hdu 2227(树状数组优化dp)

Find the nondecreasing subsequences

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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

31 2 3

 

Sample Output

7

 

解题思路：dp[i]表示前i个数里面非递增子序列个数，则dp[i] = sum{dp[k]} + 1，a[k] <= a[i]，+1表示自己单独也是一个非递增序列。O(n²)超时，这里需要用树状数组优化它。具体的做法就不多说了，前面做过几个了。排序，离散化即可。

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<set>#include<map>using namespace std;const int maxn = 100005;const int mod = 1e9+7;int n,tree[maxn<<1],a[maxn],dp[maxn];set<int> Set;map<int,int> Map;int lowbit(int x){return x & -x;}void update(int x,int c){for(int i = x; i <= n; i += lowbit(i))tree[i]  = (tree[i] + c) % mod;}int getsum(int x){int sum = 0;for(int i = x; i > 0; i -= lowbit(i))sum = (sum + tree[i]) % mod;return sum;}int main(){while(scanf("%d",&n)!=EOF){Set.clear();Map.clear();for(int i = 1; i <= n; i++){scanf("%d",&a[i]);Set.insert(a[i]);}int cnt = 0;for(set<int>::iterator it = Set.begin(); it != Set.end(); it++)Map[*it] = ++cnt; memset(tree,0,sizeof(tree));int ans = 0;for(int i = 1; i <= n; i++){int tmp = getsum(Map[a[i]]);dp[i] = (tmp + 1) % mod;update(Map[a[i]],dp[i]);}for(int i = 1; i <= n; i++)ans = (ans + dp[i]) % mod;printf("%d\n",ans);}return 0;}