LightOJ 1085 - All Possible Increasing Subsequences (离散化+树状数组+dp)

最新推荐文章于 2016-10-25 01:27:23 发布

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原文链接：http://www.cnblogs.com/wikioibai/p/4457209.html

本文解析了1085号问题——所有可能的递增子序列，这是一道经典的算法题。通过动态规划方法，文章详细介绍了如何计算给定序列中所有递增子序列的数量，并提供了完整的代码实现。

1085 - All Possible Increasing Subsequences

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Time Limit: 3 second(s)

Memory Limit: 64 MB

An increasing subsequence from a sequence A₁, A₂ ... A_n is defined by A_i1, A_i2 ... A_ik, where the following properties hold

i₁ < i₂ < i₃ < ... < i_k and
2. A_i1 < A_i2 < A_i3 < ... < A_ik

Now you are given a sequence, you have to find the number of all possible increasing subsequences.

Input

Input starts with an integer T (≤ 10), denoting the number of test cases.

Each case contains an integer n (1 ≤ n ≤ 10⁵) denoting the number of elements in the initial sequence. The next line will contain n integers separated by spaces, denoting the elements of the sequence. Each of these integers will be fit into a 32 bit signed integer.

Output

For each case of input, print the case number and the number of possible increasing subsequences modulo 1000000007.

Sample Input

Output for Sample Input

1 1 2

1 2 1000 1000 1001

1 10 11

Case 1: 5

Case 2: 23

Case 3: 7

Notes

For the first case, the increasing subsequences are (1), (1, 2), (1), (1, 2), 2.
Dataset is huge, use faster I/O methods.

题目大意：

　　就是说，给你一个长度为n的序列，然后，让你求出这个序列中所有的上升子序列的个数

解题思路：

　　直接把tree[i]当做dp[i]来用，表示的是以a[i]结尾的上升序列的最大个数

代码：

 1 # include<iostream>
 2 # include<cstdio>
 3 # include<algorithm>
 4 # include<cstring>
 5 
 6 using namespace std;
 7 
 8 # define MAX 500000+4
 9 # define MOD 1000000007
10 
11 typedef long long LL;
12 
13 LL a[MAX];
14 LL data[MAX];
15 int tree[MAX];
16 int n,cc;
17 
18 void discrete()
19 {
20     memset(data,0,sizeof(data));
21     for ( int i = 0;i < n;i++ )
22     {
23         data[i] = a[i];
24     }
25     sort(data,data+n);
26     cc = unique(data,data+n)-data;
27     for ( int i = 0;i < n;i++ )
28     {
29         a[i] = 1+lower_bound(data,data+cc,a[i])-data;
30     }
31 }
32 
33 
34 void update ( int pos,int val )
35 {
36     while ( pos <= n )
37     {
38         tree[pos]=(tree[pos]+val)%MOD;
39         pos += pos&(-pos);
40     }
41 }
42 
43 LL read ( int pos )
44 {
45     LL sum = 0;
46     while ( pos>0 )
47     {
48         sum=(sum+tree[pos])%MOD;
49         pos-=pos&(-pos);
50     }
51     return sum%MOD;
52 }
53 
54 
55 int main(void)
56 {
57     int icase = 1;
58     int t;cin>>t;
59     while ( t-- )
60     {
61         while ( cin>>n )
62         {
63         if ( n==0 )
64             break;
65         LL ans = 0;
66         for ( int i = 0;i < n;i++ )
67         {
68             cin>>a[i];
69         }
70         discrete();
71         memset(tree,0,sizeof(tree));
72         for ( int i = 0;i < n;i++ )
73         {
74             update(a[i],read(a[i]-1)+1);
75         }
76         printf("Case %d: ",icase++);
77         cout<<read(cc)%MOD<<endl;
78         }
79     }
80 
81     return 0;
82 }

转载于:https://www.cnblogs.com/wikioibai/p/4457209.html