感觉建图好妙啊。。
考虑把问题转化成选$m$次数，每次选出一个子序列， 要求两次选择的数的下标差至少为$m$。
这个很容易建图就能做。$i$向$\min\{T, i + m\}$连边，容量为1，费用为$a_i$。$i$向$i + 1$连边容量为1，费用为0.
Code
/**
* bzoj
* Problem#1283
* Accepted
* Time: 232ms
* Memory: 1436k
*/
#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <queue>
using namespace std;
typedef bool boolean; template <typename T>
void pfill(T* pst, const T* ped, T val) {
for ( ; pst != ped; *(pst++) = val);
} typedef class Edge {
public:
int ed, nx, r, c; Edge(int ed = , int nx = , int r = , int c = ) : ed(ed), nx(nx), r(r), c(c) { }
} Edge; typedef class MapManager {
public:
int* h;
vector<Edge> es; MapManager() { }
MapManager(int n) {
h = new int[(n + )];
pfill(h, h + n + , -);
} void addEdge(int u, int v, int r, int c) {
es.push_back(Edge(v, h[u], r, c));
h[u] = (signed) es.size() - ;
} void addArc(int u, int v, int cap, int c) {
addEdge(u, v, cap, c);
addEdge(v, u, , -c);
} Edge& operator [] (int p) {
return es[p];
}
} MapManager; const signed int inf = (signed) (~0u >> ); typedef class Graph {
public:
int S, T;
MapManager g; int *le;
int *f, *mf;
boolean *vis; Graph() { }
// be sure T is the last node
Graph(int S, int T) {
this->S = S;
this->T = T;
f = new int[(T + )];
le = new int[(T + )];
mf = new int[(T + )];
vis = new boolean[(T + )];
pfill(vis, vis + T, false);
} int spfa() {
queue<int> que;
pfill(f, f + T + , -inf);
que.push(S);
f[S] = , le[S] = -, mf[S] = inf;
while (!que.empty()) {
int e = que.front();
que.pop();
vis[e] = false;
for (int i = g.h[e], eu, w; ~i; i = g[i].nx) {
if (!g[i].r)
continue;
eu = g[i].ed, w = f[e] + g[i].c;
if (w > f[eu]) {
f[eu] = w, le[eu] = i, mf[eu] = min(mf[e], g[i].r);
if (!vis[eu]) {
vis[eu] = true;
que.push(eu);
}
}
}
}
if (f[T] == -inf)
return inf;
int rt = ;
for (int p = T, e; ~le[p]; p = g[e ^ ].ed) {
e = le[p];
g[e].r -= mf[T];
g[e ^ ].r += mf[T];
rt += mf[T] * g[e].c;
}
return rt;
} int max_cost() {
int rt = , delta;
while ((delta = spfa()) != inf) {
rt += delta;
// cerr << delta << '\n';
}
return rt;
}
} Graph; int n, m, k; inline void solve() {
scanf("%d%d%d", &n, &m, &k);
Graph graph(, n + );
MapManager& g = graph.g;
k = min(k, m);
g = MapManager(n + );
g.addArc(, , k, );
for (int i = , x; i <= n; i++) {
scanf("%d", &x);
g.addArc(i, min(n + , i + m), , x);
g.addArc(i, i + , k, );
}
printf("%d\n", graph.max_cost());
} int main() {
solve();
return ;
}