poj3680（*最小费用流）

/*
translation:
    见小白书p246
solution:
    最小费用流
    对于一个区间，在a,b之间连上一条容量1，费用w的边，表示选中这个得到w。区间内部点之间i,i+1之间连上一条
    容量无穷，费用0的边。每个a与s相连，容量1，费用0。每个b同理。这样建图之、后跑最小费用流即可。
note: *
*/
#include 
#include 
#include 
#include 
#include 
#include 

using namespace std;
const int maxn = 505;
const int INF = 0x3f3f3f3f;

typedef pair P;
struct Edge
{
    int to, cap, cost, rev;
    Edge(int to_, int cap_, int cost_, int rev_):to(to_),cap(cap_),cost(cost_),rev(rev_){}
};

vector G[maxn];
int V, n, k;
int a[maxn], b[maxn], w[maxn], id[100005];
int h[maxn], dist[maxn], prevv[maxn], preve[maxn];

void add_edge(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 min_cost_flow(int s, int t, int f)
{
    int res = 0;
    while(f > 0) {
        fill(dist, dist + V, INF);
        dist[s] = 0;
        bool update = true;
        while(update) {
            update = false;
            for(int v = 0; v < V; v++) {
                if(dist[v] == INF)  continue;
                for(int i = 0; i < G[v].size(); i++) {
                    Edge& e = G[v][i];
                    if(e.cap > 0 && dist[e.to] > dist[v] + e.cost) {
                        dist[e.to] = dist[v] + e.cost;
                        prevv[e.to] = v;
                        preve[e.to] = i;
                        update = true;
                    }
                }
            }
        }

        if(dist[t] == INF)  return -1;

        int d = f;
        for(int v = t; v != s; v = prevv[v]) {
            d = min(d, G[prevv[v]][preve[v]].cap);
        }
        f -= d;
        res += d * dist[t];
        for(int v = t; v != s; v = prevv[v]) {
            Edge& e = G[prevv[v]][preve[v]];
            e.cap -= d;
            G[v][e.rev].cap += d;
        }
    }
    return res;
}

int main()
{
    //freopen("in.txt", "r", stdin);
    int T;
    scanf("%d", &T);
    while(T--) {
        for(int i = 0; i < maxn; i++) G[i].clear();
        scanf("%d%d", &n, &k);

        for(int i = 0; i < n; i++) {
            scanf("%d%d%d", &a[i], &b[i], &w[i]);
        }

        vector pos;
        for(int i = 0; i < n; i++) {
            pos.push_back(a[i]);
            pos.push_back(b[i]);
        }

        sort(pos.begin(), pos.end());
        pos.erase(unique(pos.begin(), pos.end()), pos.end());

        memset(id, 0, sizeof(id));
        for(int i = 0; i < pos.size(); i++) id[pos[i]] = i;
        int s = pos.size(), t = s + 1;

        for(int i = 0; i < n; i++) {
            add_edge(id[a[i]], id[b[i]], 1, -w[i]);
            add_edge(s, id[a[i]], 1, 0);
            add_edge(id[b[i]], t, 1, 0);
        }

        for(int i = 0; i < pos.size()-1; i++) {
            add_edge(i, i + 1, INF, 0);
        }

        add_edge(s, 0, k, 0);   add_edge(pos.size() - 1, t, k, 0);

        V = t + 1;
        printf("%d\n", -min_cost_flow(s, t, k));
    }
    return 0;
}