离散化+线段树/二分查找/尺取法 HDOJ 4325 Flowers

题目传送门

题意：给出一些花开花落的时间，问某个时间花开的有几朵

分析：这题有好几种做法，正解应该是离散化坐标后用线段树成端更新和单点询问。还有排序后二分查找询问点之前总花开数和总花凋谢数，作差是当前花开的数量，放张图易理解：

还有一种做法用尺取法的思想，对暴力方法优化，对询问点排序后再扫描一遍，花开+1，花谢-1。详细看代码。

收获：一题收获很多：1. 降低复杂度可以用二分 2. 线段计数问题可以在端点标记1和-1 3. 离散化+线段树 终于会了:) (听说数据很水？)

代码1：离散化+线段树

/************************************************
* Author        :Running_Time
* Created Time  :2015-8-25 8:55:54
* File Name     :F.cpp
 ************************************************/

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
typedef pair<int, int> P;
const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;
struct ST	{
	int sum[N<<2], col[N<<2];

	void push_up(int rt)	{
        sum[rt] = sum[rt<<1] + sum[rt<<1|1];
	}
	void push_down(int rt, int len)	{
		if (col[rt])	{
			col[rt<<1] += col[rt];
			col[rt<<1|1] += col[rt];
			sum[rt<<1] += col[rt] * (len - (len >> 1));
			sum[rt<<1|1] += col[rt] * (len >> 1);
			col[rt] = 0;
		}
	}
	void build(int l, int r, int rt)	{
		col[rt] = 0;	sum[rt] = 0;
		if (l == r)	return ;
		int mid = (l + r) >> 1;
		build (lson);	build (rson);
//		push_up (rt);
	}
	void updata(int ql, int qr, int l, int r, int rt)	{
		if (ql <= l && r <= qr)	{
			sum[rt] += (r - l + 1);	col[rt] += 1;	return ;
		}
		push_down (rt, r - l + 1);
		int mid = (l + r) >> 1;
		if (ql <= mid)	updata (ql, qr, lson);
		if (qr > mid)	updata (ql, qr, rson);
//		push_up (rt);
	}
	int query(int ql, int qr, int l, int r, int rt)	{
		if (ql <= l && r <= qr)	return sum[rt];
		push_down (rt, r - l + 1);
		int mid = (l + r) >> 1;
		if (ql <= mid)	return query (ql, qr, lson);
		if (qr > mid)	return query (ql, qr, rson);
	}
}st;
int x1[N], x2[N], q[N];
int X[N*3];

int my_binary_search(int l, int r, int key)	{
	while (l <= r)	{
		int mid = (l + r) >> 1;
		if (X[mid] == key)	return mid;
		if (X[mid] < key)	l = mid + 1;
		else	r = mid - 1;
	}
	return -1;
}

int main(void)	{
	int T, cas = 0;	scanf ("%d", &T);
	while (T--)	{
		int n, m;
		scanf ("%d%d", &n, &m);
		int tot = 0;
		for (int i=1; i<=n; ++i)	{
			scanf ("%d%d", &x1[i], &x2[i]);
			X[tot++] = x1[i];	X[tot++] = x2[i];
		}
		for (int i=1; i<=m; ++i)	{
			scanf ("%d", &q[i]);
			X[tot++] = q[i];
		}
		sort (X, X + tot);
		int k = 1;
		for (int i=1; i<tot; ++i)	{
			if (X[i] != X[i-1])	X[k++] = X[i];
		}

		st.build (0, k, 1);
		for (int ql, qr, i=1; i<=n; ++i)	{
			ql = lower_bound (X, X+k, x1[i]) - X;
			qr = lower_bound (X, X+k, x2[i]) - X;
//			ql = my_binary_search (0, k-1, x1[i]);
//			qr = my_binary_search (0, k-1, x2[i]);
			st.updata (ql, qr, 0, k, 1);
		}

		printf ("Case #%d:
", ++cas);
		for (int p, i=1; i<=m; ++i)	{
			p = lower_bound (X, X+k, q[i]) - X;
//			p = my_binary_search (0, k-1, q[i]);
			printf ("%d
", st.query (p, p, 0, k, 1));
		}
	}

    return 0;
}

代码2：二分查找

/************************************************
* Author        :Running_Time
* Created Time  :2015-8-25 8:55:54
* File Name     :F.cpp
 ************************************************/

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
int x1[N], x2[N];
int n, m;

int main(void)    {
	int T, cas = 0;	scanf ("%d", &T);
	while (T--)	{
		scanf ("%d%d", &n, &m);
		for (int i=1; i<=n; ++i)	{
			scanf ("%d%d", &x1[i], &x2[i]);
		}
		sort (x1+1, x1+1+n);	sort (x2+1, x2+1+n);

		printf ("Case #%d:
", ++cas);
		for (int i=1; i<=m; ++i)	{
			int p;	scanf ("%d", &p);
			printf ("%d
", lower_bound (x1+1, x1+1+n, p) - (x1 + 1) - (lower_bound (x2+1, x2+1+n, p) - (x2 + 1)));
		}
	}

    return 0;
}

代码3：尺取法

#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;

typedef long long ll;
typedef pair<int, int> P;
const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;
P p[N], q[N];
int ans[N];
int n, m;

int main(void)    {
	int T, cas = 0;	scanf ("%d", &T);
	while (T--)	{
		scanf ("%d%d", &n, &m);
		int tot = 0;
		for (int a, b, i=1; i<=n; ++i)	{
			scanf ("%d%d", &a, &b);
			p[++tot] = P (a, 1);
			p[++tot] = P (b + 1, -1);
		}
		sort (p+1, p+1+tot);
		for (int i=1; i<=m; ++i)	{
			int x;	scanf ("%d", &x);
			q[i] = P (x, i);
		}
		sort (q+1, q+1+m);

		printf ("Case #%d:
", ++cas);
		for (int j=1, cnt=0, i=1; i<=m; ++i)	{
			while (j <= tot && p[j].first <= q[i].first)	{
				cnt += p[j].second;	++j;
			}
			ans[q[i].second] = cnt;
		}
		for (int i=1; i<=m; ++i)	printf ("%d
", ans[i]);
	}

    return 0;
}