HDU 2227 Find the nondecreasing subsequences(dp+线段树维护)

Find the nondecreasing subsequences

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2225    Accepted Submission(s): 862

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

 

题目意思很简单，一句话，对于问题规模为10w求出所有非递减子序列的数量。

dp的状态转移方程即为 dp[j] = (Σ(i from 1 ~ j-1 && h[i]<=h[j])dp[i])+1

然后得到一个 O(n²)的算法，这样肯定是过不去的。

然后队友教我用线段树去维护。

求dp[j]时，需要求之前所有下标高度小于h[j] 的dp[i]之和。

可以发现 dp[i] i ∈ {i|i<j and h[i] <= h[j]}

所以只要对h[N]进行一次排序，得到h[i]到新下标的映射， (他们管这个叫离散化)

再从前向后扫一边h[N]即可保证i集合的条件。

说了这么多不如直接看代码。

由于有线段树维护 复杂度削减至 O(n log(n))

#include <algorithm>
#include <iostream>
#include <cstring>
#include <queue>
#include <unordered_map>
#include <set>
#include <map>

using namespace std;
const unsigned int N = 100005;
typedef long long ll;
typedef unsigned int iu;
const int mod = 1000000007;
ll dp[N<<2u];
void build(iu l,iu r,iu rt){
    dp[rt] = 0;
    if(l==r){
        return;
    }
    iu m = (l+r)>>1u;
    build(l,m,rt<<1u);
    build(m+1,r,rt<<1u|1u);
}
void pushUp(iu rt){
    dp[rt] = (dp[rt<<1u]+dp[rt<<1u|1u])%mod;
}
ll query(iu L,iu R ,iu l,iu r,iu rt){
    if(L <= l and r <= R){
        return dp[rt];
    }
    iu m = (l+r)>>1u;
    ll ans = 0;
    if(m>=L)ans=query(L,R,l,m,rt<<1u);
    if(m<R)ans+=query(L,R,m+1,r,rt<<1u|1u);
    return ans%mod;
}
void upDate(iu loc,ll val,iu l,iu r,iu rt){
    if(l==r){
        dp[rt]+=val;
        dp[rt]%=mod;
        return;
    }
    iu m = (l+r)>>1u;
    if(m>=loc)upDate(loc,val,l,m,rt<<1u);
    if(m<loc)upDate(loc,val,m+1,r,rt<<1u|1u);
    pushUp(rt);
}
map<ll,iu>reflect;///unordered_map 用迭代器访问不一定有序;
ll h[N];
int main(){
    int n;
    while (cin>>n){
        reflect.clear();
        for(int i = 1 ; i <= n ; ++i){
            scanf("%lld",h+i);
            reflect[h[i]] = 0;
        }
        iu cnt = 0;
        for(auto &i :reflect){///细节 auto i : reflect 无法更改其中数值
            i.second = ++cnt;
        }
        iu k = reflect.size();
        build(1,k,1);
        long long ans ;
        for(int i = 1 ; i <= n ; ++i){
            iu pos = reflect[h[i]];
            ans = query(1,pos,1,k,1);
            upDate(pos,ans+1,1,k,1);
        }
        printf("%lld\n",query(1,k,1,k,1));
    }
}

数据结构真有意思;

另外HDU 3607和本题类似不妨一做

HDU 3607 题解传送门

https://www.cnblogs.com/xusirui/p/9396145.html