제출 #566190

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
566190		quicksloth	수열 (APIO14_sequence)	C++17	71 / 100	374 ms	82536 KiB

sequence

#include <iostream>
using namespace std;
#define maxn 100000
#define maxk 200
#define ll long long

int best[maxn][maxk];
//arr is prefix sums

int main()
{
    ll dp[maxn], prv[maxn], arr[maxn];
    //prv is not quite necessary
    int n, k;
    cin >> n >> k;
    k++;
    //array endgth n, split into k+1 intervals, each time points = product of sums of two sides
    //if decided how intervals will be picked: then how to arrange which to select?
    //answer for a set of intervals = sum ab where a, b are distinct in it
    //true by induction on set's size
    //= ((sum a)^2 - sum a^2)/2 where first sum is known (sum of array) and second is split up
    //divide up the array optimally to minimize sum (segment sum)^2? dp with prefix sums
    for (int i = 0; i < n; i++)
    {
        cin >> arr[i];
        if (i) arr[i] += arr[i-1];
        //1 interval
        dp[i] = arr[i]*arr[i];
        best[i][0] = -1;
        //cut at -1 means this is final interval
    }
    int stack[maxn];
    ll cut[maxn];
    cut[0] = 0;
    //stack[0] is best from cut[0] to cut[1]-1, ...
    int end, start;
    for (int t = 1; t < k; t++)
    {
        //t+1 intervals to make 0..i-1; if t = 0 then just interval sum ^ 2
        //else: dp[i] = min_x<i arr[i]^2 - 2*arr[x]*arr[i] + arr[x]^2 + prv[x]
        //with x >= t-1 (i >= t to make it well defined)
        //do all of dp[?] in O(n)? need to insert another x or query best x for given i
        //monotonic stack: record pair<int,ll>: best value of x for arr[i] <= that ll
        for (int i = 0; i < n; i++) prv[i] = dp[i];
        end = 0;
        start = 0;
        for (int i = t; i < n; i++)
        {
            //x = i-1: insert it into the monotonic stack
            ll x = 0;
            bool special = false;
            while (start != end)
            {
                //see when it beats the last function, then truncate
                //definitely should work at some value of arr[i]: prv is increasing not too fast
                //just assume it ought to work
                //solve for smallest int x:
//arr[stack[end-1]]*arr[stack[end-1]]+prv[stack[end-1]]-2*arr[stack[end-1]]*x >
//arr[i-1]*arr[i-1]+prv[i-1]-2*arr[i-1]*x
                //or tmp2x > tmp1 or x = ceil of (tmp1+1)/tmp2
                //just take floor and +1 if needed
                if (arr[i-1] == arr[stack[end-1]])
                {
                    //impossible to beat previous function, so done
                    special = true;
                    break;
                }
ll tmp1 = arr[i-1]*arr[i-1]+prv[i-1]-arr[stack[end-1]]*arr[stack[end-1]]-prv[stack[end-1]],
   tmp2 = 2*arr[i-1]-2*arr[stack[end-1]];
                x = (tmp1+1)/tmp2;
                if ((tmp1+1)%tmp2) x++;
                //if at some point it is optimal, it will continue to be optimal later
                if (x > cut[end-1])
                {
                    //previous one is sometimes optimal so cannot discard
                    break;
                }
                end--;
            }
            if (!special)
            {
                stack[end] = i-1;
                cut[end] = x;
                end++;
                cut[end] = 1e18;
            }
            //the last curve is always optimal, but previous one is up to x-1 only
            //now find which curve is best for arr[i]: pop from front of the stack
            while (start != end)
            {
                if (cut[start+1] > arr[i]) break;
                start++;
            }
            best[i][t] = stack[start];
            dp[i] = (arr[i]-arr[stack[start]])*(arr[i]-arr[stack[start]])+prv[stack[start]];
        }
    }
    //now dp[n-1] is the answer
    cout << (arr[n-1]*arr[n-1]-dp[n-1])/2 << '\n';
    int cur = n-1;
    while (k)
    {
        cur = best[cur][--k];
        if (cur != -1) cout << cur+1 << ' ';
    }
}

• Subtask 1: 11 / 11
• Subtask 2: 11 / 11
• Subtask 3: 11 / 11
• Subtask 4: 17 / 17
• Subtask 5: 21 / 21
• Subtask 6: 0 / 29