POJ 3686 最小费用流

题意

传送门 POJ 3686

题解

参考白书解法。考虑只有一个工厂情况，按照 $a_{1},a_{2},\dots,a_{N}$ 顺序加工玩具，观察总时间最小值 $T$

$T=Z_{a_{1}}+(Z_{a{1}}+Z_{a_{2}})+\dots +(Z_{a{1}}+Z_{a_{2}}+\dots+Z_{a_{N}})$

变形得到

$T=N\times Z_{a_{1}}+(N-1)\times Z_{a_{2}}+\dots +1\times Z_{a_{N}}$

可以看做在不同加工次序下，各自需要花费 $1$ 倍到 $N$ 倍的时间。每个加工次序和工厂的二元组都对应玩具的集合，求最小费用流即可。

#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 2555
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 50
#define MAX_M 50
int N, M;
int Z[MAX_N][MAX_M];

void solve(){
	int s = N + M * N, t = s + 1;
	V = t + 1;
	// 清空邻接表元素
	for(int v = 0; v < V; v++) G[v].clear();
	for(int i = 0; i < N; i++) add_edge(s, i, 1, 0);
	// 对每个加工次序和工厂二元组连边
	for(int i = 0; i < M; i++){
		for(int j = 0; j < N; j++){
			add_edge(N + j * M + i, t, 1, 0);
			for(int k = 0; k < N; k++){
				add_edge(k, N + j * M + i, 1, Z[k][i] * (j + 1));
			}
		}
	}
	printf("%.6f\n", (double)min_cost_flow(s, t, N) / N);
}

int main(){
	int t;
	scanf("%d", &t);
	while(t--){
		scanf("%d%d", &N, &M);
		for(int i = 0; i < N; i++){
			for(int j = 0; j < M; j++){
				scanf("%d", &Z[i][j]);
			}
		}
		solve();
	}
	return 0;
}