问题是:
Given an integer n and an array v of n integers, count how many ascending subsets can be formed with these numbers.
有一些限制:
- 1 ≤ n ≤ 300
- v[i] ≤ 1.000.000, whatever 1 ≤ i ≤ n
- S ≤ 10^18
例如,这里有一个例子:
Input :
6
5 2 6 1 1 8
Output:
15
解释:有 15 个升序子集。 {5}, {2}, {6}, {1}, {1}, {8}, {5, 6}, {5, 8}, {2, 6}, {2, 8}, {6 , 8}, {1, 8}, {1, 8}, {5, 6, 8}, {2, 6, 8}。
我把这个问题作为作业。我搜索了堆栈溢出、数学堆栈等,但找不到任何想法。
如果您能给我提示如何解决这个问题,那将非常有帮助。
编辑: 所以我想出了这个解决方案,但显然它在某处溢出了?你能帮帮我吗
#include <iostream>
#include <fstream>
#include <queue>
using namespace std;
ifstream fin("nrsubsircresc.in");
ofstream fout("nrsubsircresc.out");
int v[301];
int n;
long long s;
queue <int> Coada;
int main()
{
fin >> n;
for(int i = 0; i < n; i++)
{
fin >> v[i]; // reading the array
s++;
Coada.push(i);
}
while(!Coada.empty()) // if the queue is not empty
{
for(int k = Coada.front() + 1; k < n; k++) //
if( v[k] > v[Coada.front()] )
{
s++;
Coada.push(k);
}
Coada.pop();
}
fout << s;
return 0;
}
最佳答案
我在这里实现了贪婪的方法:
#include <algorithm>
#include <vector>
#include <array>
using namespace std;
using ll = long long;
const int N = 6;
array<int, N> numbers = { 5, 2, 6, 1, 1, 8 }; // provided data
vector<ll> seqends(N);
int main() {
for (int i = 0; i < N; ++i) {
// when we adding new number to subarray so far,
// we add at least one ascending sequence, which consists of this number
seqends[i] = 1;
// next we iterate via all previous numbers and see if number is less than the last one,
// we add number of sequences which end at this number to number of sequences which end at the last number
for (int j = 0; j < i; ++j) {
if (numbers[j] < numbers[i]) seqends[i] += seqends[j];
}
}
// Finally we sum over all possible ends
ll res = accumulate(seqends.cbegin(), seqends.cend(), (ll)0);
cout << res;
}
该算法需要O(N)空间和O(N2)时间。
关于c++ - 动态规划 : Count how many ascending subsets exists in a set,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/48148951/