hdoj 2227 Find the nondecreasing subsequences 【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)
Total Submission(s): 1711    Accepted Submission(s): 625


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
 



题意:给你n个数,问你不下降子序列的个数%1000000007。


分析:明显的dp,用dp[i]表示以a[i]结尾的不下降子序列的个数。显然有

dp[i] = sigama(dp[j]) + 1 其中(1 <= j < i && a[j] <= a[i])。 最后ans = sigma(dp[i]) 其中(1 <= i <= n)


代码:贴出一部分 没有取余操作。


LL a[MAXN];
LL dp[MAXN];
int main()
{
    int n;
    while(~Ri(n))
    {
        for(int i = 1; i <= n; i++)
            Ri(a[i]);
        CLR(dp, 0);
        LL ans = 0;
        for(int i = 1; i <= n; i++)
        {
            dp[i] = 1;
            for(int j = 1; j < i; j++)
                if(a[i] >= a[j])
                    dp[i] += dp[j];
            ans += dp[i];
        }
        Pl(ans);
    }
    return 0;
}

提交上去—— TLE,O(n*n)的时间复杂度太高。 

由于a[j] <= a[i] 且 (i <= j < i)我们可以考虑利用逆序数的性质来求解,而树状数组是求解逆序数的最佳工具。


AC代码:树状数组优化dp 求出每个数前面有多少个数小于或等于它 就可以了。


#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <map>
#include <string>
#include <vector>
#define lson o<<1|1, l, mid
#define rson o<<1, mid+1, r
#define ll o<<1
#define rr o<<1|1
#define INF 0x3f3f3f3f
#define eps 1e-8
#define debug printf("1\n")
#define MAXN 101000
#define MAXM 100000
#define LL long long
#define CLR(a, b) memset(a, (b), sizeof(a))
#define W(a) while(a--)
#define Ri(a) scanf("%d", &a)
#define Pi(a) printf("%d\n", (a))
#define Rl(a) scanf("%lld", &a)
#define Pl(a) printf("%lld\n", (a))
#define Rs(a) scanf("%s", a)
#define Ps(a) printf("%s\n", (a))
#define MOD 1000000007
using namespace std;
int a[MAXN]; int dp[MAXN];
int n;
int lowbit(int x){
    return x & (-x);
}
int update(int x, int d)
{
    while(x <= n)
    {
        dp[x] += d;
        dp[x] %= MOD;
        x += lowbit(x);
    }
}
int sum(int x)
{
    int s = 0;
    while(x > 0)
    {
        s += dp[x];
        s %= MOD;
        x -= lowbit(x);
    }
    return s;
}
struct MES{
    int val, id;
};
MES num[MAXN];
bool cmp(MES a, MES b){
    if(a.val != b.val)
        return a.val < b.val;
    else
        return a.id < b.id;
}
int main()
{
    while(~Ri(n))
    {
        CLR(dp, 0);
        for(int i = 1; i <= n; i++)
            Ri(num[i].val), num[i].id = i;
        sort(num+1, num+n+1, cmp);
        num[0].id = 0;
        for(int i = 1; i <= n; i++)
        {
            int add = sum(num[i].id-1);
            add %= MOD;
            update(num[i].id, add+1);
        }
        Pi(sum(n)%MOD);
    }
    return 0;
}


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