#HDU 4325 Flowers 【离散化、区间加减法】

最新推荐文章于 2024-01-02 13:59:08 发布

ReskiP

最新推荐文章于 2024-01-02 13:59:08 发布

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分类专栏：解题报告文章标签： ACM 算法线段树区间处理快速查询

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本文介绍了一种利用线段树解决特定时间段内花朵重叠数量的问题。通过线段树实现区域累加，有效地处理了大量时间区间查询的需求，并通过离散化简化了处理过程。

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题目：

Flowers

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2866 Accepted Submission(s): 1404

Problem Description

As is known to all, the blooming time and duration varies between different kinds of flowers. Now there is a garden planted full of flowers. The gardener wants to know how many flowers will bloom in the garden in a specific time. But there are too many flowers in the garden, so he wants you to help him.

Input

The first line contains a single integer t (1 <= t <= 10), the number of test cases.
For each case, the first line contains two integer N and M, where N (1 <= N <= 10^5) is the number of flowers, and M (1 <= M <= 10^5) is the query times.
In the next N lines, each line contains two integer S _i and T _i (1 <= S _i <= T _i <= 10^9), means i-th flower will be blooming at time [S _i, T _i].
In the next M lines, each line contains an integer T _i, means the time of i-th query.

Output

For each case, output the case number as shown and then print M lines. Each line contains an integer, meaning the number of blooming flowers.
Sample outputs are available for more details.

Sample Input

Sample Output

   
   
    
    Case #1:
0
Case #2:
1
2
1

Author

BJTU

Source

2012 Multi-University Training Contest 3

Recommend

zhoujiaqi2010

思路：先给出N个时间区间，问在时间点 Xi 上共有多少层时间区间重叠。

线段树区域累加，由于区域过大，在处理前先进行离散化。

#define _CRT_SECURE_NO_WARNINGS

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <vector>
#include <algorithm>

#define MAXN 1000000
int lazy[2 * MAXN];
int t[2 * MAXN], a[2 * MAXN];

//int be[MAXN], ed[MAXN];
//int hips[2 * MAXN];
//double endhips[4 * MAXN];
//int maxn;

using namespace std;

void BuildTree(int l, int r, int x){
	if (l == r)
	{
		t[x] = a[l];
		return;
	}
	int m = (l + r) >> 1;
	BuildTree(l, m, x << 1);
	BuildTree(m + 1, r, x << 1 | 1);
	t[x] = t[x << 1] + t[x << 1 | 1];
	return;
}

void PushDown(int l, int r, int x){
	int m = (l + r) >> 1;
	if (lazy[x]){
		t[x << 1] += lazy[x] * (m - l + 1);
		t[x << 1 | 1] += lazy[x] * (r - m);
		lazy[x << 1 | 1] += lazy[x];
		lazy[x << 1] += lazy[x];
		lazy[x] = 0;
	}
	return;
}

void Modify(int pos, int val, int l, int r, int x){
	if (pos == l && r == l)
	{
		t[x] = val;
		return;
	}
	int m = (l + r) >> 1;
	if (pos <= m)
	{
		Modify(pos, val, l, m, x << 1);
	}
	else Modify(pos, val, m + 1, r, x << 1 | 1);
	t[x] = t[x << 1] + t[x << 1 | 1];
	return;
}

void SegModify(int L, int R, int val, int l, int r, int x){
	if (l == L&&r == R){
		t[x] += (R - L + 1)*val;
		lazy[x] += val;
		return;
	}
	PushDown(l, r, x);
	int m = (l + r) >> 1;
	if (R <= m)SegModify(L, R, val, l, m, x << 1);
	else if (L>m)SegModify(L, R, val, m + 1, r, x << 1 | 1);
	else {
		SegModify(L, m, val, l, m, x << 1);
		SegModify(m + 1, R, val, m + 1, r, x << 1 | 1);
	}
	t[x] = t[x << 1] + t[x << 1 | 1];
	return;
}

int Query(int L, int R, int l, int r, int x){
	if (L == l && R == r)
		return t[x];
	PushDown(l, r, x);
	int m = (l + r) >> 1;
	if (R <= m)return Query(L, R, l, m, x << 1);
	else if (L>m)return Query(L, R, m + 1, r, x << 1 | 1);
	else return Query(L, m, l, m, x << 1) + Query(m + 1, R, m + 1, r, x << 1 | 1);
}

int main()
{
	int q;
	cin >> q;
	for (size_t i = 0; i < q; i++)
	{
		memset(t, 0, sizeof(t));
		//memset(hips, 0, sizeof(hips));
		//memset(endhips, 0, sizeof(endhips));
		memset(lazy, 0, sizeof(lazy));

		cout << "Case #" << i + 1 << ":\n";
		int nn, mm;
		cin >> nn >> mm;
		//maxn = 0;

		int n = 1;
		while (n<MAXN / 2)
		{
			n *= 2;
		}

		for (size_t i = 0; i < nn; i++)
		{
			int be, ed;
			cin >> be >> ed;
			SegModify(be, ed, 1, 1, n, 1);
		}
		for (size_t i = 0; i < mm; i++)
		{
			int quer;
			cin >> quer;
			cout << Query(quer, quer, 1, n, 1) << "\n";
		}


		/*for (size_t i = 0; i < nn; i++)
		{
		cin >> be[i] >> ed[i];
		hips[maxn] = be[i];
		maxn++;
		hips[maxn] = ed[i];
		maxn++;
		}
		int temp = 0;
		sort(hips, hips + maxn);
		for (size_t i = 0; i < maxn; i++)
		{
		if (endhips[temp] != hips[i])
		{
		temp += 2;
		endhips[temp] = hips[i];
		}
		}
		temp++;

		for (size_t i = 1; i < temp; i++)
		{
		if (!endhips[i])
		{
		endhips[i] = (endhips[i - 1] + endhips[i + 1]) / 2;
		}
		}

		int n = 1;
		while (n<10)
		{
		n *= 2;
		}

		for (size_t i = 0; i < nn; i++)
		{
		SegModify(*upper_bound(endhips, endhips + temp, be[i]), *lower_bound(endhips, endhips + temp, ed[i]), 1, 1, n, 1);
		}

		for (size_t i = 0; i < mm; i++)
		{
		int query;
		cin >> query;
		query = lower_bound(endhips, endhips + temp, query) - endhips;
		cout << Query(query,query,1,n,1)<<"\n";
		}*/
	}
	return 0;
}