题意：

有\(n\)个数\(a_1\cdots a_n\)，现要你给出\(k\)个不相交的非降子序列，使得和最大。

思路：

费用流建图，每个点拆点，费用为\(-a[i]\)，然后和源点连边，和后面非降的数连边，源点和超级源点连一条容量\(k\)的边，跑费用流。
用\(spfa\)费用流\(TLE\)，这里因为不会出现负环，所以用\(Dijkstra\)优化。

代码：

/*******
dijkstra优化费用流模板
*******///不能有负环
#include<functional>    //C++编译需要头文件
typedef pair<int, int> pii;//first保存最短距离，second保存顶点的编号
struct edge {int to, cap, cost, rev; //终点,容量（指残量网络中的）,费用,反向边编号edge() {}edge(int to, int _cap, int _cost, int _rev):to(to), cap(_cap), cost(_cost), rev(_rev) {}
};
int V;          //顶点数
int h[maxn];    //顶点的势
int dis[maxn];  //最短距离
int prevv[maxn];//最短路中的父结点
int pree[maxn]; //最短路中的父边
vector<edge> G[maxn];   //图的邻接表void init(int n) {V = n;for(int i = 0; i <= V; ++i) G[i].clear();
}
void addEdge(int from, int to, int cap, int cost){G[from].push_back(edge(to, cap, cost, G[to].size()));G[to].push_back(edge(from, 0, -cost, G[from].size() - 1));
}
int MCMF(int s, int t, int f, int &flow){   //f为最多能流多少int res = 0;for(int i = 0; i < 1 + V; i++) h[i] = 0;while(f){priority_queue<pii, vector<pii>, greater<pii> > q;for(int i = 0; i < 1 + V; i++) dis[i] = INF;dis[s] = 0; q.push(pii(0, s));while(!q.empty()) {pii now = q.top(); q.pop();int v = now.second;if(dis[v] < now.first) continue;for(int i = 0; i < G[v].size(); ++i) {edge &e = G[v][i];if(e.cap > 0 && dis[e.to] > dis[v] + e.cost + h[v] - h[e.to]){dis[e.to] = dis[v] + e.cost + h[v] - h[e.to];prevv[e.to] = v;pree[e.to] = i;q.push(pii(dis[e.to], e.to));}}}if(dis[t] == INF) break;for(int i = 0; i <= V; ++i) h[i] += dis[i];int d = f;for(int v = t; v != s; v = prevv[v]) d = min(d, G[prevv[v]][pree[v]].cap);f -= d; flow += d; res += d * h[t];for(int v = t; v != s; v = prevv[v]) {edge &e = G[prevv[v]][pree[v]];e.cap -= d;G[v][e.rev].cap += d;}}return res;
}

#include<map>
#include<set>
#include<cmath>
#include<cstdio>
#include<stack>
#include<ctime>
#include<vector>
#include<queue>
#include<cstring>
#include<string>
#include<sstream>
#include<iostream>
#include<algorithm>
typedef long long ll;
using namespace std;
const int maxn = 5000 + 5;
const int INF = 0x3f3f3f3f;
const ll MOD = 1e9 + 7;
using namespace std;//不能有负环
#include<functional>    //C++编译需要头文件
typedef pair<int, int> pii;//first保存最短距离，second保存顶点的编号
struct edge {int to, cap, cost, rev; //终点,容量（指残量网络中的）,费用,反向边编号edge() {}edge(int to, int _cap, int _cost, int _rev):to(to), cap(_cap), cost(_cost), rev(_rev) {}
};
int V;          //顶点数
int h[maxn];    //顶点的势
int dis[maxn];  //最短距离
int prevv[maxn];//最短路中的父结点
int pree[maxn]; //最短路中的父边
vector<edge> G[maxn];   //图的邻接表void init(int n) {V = n;for(int i = 0; i <= V; ++i) G[i].clear();
}
void addEdge(int from, int to, int cap, int cost){G[from].push_back(edge(to, cap, cost, G[to].size()));G[to].push_back(edge(from, 0, -cost, G[from].size() - 1));
}
int MCMF(int s, int t, int f, int &flow){   //f为最多能流多少int res = 0;for(int i = 0; i < 1 + V; i++) h[i] = 0;while(f){priority_queue<pii, vector<pii>, greater<pii> > q;for(int i = 0; i < 1 + V; i++) dis[i] = INF;dis[s] = 0; q.push(pii(0, s));while(!q.empty()) {pii now = q.top(); q.pop();int v = now.second;if(dis[v] < now.first) continue;for(int i = 0; i < G[v].size(); ++i) {edge &e = G[v][i];if(e.cap > 0 && dis[e.to] > dis[v] + e.cost + h[v] - h[e.to]){dis[e.to] = dis[v] + e.cost + h[v] - h[e.to];prevv[e.to] = v;pree[e.to] = i;q.push(pii(dis[e.to], e.to));}}}if(dis[t] == INF) break;for(int i = 0; i <= V; ++i) h[i] += dis[i];int d = f;for(int v = t; v != s; v = prevv[v]) d = min(d, G[prevv[v]][pree[v]].cap);f -= d; flow += d; res += d * h[t];for(int v = t; v != s; v = prevv[v]) {edge &e = G[prevv[v]][pree[v]];e.cap -= d;G[v][e.rev].cap += d;}}return res;
}
int a[maxn];
int main(){int T;scanf("%d", &T);while(T--){int n, k;scanf("%d%d", &n, &k);init(n * 2 + 5);int s = n * 2 + 1, e = n * 2 + 2;int ss = n * 2 + 3, se = n * 2 + 4;for(int i = 1; i <= n; i++){scanf("%d", &a[i]);addEdge(i, i + n, 1, -a[i]);addEdge(s, i, 1, 0);addEdge(i + n, e, 1, 0);}for(int i = 1; i <= n; i++){for(int j = i + 1; j <= n; j++){if(a[i] <= a[j])addEdge(i + n, j, 1, 0);}}addEdge(ss, s, k, 0);addEdge(e, se, k, 0);int flow;printf("%d\n", -MCMF(ss, se, INF, flow));}return 0;
}