前人之述备矣、、、
树套树即BIT套treap 和 CQD分治 + BIT的方法都有了
于是就做好了233
1 /************************************************************** 2 Problem: 3262 3 User: rausen 4 Language: C++ 5 Result: Accepted 6 Time:1356 ms 7 Memory:5888 kb 8 ****************************************************************/ 9 10 #include <cstdio> 11 #include <algorithm> 12 13 #define lowbit(x) x & -x 14 using namespace std; 15 const int N = 100005; 16 17 struct data { 18 int a, b, c, s, ans; 19 }a[N], p[N]; 20 inline bool sort1_cmp (const data &a, const data &b) { 21 return a.a == b.a ? (a.b == b.b ? a.c < b.c : a.b < b.b) : a.a < b.a; 22 } 23 inline bool operator < (const data &a, const data &b) { 24 return a.b == b.b ? a.c < b.c : a.b < b.b; 25 } 26 inline bool operator == (const data &a, const data &b) { 27 return a.a == b.a && a.b == b.b && a.c == b.c; 28 } 29 inline bool operator != (const data &a, const data &b) { 30 return !(a == b); 31 } 32 33 int tot, n, m, BIT[N << 1], ans[N]; 34 35 inline int read() { 36 int x = 0; 37 char ch = getchar(); 38 while (ch < '0' || '9' < ch) 39 ch = getchar(); 40 while ('0' <= ch && ch <= '9') { 41 x = x * 10 + ch - '0'; 42 ch = getchar(); 43 } 44 return x; 45 } 46 47 inline void update(int x, int del) { 48 while (x <= m) 49 BIT[x] += del, x += lowbit(x); 50 } 51 52 inline int query(int x) { 53 int res = 0; 54 while (x) 55 res += BIT[x], x -= lowbit(x); 56 return res; 57 } 58 59 void work(int l, int r) { 60 if (l == r) return; 61 int mid = l + r >> 1, i, j; 62 work(l, mid), work(mid + 1, r); 63 sort(p + l, p + mid + 1), sort(p + mid + 1, p + r + 1); 64 for (i = l, j = mid + 1; j <= r; ++j) { 65 for (; i <= mid && p[i].b <= p[j].b; ++i) 66 update(p[i].c, p[i].s); 67 p[j].ans += query(p[j].c); 68 } 69 for (j = l; j < i; ++j) 70 update(p[j].c, -p[j].s); 71 } 72 73 int main() { 74 int i, cnt; 75 tot = read(), m = read(); 76 for (i = 1; i <= tot; ++i) 77 a[i].a = read(), a[i].b = read(), a[i].c = read(); 78 sort(a + 1, a + tot + 1, sort1_cmp); 79 for (cnt = 1, i = 1; i <= tot; ++i, ++cnt) 80 if (a[i] != a[i + 1]) { 81 p[++n] = a[i]; 82 p[n].s = cnt; 83 cnt = 0; 84 } 85 work(1, n); 86 for (i = 1; i <= n; ++i) 87 ans[p[i].ans + p[i].s - 1] += p[i].s; 88 for (i = 0; i < tot; ++i) 89 printf("%d ", ans[i]); 90 return 0; 91 }