分层图 单调决策性DP

easy 写法。

#include<bits/stdc++.h>
using namespace std;
#define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout);
#define LL long long
#define ULL unsigned LL
#define fi first
#define se second
#define pb push_back
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define lch(x) tr[x].son[0]
#define rch(x) tr[x].son[1]
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
typedef pair<int,int> pll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const LL _INF = 0xc0c0c0c0c0c0c0c0;
const LL mod =  (int)1e9+;
const int N = 2e5 + ;
int h[N], n, k, m;
LL pre[N], sum[N];
LL cost[][N];
LL Get(int l, int r){
    return pre[r] - pre[l] - (r-l) * sum[l];
}
priority_queue<LL, vector<LL>, greater<LL> > pq;
void solve(LL pre[], LL now[], int L, int R, int l, int r){
    if(l > r) return ;
    int mid = l+r >> ;
    now[mid] = _INF;
    int p = L;
    for(int i = min(mid, R); i >= L; --i){
        if(pre[i] - Get(i, mid) > now[mid]){
            now[mid] = pre[i] - Get(i, mid);
            p = i;
        }
    }
    solve(pre, now, L, p, l, mid - );
    solve(pre, now, p, R, mid + , r);
}
int main(){
    int T;
    scanf("%d", &T);
    while(T--){
        scanf("%d%d%d", &n, &k, &m);
        for(int i = ; i <= n; ++i){
            scanf("%d", &h[i]);
            sum[i] = sum[i-] + h[i];
            pre[i] = pre[i-] + sum[i];
            cost[][i] = -pre[i];
            pq.push(1ll * (n - i + ) * h[i]);
            if(pq.size() > m) pq.pop();
        }
        int now = , pre = ;
        for(int i = ; i <= k; ++i){
            swap(now, pre);
            solve(cost[pre], cost[now], , n, , n);
        }
        LL ans = cost[now][n];
        while(!pq.empty()) {ans += pq.top(); pq.pop();}
        printf("%lld\n", ans);
    }
    return ;
}