題目連結

題意:給出1~n點和m條無向邊，要求從1走到n再從n回到1處的最小費用，要求每條邊走過不超過1次

題解:可轉換為求從1到n的流量f=2的最小費用流問題（不可先從左到右掃一遍最短路再刪去使用過的邊最後再次最短路回起點，只做到區域性最優而非全域性）最小費用最大流（Dijkstra+最大流演算法）O(F|E|log|V|)或O(F|V|^2)

程式碼如下:

#include<iostream>
#include<cstring>
#include<string>
#include<cstdio>
#include<cmath>
#include<vector>
#include<queue>
#include<map>
#include<functional>
#include<algorithm>
using namespace std;
#define ll long long
#define inf 0x3f3f3f3f
const int maxn = (int)2e5 + 50;
typedef pair<int, int>P; //first->最短路距離  second->頂點編號
struct edge {
	int to, cap, cost, rev;
};
int V;//頂點數
vector<edge>G[maxn];
int h[maxn];  //頂點的勢，用來避免負環的出現
int dist[maxn];//最短距離
int prevv[maxn], preve[maxn];//最短路的前驅結點和對應的邊
void add(int from, int to, int cap, int cost) {
	edge ee;
	ee.to = to, ee.cap = cap, ee.rev = G[to].size(), ee.cost = cost;
	G[from].push_back(ee);
	ee.to = from, ee.cap = 0, ee.rev = G[from].size() - 1, ee.cost = -cost;
	G[to].push_back(ee);
}
int min_cost_flow(int s, int t, int f) {
	int res = 0;
	fill(h, h + V, 0);//初始化h
	while (f > 0) {
		//使用Dijkstra演算法更新h
		priority_queue<P, vector<P>, greater<P> >q; //greater需要<functional>標頭檔案,注意> >中間要加" "
		fill(dist, dist + V, inf);
		dist[s] = 0;
		q.push(P(0, s));
		while (!q.empty()) {
			P p = q.top(); q.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];
				if (e.cap > 0 && dist[e.to] > dist[v] + e.cost + h[v] - h[e.to]) {
					dist[e.to] = dist[v] + e.cost + h[v] - h[e.to];
					prevv[e.to] = v;
					preve[e.to] = i;
					q.push(P(dist[e.to], e.to));
				}
			}
		}
		if (dist[t] == inf)//不能再增廣
			return -1;
		for (int v = 0; v < V; v++)
			h[v] += dist[v];
		//沿s->t的最短路儘量增廣
		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;
}
int main() {
	int m, u, v, w;
	while (~scanf("%d%d", &V, &m)) {
		while (m--) {
			scanf("%d%d%d", &u, &v, &w);
			u--, v--;
			add(u, v, 1, w);
			add(v, u, 1, w);
		}
		printf("%d\n", min_cost_flow(0, V - 1, 2));//這個是從0到v-1的
	}
	return 0;
}