【bzoj3262】陌上花开

题目描述：

有n朵花，每朵花有三个属性：花形(s)、颜色(c)、气味(m)，又三个整数表示。现要对每朵花评级，一朵花的级别是它拥有的美丽能超过的花的数量。定义一朵花A比另一朵花B要美丽，当且仅当Sa>=Sb,Ca>=Cb,Ma>=Mb。显然，两朵花可能有同样的属性。需要统计出评出每个等级的花的数量。

输入：
第一行为N,K (1 <= N <= 100,000, 1 <= K <= 200,000 ), 分别表示花的数量和最大属性值。
以下N行，每行三个整数si, ci, mi (1 <= si, ci, mi <= K)，表示第i朵花的属性

输出：

包含N行，分别表示评级为0...N-1的每级花的数量。

样例输入：
10 3
3 3 3
2 3 3
2 3 1
3 1 1
3 1 2
1 3 1
1 1 2
1 2 2
1 3 2
1 2 1

样例输出：
3
1
3
0
1
0
1
0
0
1

题解：

三维偏序问题。排序一维，cdq分治一维，最后一维用树状数组就可以啦～最后两维比起树套树，分治+树状数组果然好写得多～

代码：

#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#ifdef WIN32#define LL "%I64d"#else#define LL "%lld"#endif#ifdef CT#define debug(...) printf(__VA_ARGS__)#define setfile() #else#define debug(...)#define filename ""#define setfile() freopen(filename".in", "r", stdin); freopen(filename".out", "w", stdout);#endif#define R register#define getc() (S == T && (T = (S = B) + fread(B, 1, 1 << 15, stdin), S == T) ? EOF : *S++)#define dmax(_a, _b) ((_a) > (_b) ? (_a) : (_b))#define dmin(_a, _b) ((_a) < (_b) ? (_a) : (_b))#define cmax(_a, _b) (_a < (_b) ? _a = (_b) : 0)#define cmin(_a, _b) (_a > (_b) ? _a = (_b) : 0)char B[1 << 15], *S = B, *T = B;inline int FastIn(){R char ch; R int cnt = 0; R bool minus = 0;while (ch = getc(), (ch < '0' || ch > '9') && ch != '-') ;ch == '-' ? minus = 1 : cnt = ch - '0';while (ch = getc(), ch >= '0' && ch <= '9') cnt = cnt * 10 + ch - '0';return minus ? -cnt : cnt;}#define maxn 100010struct Point{int x, y, z, ans, tim;inline bool operator < (const Point &that) const{if (x != that.x) return x < that.x;if (y != that.y) return y < that.y;return z < that.z;}inline bool operator == (const Point &that) const{return x == that.x && y == that.y && z == that.z;}}p[maxn], t[maxn];int ans[maxn], now, n, k;#define maxm 200010int bit[maxm], last[maxm];#define lowbit(_x) ((_x) & -(_x))inline void add(R int x, R int val){for (; x <= k; x += lowbit(x)){if (last[x] != now)bit[x] = 0;bit[x] += val;last[x] = now;}}inline int query(R int x){R int ret = 0;for (; x; x -= lowbit(x)){if (last[x] == now) ret += bit[x];}return ret;}void cdq(R int left, R int right){if (left == right) return;R int mid = left + right >> 1;cdq(left, mid);cdq(mid + 1, right);++now;for (R int i = left, j = mid + 1; j <= right; ++j){for (; i <= mid && p[i].y <= p[j].y; ++i)add(p[i].z, p[i].tim);p[j].ans += query(p[j].z);}R int i, j, kk = 0;for (i = left, j = mid + 1; i <= mid && j <= right; ){if (p[i].y <= p[j].y)t[kk++] = p[i++];elset[kk++] = p[j++];}for (; i <= mid; )t[kk++] = p[i++];for (; j <= right; )t[kk++] = p[j++];for (R int i = 0; i < kk; ++i)p[left + i] = t[i];}int main(){//setfile();n = FastIn(); k = FastIn();for (R int i = 1; i <= n; ++i){R int x = FastIn(), y = FastIn(), z = FastIn();p[i] = (Point) {x, y, z, 0, 1};}std::sort(p + 1, p + n + 1);R int cnt = 0;for (R int i = 1; i <= n; ++i){if (p[i] == p[i - 1])p[cnt].tim++;else p[++cnt] = p[i];}for(R int i = 1; i <= cnt; ++i) p[i].ans = p[i].tim - 1;cdq(1, cnt);for (R int i = 1; i <= cnt; ++i)ans[p[i].ans] += p[i].tim;for (R int i = 0; i < n; ++i) printf("%d\n",ans[i] );return 0;}/*5 32 2 21 1 11 1 11 1 12 2 2*/