題意：

  一個工廠有N個部分，M個部件。每個部件分別需要從Li~Ri部分進行加工，獲得收益Wi。

  限制每個部分最多使用K次，並且每個部件最多隻能加工一次，問最大收益。

思路：

  因為最多隻有200*2=400個點，但N的範圍為1e5，因此需要先將點離散化，從小到大排序。

  之後建立一個源點S，向第一個點連一條流量為K，費用為0的邊。

  對於每一對L、R，從L向R+1連一條流量為1，費用為-Wi的邊。

  最後對於所有點，從i向i+1連一條流量為INF，費用為0的邊。

  之後跑一遍最小費用流再將結果取相反數就可以了。

AC程式碼：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cstdlib>
#include<utility>
#include<algorithm>
#include<utility>
#include<queue>
#include<vector>
#include<set>
#include<stack>
#include<cmath>
#include<map>
#include<ctime>
#include<functional>
#include<bitset>
#define P pair<int,int>
#define ll long long
#define ull unsigned long long
#define lson id*2,l,mid
#define rson id*2+1,mid+1,r
#define ls id*2
#define rs (id*2+1)
#define Mod(a,b) a<b?a:a%b+b
#define cl0(a) memset(a,0,sizeof(a))
#define cl1(a) memset(a,-1,sizeof(a))
using namespace std;

const ll M = 1e9 + 7;
const ll INF = 1e9;
const int N = 410;
const double _e = 10e-6;
const int maxn = 5e4 + 10;
const int matSize = 9;
const int dx[4] = { 0,0,1,-1 }, dy[4] = { 1,-1,0,0 };
const int _dx[8] = { -1,-1,-1,0,0,1,1,1 }, _dy[8] = { -1,0,1,-1,1,-1,0,1 };

struct edge { int to, next, cap, flow, cost; }G[N * 2];
int head[N], tol, pre[N], dis[N], V;
bool vis[N];
void init(int n)
{
	V = n; tol = 0;
	memset(head, -1, sizeof(head));
}
void add_edge(int u, int v, int cap, int cost)
{
	G[tol].to = v; G[tol].cap = cap; G[tol].cost = cost; G[tol].flow = 0; G[tol].next = head[u];
	head[u] = tol++;
	G[tol].to = u; G[tol].cap = 0; G[tol].cost = -cost; G[tol].flow = 0; G[tol].next = head[v];
	head[v] = tol++;
}
bool spfa(int s, int t)
{
	queue<int>q;
	for (int i = 0; i < V; i++) {
		dis[i] = INF; vis[i] = false; pre[i] = -1;
	}
	dis[s] = 0; vis[s] = true; q.push(s);
	while (!q.empty()) {
		int u = q.front(); q.pop();
		vis[u] = false;
		for (int i = head[u]; i != -1; i = G[i].next) {
			int v = G[i].to;
			if (G[i].cap>G[i].flow&&dis[v]>dis[u] + G[i].cost) {
				dis[v] = dis[u] + G[i].cost;
				pre[v] = i;
				if (!vis[v]) {
					vis[v] = true;
					q.push(v);
				}
			}
		}
	}
	if (pre[t] == -1) return false;
	else return true;
}
int mcmf(int s, int t, int &cost)
{
	int flow = 0; cost = 0;
	while (spfa(s, t)) {
		int minn = INF;
		for (int i = pre[t]; i != -1; i = pre[G[i ^ 1].to]) {
			if (minn>G[i].cap - G[i].flow)
				minn = G[i].cap - G[i].flow;
		}
		for (int i = pre[t]; i != -1; i = pre[G[i ^ 1].to]) {
			G[i].flow += minn; G[i ^ 1].flow -= minn;
			cost += G[i].cost*minn;
		}
		flow += minn;
	}
	return flow;
}

struct node
{
	int l, r;
	int val;
};

int x, y, n, m;

int k, v;
node a[210];
int b[410];
map<int, int> mp;

int main()
{
	int t;
	scanf("%d", &t);
	while (t--) {
		scanf("%d%d%d", &n, &k, &m);
		int cnt = 1; mp.clear();
		for (int i = 0; i < m; i++) {
			scanf("%d%d%d", &a[i].l, &a[i].r, &a[i].val);
			b[i * 2] = a[i].l; b[i * 2 + 1] = a[i].r;
		}
		sort(b, b + m * 2); mp[b[0]] = cnt;
		for (int i = 1; i < m * 2; i++) {
			if (b[i] != b[i - 1])
				mp[b[i]] = cnt++;
		}
		init(cnt + 2);
		int S = cnt + 1, T = cnt;
		for (int i = 0; i < m; i++)
			add_edge(mp[a[i].l] - 1, mp[a[i].r], 1, -a[i].val);
		add_edge(S, 0, k, 0);
		for (int i = 0; i < cnt; i++)
			add_edge(i, i + 1, INF, 0);
		int ans = 0; mcmf(S, T, ans);
		printf("%d\n", -ans);
	}
	return 0;
}