【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 100010
struct 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 200010
int 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++];
		else
			t[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 3
2 2 2
1 1 1
1 1 1
1 1 1
2 2 2
*/



 

posted @ 2016-04-19 09:45  cot  阅读(104)  评论(0编辑  收藏  举报