HDU 3450 Counting Sequences（线段树）

Counting Sequences

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/65536 K (Java/Others)

Total Submission(s): 2335    Accepted Submission(s): 820

Problem Description

For a set of sequences of integers｛a1，a2，a3，...an｝, we define a sequence｛ai1,ai2,ai3...aik｝in which 1<=i1<i2<i3<...<ik<=n, as the sub-sequence of ｛a1，a2，a3，...an｝. It is quite obvious that a sequence with the length n has 2^n sub-sequences. And for a sub-sequence｛ai1,ai2,ai3...aik｝，if
it matches the following qualities: k >= 2, and the neighboring 2 elements have the difference not larger than d, it will be defined as a Perfect Sub-sequence. Now given an integer sequence, calculate the number of its perfect sub-sequence.

 

Input

Multiple test cases The first line will contain 2 integers n, d（2<=n<=100000，1<=d=<=10000000） The second line n integers, representing the suquence

 

Output

The number of Perfect Sub-sequences mod 9901

 

Sample Input

4 2
1 3 7 5

 

Sample Output

4

思路：题目求满足要求的子串有多少个，那么我们应该固定序列的结尾点，dp[i],表示以i点为结尾的满足条件的序列有多少个，对于每一个点作为序列的最后一个点，都去找它前面所有满足条件的点k,然后加上dp[k]，类似于动态规划，那么如何高效的求满足条件的点，即这个点前面的所有点，哪些是满足在a[i]-d,和a[i]+d.我们可以用线段树

另外，需要离散化一下，这里时间只有1秒钟，如果用map离散会超时，所以我们可以用二分法离散
#include <iostream>
#include <string.h>
#include <algorithm>
#include <math.h>
#include <stdlib.h>
#include <stdio.h>
#include <map>

using namespace std;
const int maxn=1e5;
typedef long long int LL;
LL sum[maxn*8+5];
int n,d;
void pushup(int node)
{
	sum[node]=sum[node<<1]+sum[node<<1|1];
	sum[node]%=9901;
}
void update(int node,int l,int r,int val,LL num)
{
	if(l==r)
	{
		sum[node]+=num;
		sum[node]%=9901;
		return;
	}
	int mid=(l+r)>>1;
	if(val<=mid)
		update(node<<1,l,mid,val,num);
	else
		update(node<<1|1,mid+1,r,val,num);
	pushup(node);
}
LL query(int node,int l,int r,int L,int R)
{
	if(L<=l&&r<=R)
		return sum[node]%9901;
	int mid=(l+r)>>1;
	LL ret=0;
	if(L<=mid) ret+=query(node<<1,l,mid,L,R);
	if(R>mid) ret+=query(node<<1|1,mid+1,r,L,R);
	return ret%9901;
}
int a[maxn+5];
int c[maxn+5];
int b[maxn+5];
int cot;
int ans;

int fun1(int k,int tag)
{
	int l=1,r=cot;
	while(l<=r)
	{
		int mid=(l+r)/2;
		if(k<a[mid])
			r=mid-1;
		else
			l=mid+1;
	}
	if(tag) return l;
	return r;
}
int fun2(int k)
{
	int l=1,r=cot;
	while(l<=r)
	{
		int mid=(l+r)/2;
		if(k==a[mid])
			return mid;
		else if(k<a[mid])
			r=mid-1;
		else
			l=mid+1;
	}
	return l;
}
int main()
{
	while(scanf("%d%d",&n,&d)!=EOF)
    {

		memset(sum,0,sizeof(sum));

		for( int i=1;i<=n;i++)
		{
			scanf("%d",&b[i]);
			c[i]=b[i];
		}
		sort(b+1,b+n+1);
		cot=1;a[1]=b[1];
		for(int i=2;i<=n;i++)
		{
			if(b[i]!=b[i-1])
				a[++cot]=b[i];
		}
		for(int i=1;i<=n;i++)
		{
			int r=fun1(c[i],0)-1;
			int l=fun1(c[i],1)+1;
			int k=fun2(c[i]);

			LL num=query(1,1,n,l,r);
			ans+=num;
			ans%=9901;
			update(1,1,n,k,(num+1)%9901);
		}
		printf("%d\n",ans);
    }

	return 0;
}