Codeforces Round #750 (Div. 2) E(dp)

Posted 2021-11-22 吃花椒的妙酱

E. Pchelyonok and Segments

传送门

题目大意：一个数组，将其划分为k个不想交的区间s1,s2…sk，使得sum1 < sum2 < …<sumk，且区间长度满足len1 = k,len2 = k-1,…lenk = 1,求最大的k

思路：dp
倒着考虑，维护的区间和需要减小，且区间长度每次+1
设dp[i][j]表示第i到n个数划分成了j个区间，则有转移方程
dp[i][j] = max(sum[i+j-1] - sum[i-1])
(sum[i+j-1] - sum[i-1] <dp[i+j][j-1])
最后看dp[1][i]，最后输出i最大且dp[1][i]有值的i即可

#define _CRT_SECURE_NO_WARNINGS
#include <cmath>
#include <cstring>
#include <string>
#include <algorithm>
#include <map>
#include <list>
#include <queue>
#include <vector>
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <deque>
#include <set>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
#define _for(i,a,b) for(int i=(a) ;i<=(b) ;i++)
#define _rep(i,a,b) for(int i=(a) ;i>=(b) ;i--)
#define scd(v) scanf("%d",&v)
#define scdd(a,b) scanf("%d %d",&a,&b)
#define endl "\\n"
#define IOS ios::sync_with_stdio(false),cin.tie(0)
#define pb push_back
#define all(v) v.begin(),v.end()
#define mst(v,a) memset(v,a,sizeof(v))
#define ls p<<1
#define rs p<<1|1
#define int long long
#define inf 0x7f7f7f7f
#define fi first
#define se second
#define pii pair<int , int >
#define ls p<<1
#define rs p<<1|1
#define lson p<<1,l,mid
#define rson p<<1|1,mid+1,r
#define AC return 0
const int N = 1e5 + 100;
const int mod = 1e9 + 9;
const double eps = 1e-7;
int n, m, k;
int a[N], b[N];
int dp[N][505];
int sum[N];
void de()
{
}
void solve()
{
    int k = 1;
    while (k * (1 + k)/2 <= n) k++;
    k--;
    _for(i, 1, n) sum[i] = sum[i - 1] + a[i];
    _rep(i, n, 1)
    {
        _for(j, 1, k) dp[i][j] = dp[i+1][j];
        dp[i][1] = max(dp[i + 1][1], a[i]);
        for (int j = 2; i + j <= n && j<=k; j++)
        {
            if (dp[i+j][j-1] && sum[i+j-1] - sum[i-1] < dp[i+j][j-1]  )
            {
                dp[i][j] = max(dp[i][j], sum[i + j -1] - sum[i - 1]);
            }
        }
    }
    int ans = -1;
    _rep(i, k, 1)
    {
        if (dp[1][i])
        {
            ans = i;
            //cout << i << endl;
            //return ;
            break;
        }
    }
    _for(i, 1, n)
    {
        _for(j, 1, k) dp[i][j] = 0;
    }
    cout << ans << endl;
}
signed main()
{
   //  freopen("data.txt","r",stdin);
    IOS;
    int T; cin>>T;
    while (T--)
    {
        cin >> n;
        _for(i, 1, n) cin >> a[i];
        solve();
    }
}