HDU 2227 Find the nondecreasing subsequences （线段树）

 

Find the nondecreasing subsequences

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1235    Accepted Submission(s): 431

Problem Description

How many nondecreasing subsequences can you find in the sequence S = {s1, s2, s3, ...., sn} ? For example, we assume that S = {1, 2, 3}, and you can find seven nondecreasing subsequences, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}.

 

 

Input

The input consists of multiple test cases. Each case begins with a line containing a positive integer n that is the length of the sequence S, the next line contains n integers {s1, s2, s3, ...., sn}, 1 <= n <= 100000, 0 <= si <= 2^31.

 

 

Output

For each test case, output one line containing the number of nondecreasing subsequences you can find from the sequence S, the answer should % 1000000007.

 

 

Sample Input

3 1 2 3

 

 

Sample Output

7

 

 

Author

8600

 

 

Recommend

lcy

 

 

 

解题思路： 每进来一个数，方法数是原来所有小于等于它的数的方法数之和+1，然后再把这个数添加进为原来的数，下面依次循环。

100000个数，但是数值比较大，所以离散化一下，线段树来维护，每次lg的操作

因此总效率 O（n^lgn）

 

 

#include <iostream>
#include <cstdio>
#include <map>
#include <algorithm>
using namespace std;

const int maxn=100010;
const int yu= 1000000007;
int n;
unsigned int data[maxn];

map <unsigned int,int> mp;

struct node{
	int l,r;
	unsigned c;	
}a[maxn*4];

void build(int l,int r,int k){
	a[k].l=l;a[k].r=r;a[k].c=0;
	if(l<r){
		int mid=(l+r)/2;
		build(l,mid,2*k);
		build(mid+1,r,2*k+1);
	}
}

void insert(int l,int r,int c,int k){
	a[k].c=(a[k].c+c)%yu;
	if(l<=a[k].l && a[k].r<=r) return;
	else{
		int mid=(a[k].l+a[k].r)/2;
		if(l>=mid+1) insert(l,r,c,2*k+1);
		else if(r<=mid) insert(l,r,c,2*k);
		else{
			insert(l,mid,c,2*k);
			insert(mid+1,r,c,2*k+1);
		}
	}
}

unsigned query(int l,int r,int k){
	if(l<=a[k].l && a[k].r<=r){
		return a[k].c;
	}else{
		int mid=(a[k].l+a[k].r)/2;
		if(l>=mid+1) return query(l,r,2*k+1);
		else if(r<=mid) return query(l,r,2*k);
		else{
			 return (query(l,mid,2*k)+query(mid+1,r,2*k+1))%yu;
		}
	}
}

void initial(){
	mp.clear();
	build(1,n,1);
}

void input(){
	int cnt=1;
	map <unsigned int,int>::iterator it;
	for(int i=1;i<=n;i++){
		scanf("%d",&data[i]);
		mp[data[i]]=0;
	}
	for(it=mp.begin();it!=mp.end();it++){
		it->second=cnt++;
	}
}

void computing(){
	int ans=0,tmp,pos;
	for(int i=1;i<=n;i++){
		pos=mp[data[i]];
		tmp=query(1,pos,1)+1;
		ans=(ans+tmp)%yu;
		insert(pos,pos,tmp,1);
	}
	cout<<ans<<endl;
}

int main(){
	while(scanf("%d",&n)!=EOF){
		initial();
		input();
		computing();
	}
	return 0;
}