POJ 2135 最小费用流

题意

传送门 POJ 2135

题解

把起点作为源点，终点作为汇点，求一条流量为 $2$ 的最小费用流即可。

#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 1000
typedef pair<int, int> P;
// 终点、容量、费用、反向边索引
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<P, vector<P>, greater<P> > que;
		memset(dist, 0x3f, sizeof(dist));
		dist[s] = 0;
		que.push(P(0, s));
		while(!que.empty()){
			P 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(P(dist[e.to], e.to));
				}
			}
		}
		if(dist[t] == INF){
			// 不能再增广
			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 1000
#define MAX_M 10000
int N, M;
int a[MAX_M], b[MAX_M], c[MAX_M];

void solve(){
	int s = 0, t = N - 1;
	V = N;
	for(int i = 0; i < M; i++){
		// 双向通行
		add_edge(a[i], b[i], 1, c[i]);
		add_edge(b[i], a[i], 1, c[i]);
	}
	printf("%d\n", min_cost_flow(s, t, 2));
}

int main(){
	while(~scanf("%d%d", &N, &M)){
		for(int i = 0; i < N; i++) G[i].clear();
		for(int i = 0; i < M; i++){
			scanf("%d%d%d", a + i, b + i, c + i);
			--a[i], --b[i];
		}
		solve();
	}
	return 0;
}