return x == a.x&&y == a.y&&z == a.z; } }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 { ans += left + temp->inc; if (x == temp->v) return ans; 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, int u, int v) { f[x].add(v); if (l == r) return; int mid = l + r >> 1; if (u <= mid) insert(x << 1, l, mid, u, v); else insert(x << 1 | 1, mid + 1, r, u, v); } int get(int x, int l, int r, int u, int v) { if (r <= u) return f[x].find(v); int mid = l + r >> 1; if (u <= mid) return get(x << 1, l, mid, u, v); else return get(x << 1, l, mid, u, v) + get(x << 1 | 1, mid + 1, r, u, v); } void release(int x, int l, int r) { f[x].clear(); if (l == r) return; int mid = l + r >> 1; release(x << 1, l, mid); release(x << 1 | 1, mid + 1, r); } }st; int main() { scanf("%d", &T); while (T--) { scanf("%d", &n); int maxy = 0; for (int i = 0; i < n; i++) { a[i].read(i); maxy = max(maxy, a[i].y); ans[i] = 0; } sort(a, a + n); for (int i = n - 2; i >= 0; i--) if (a[i] == a[i + 1]) ans[a[i].id] = ans[a[i + 1].id] + 1; st.release(1, 1, maxy); for (int i = 0; i < n; i++) { ans[a[i].id] += st.get(1, 1, maxy, a[i].y, a[i].z); st.insert(1, 1, maxy, a[i].y, a[i].z); } for (int i = 0; i < n; i++) printf("%d\n", ans[i]); } return 0; }
排序+树状数组套treap
#include<cstdio> #include<cstring> #include<cmath> #include<vector> #include<algorithm> #include<queue> #include<iostream> using namespace std; typedef long long LL; const int low(int x){ return x&-x; } const int maxn = 1e5 + 10; const int mod = 1e9 + 7; int T, n, ans[maxn], maxy; struct point { int x, y, z, id; void read(int d){ scanf("%d%d%d", &x, &y, &z); id = d; } bool operator<(const point& a) const { return x == a.x ? y == a.y ? z < a.z : y < a.y : x < a.x; } bool operator==(const point& a) const { return x == a.x&&y == a.y&&z == a.z; } }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 { ans += left + temp->inc; if (x == temp->v) return ans; temp = temp->r; } } return ans; } void clear(){ if (root) root->clear(); root = NULL; } }; Treap f[maxn]; void insert(int x, int y) { for (int i = x; i <= maxy; i += low(i)) f[i].add(y); } int get(int x, int y) { int ans = 0; for (int i = x; i; i -= low(i)) ans += f[i].find(y); return ans; } int main() { scanf("%d", &T); while (T--) { scanf("%d", &n); maxy = 0; for (int i = 0; i < n; i++) a[i].read(i), maxy = max(maxy, a[i].y), ans[i] = 0; sort(a, a + n); for (int i = 1; i <= maxy; i++) f[i].clear(); for (int i = n - 2; i >= 0; i--) if (a[i] == a[i + 1]) ans[a[i].id] = ans[a[i + 1].id] + 1; for (int i = 0; i < n; i++) { ans[a[i].id] += get(a[i].y, a[i].z); insert(a[i].y, a[i].z); } for (int i = 0; i < n; i++) printf("%d\n", ans[i]); } return 0; }
排序+树状数组套treap，用数组写的比指针要快点，不需要去释放空间
#include<cstdio> #include<cstring> #include<cmath> #include<vector> #include<algorithm> #include<queue> #include<iostream> using namespace std; typedef long long LL; const int low(int x){ return x&-x; } const int maxn = 1e5 + 10; int T, n, ans[maxn], maxy, f[maxn]; struct point { int x, y, z, id; void read(int d){ scanf("%d%d%d", &x, &y, &z); id = d; } bool operator<(const point& a) const { return x == a.x ? y == a.y ? z < a.z : y < a.y : x < a.x; } bool operator==(const point& a) const { return x == a.x&&y == a.y&&z == a.z; } }a[maxn]; struct Treaps { const static int maxn = 1e6 + 5e5; const static int mod = 1e9 + 7; int L[maxn], R[maxn], v[maxn], p[maxn], A[maxn], C[maxn], tot; void clear(){ 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++; } int NewTree(int x){ return Node(x, rand() % mod); } 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; x = y; Count(R[x]); } void rotate_left(int &x) { int y = R[x]; R[x] = L[y]; L[y] = x; x = y; Count(L[x]); } 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() % mod); } 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] ? 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); } int find(int x, int V) { int ans = 0; for (int i = x; i;) { if (V < v[i]) i = L[i]; else { ans += C[L[i]] + A[i]; if (V == v[i]) break; i = R[i]; } } return ans; } }t; void insert(int x, int y) { for (int i = x; i <= maxy; i += low(i)) if (f[i]) t.add(f[i], y); else f[i] = t.NewTree(y); } int get(int x, int y) { int ans = 0; for (int i = x; i; i -= low(i)) ans+=t.find(f[i], y); return ans; } int main() { scanf("%d", &T); while (T--) { scanf("%d", &n); maxy = 0; t.clear(); memset(f, 0, sizeof(f)); for (int i = 0; i < n; i++) a[i].read(i), maxy = max(maxy, a[i].y), ans[i] = 0; sort(a, a + n); for (int i = n - 2; i >= 0; i--) if (a[i] == a[i + 1]) ans[a[i].id] = ans[a[i + 1].id] + 1; for (int i = 0; i < n; i++) { ans[a[i].id] += get(a[i].y, a[i].z); insert(a[i].y, a[i].z); } for (int i = 0; i < n; i++) printf("%d\n", ans[i]); } return 0; }