POJ 3680 coordinate discretization + minimum cost flow

Title

Portal POJ 3680

answer

Discretized coordinates, connected to the end of the interval, the capacity is $1$ , the weight is $-w_{i}$ ; For all adjacent coordinate values, the capacity is INF, and the weight is $0$ ; this corresponds to a DAG. There are negative weighted edges in the graph, so that the initial full flow processing, you can directly use dijkstra to find the minimum cost flow.

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <queue>
#include <vector>
#define min(a,b)    (((a) < (b)) ? (a) : (b))
#define max(a,b)    (((a) > (b)) ? (a) : (b))
#define abs(x)    ((x) < 0 ? -(x) : (x))
#define INF 0x3f3f3f3f
#define delta 0.85
#define eps 1e-5
#define PI 3.14159265358979323846
using namespace std;

#define MAX_V 405
struct edge{
	int to, cap, cost, rev;
	edge(int to, int cap, int cost, int rev):to(to), cap(cap), cost(cost), rev(rev){}
};

int V;
vector<edge> G[MAX_V];
int h[MAX_V], dist[MAX_V];
int prevv[MAX_V], preve[MAX_V];

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;
	memset(h, 0, sizeof(h));
	while(f > 0){
		// Dijkstra
		priority_queue<pair<int, int>, vector<pair<int, int> >, greater<pair<int, int> > > que;
		memset(dist, 0x3f, sizeof(dist));
		dist[s] = 0;
		que.push(pair<int, int>(0, s));
		while(!que.empty()){
			pair<int, int> p = que.top(); que.pop();
			int v = p.second;
			if(dist[v] < p.first) continue;
			for(int i = 0; i < G[v].size(); i++){
				edge &e = G[v][i];
				int d2 = dist[v] + e.cost + h[v] - h[e.to];
				if(e.cap > 0 && d2 < dist[e.to]){
					dist[e.to] = d2;
					prevv[e.to] = v;
					preve[e.to] = i;
					que.push(pair<int, int>(dist[e.to], e.to));
				}
			}
		}
		if(dist[t] == INF){
			return -1;
		}
		for(int v = 0; v < V; v++) h[v] += dist[v];
		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 * h[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;
}

#define MAX_N 200
#define MAX_K 200
int N, K;
int a[MAX_N], b[MAX_N], w[MAX_N];
// 坐标离散化
int compress(int *x, int *y){
	vector<int> xs(N * 2);
	for(int i = 0, i2 = N; i < N; i++, i2++) xs[i] = x[i], xs[i2] = y[i];
	sort(xs.begin(), xs.end());
	xs.erase(unique(xs.begin(), xs.end()), xs.end());
	for(int i = 0; i < N; i++){
		x[i] = distance(xs.begin(), lower_bound(xs.begin(), xs.end(), x[i]));
		y[i] = distance(xs.begin(), lower_bound(xs.begin(), xs.end(), y[i]));
	}
	return xs.size();
}

void solve(){
	int sz = compress(a, b), s = sz, t = s + 1;
	V = t + 1;
	for(int v = 0; v < V; v++) G[v].clear();
	add_edge(s, 0, K, 0);
	add_edge(sz - 1, t, K, 0);
	for(int i = 0; i + 1 < sz; i++) add_edge(i, i + 1, INF, 0);
	int sum = 0;
	for(int i = 0; i < N; i++){
		// 令所有负权边初始时满流
		add_edge(b[i], a[i], 1, w[i]);
		add_edge(s, b[i], 1, 0);
		add_edge(a[i], t, 1, 0);
		sum += w[i];
	}
	// 答案为最小费用流的相反数
	printf("%d\n", sum - min_cost_flow(s, t, K + N));
}

int main(){
	int t;
	scanf("%d", &t);
	while(t--){
		scanf("%d%d", &N, &K);
		for(int i = 0; i < N; i++) scanf("%d%d%d", a + i, b + i, w + i);
		solve();
	}
	return 0;
}