【HDOJ 5654】 xiaoxin and his watermelon candy（离线+树状数组）

pid=5654">【HDOJ 5654】 xiaoxin and his watermelon candy（离线+树状数组）

xiaoxin and his watermelon candy

Time Limit: 4000/4000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 233    Accepted Submission(s): 61

Problem Description

During his six grade summer vacation, xiaoxin got lots of watermelon candies from his leader when he did his internship at Tencent. Each watermelon candy has it's sweetness which denoted by an integer number.



xiaoxin is very smart since he was a child. He arrange these candies in a line and at each time before eating candies, he selects three continuous watermelon candies from a specific range [L, R] to eat and the chosen triplet must satisfies:



if he chooses a triplet (ai,aj,ak)
then:

1. j=i+1,k=j+1

2.  ai≤aj≤ak



Your task is to calculate how many different ways xiaoxin can choose a triplet in range [L, R]?

two triplets (a0,a1,a2)
and (b0,b1,b2)
are thought as different if and only if:

a0≠b0
or a1≠b1
or a2≠b2

 

Input

This problem has multi test cases. First line contains a single integer
T(T≤10)
which represents the number of test cases.



For each test case, the first line contains a single integer
n(1≤n≤200,000)which
represents number of watermelon candies and the following line contains
n
integer numbers which are given in the order same with xiaoxin arranged them from left to right.

The third line is an integer Q(1≤200,000)
which is the number of queries. In the following Q
lines, each line contains two space seperated integers
l,r(1≤l≤r≤n)
which represents the range [l, r].

 

Output

For each query, print an integer which represents the number of ways xiaoxin can choose a triplet.

 

Sample Input

1
5
1 2 3 4 5
3
1 3
1 4
1 5

 

Sample Output

1
2
3

 

Source

BestCoder Round #77 (div.1)

 

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题目大意:有n个糖果。从左到右列出每一个糖果的甜度

之后有Q次查询，每次查询[L,R]中三元组的个数

这个三元组要求满足为连续的三个值，然后这三个值为非递减。

问[L,R]中不反复的三元组的个数。

反复表示三元组中三个数均对影响等。

假设反复则仅仅记一次

正解为主席树………………额………………恩…………

搜到这个离线+树状数组的写法，给大家分享下

思路非常巧妙。先离线存储全部查询。

然后对查询进行排序，以右边界从小到大排序。

然后从1到n開始枚举位置pos

每到一个位置，看一下从当前往后连续的三个满不满足题目中三元组的要求，假设满足。在树状数组中1的位置+1 pos+1处-1

如今让我们先不考虑反复。

经过如上处理，对于全部右边界等于pos+2的区间，树状数组中0~L的值即为三元组个数！

由于假设满足R等于pos+2 事实上就是找全部出现过的l >= L的三元组，在遍历的过程中。每一个满足要求的三元组pos+1处-1了，事实上就是求0~L的区间和了

想想看～～

至于R 等于pos+2 因为对查询依照R排序了，所以在遍历的过程中对于每一个pos都把查询遍历到右区间pos+2就好啦

这里没有去重，关于去重。我是琢磨了好久……做法就是哈希，哈希三元组。

假设没出现过。跟上面一样，在线段树1处+1

假设出现过，须要在上次出现的位置+1处 累加上一个1

这样就能够避免反复统计了

做这道题真长记性……因为要哈希，排序必须对全部元素都进行比較。

否则会出现重叠！

这也是跟其内部实现相关。

代码例如以下:

#include <iostream>
#include <cmath>
#include <vector>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <queue>
#include <stack>
#include <list>
#include <algorithm>
#include <map>
#include <set>
#define LL long long
#define Pr pair<int,int>
#define fread() freopen("in.in","r",stdin)
#define fwrite() freopen("out.out","w",stdout)

using namespace std;
const int INF = 0x3f3f3f3f;
const int msz = 10000;
const int mod = 1e9+7;
const double eps = 1e-8;

int bit[233333];
int n;

int Lowbit(int x)
{
	return x&(-x);
}

int sum(int x)
{
	int ans = 0;
	while(x)
	{
		ans += bit[x];
		x -= Lowbit(x);
	}
	return ans;
}

void add(int x,int d)
{
	while(x <= n)
	{
		bit[x] += d;
		x += Lowbit(x);
	}
}

struct Point
{
	int l,r,id;
	bool operator <(const struct Point a)const
	{
		return r == a.r?

l == a.l? id < a.id: l < a.l: r < a.r;
	}
	Point(int _l = 0,int _r = 0,int _id = 0):l(_l),r(_r),id(_id){};
};

Point pt[233333];
int num[233333];
int ans[233333];
int p[233333];

int main()
{
	int t,m;

	scanf("%d",&t);
	while(t--)
	{
		scanf("%d",&n);
		for(int i = 1; i <= n; ++i)
			scanf("%d",&num[i]);
		scanf("%d",&m);
		for(int i = 0; i < m; ++i)
		{
			scanf("%d%d",&pt[i].l,&pt[i].r);
			pt[i].id = i;
		}

		memset(ans,0,sizeof(ans));
		memset(bit,0,sizeof(bit));
		sort(pt,pt+m);

		map <Point,int> mp;
		int j = 0;
		int tp = 1;
		for(int i = 1; i <= n-2; ++i)
		{
			if(num[i] <= num[i+1] && num[i+1] <= num[i+2])
			{
				if(!mp[Point(num[i],num[i+1],num[i+2])])
				{
					mp[Point(num[i],num[i+1],num[i+2])] = tp;
					p[tp++] = 0;
				}
				int x = mp[Point(num[i],num[i+1],num[i+2])];
				add(p[x]+1,1);
				add(i+1,-1);
				p[x] = i;
			}
			for(; j < m && pt[j].r <= i+2; ++j)
			{
				if(pt[j].l+2 > pt[j].r) continue;
				ans[pt[j].id] = sum(pt[j].l);
			}
		}

		for(int i = 0; i < m; ++i)
			printf("%d\n",ans[i]);
	}

	return 0;
}