UVA 10806 10806 - Dijkstra, Dijkstra(費用流)

解題思路：

從起點走到終點，再從終點走回來，每個邊隻能走一次，求最短路。 因為每個邊隻能走一次，是以将邊看成一個點，向它連的兩個頂點連邊，容量為1，費用隻加在一條邊上，源點向起點連容量為2費用為0的邊，終點向彙點連容量為2費用為0的邊，跑一遍費用流即可。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <cmath>
#define LL long long
using namespace std;
const int maxn = 10000 + 10;
const int INF = 0x3f3f3f3f;
struct Edge
{
	int from, to, cap, flow, cost;
	Edge(int u, int v, int c, int f, int w) : from(u), to(v), cap(c), flow(f), cost(w) { }
};
int n;
vector<Edge>edges;
vector<int>G[maxn];
int inq[maxn];
int d[maxn];
int p[maxn];
int a[maxn];
void init()
{
	for(int i=0;i<maxn;i++)
		G[i].clear();
	edges.clear();
}
void AddEdge(int from, int to, int cap, int cost)
{
	edges.push_back(Edge(from,to,cap,0,cost));
	edges.push_back(Edge(to,from,0,0,-cost));
	int M = edges.size();
	G[from].push_back(M-2);
	G[to].push_back(M-1);
}
bool SPFA(int s, int t, int& flow, LL& cost)
{
	for(int i=0;i<=n+1;i++)
 		d[i] = INF;
	memset(inq, 0, sizeof(inq));
	d[s] = 0; inq[s] = 1; p[s] = 0; a[s] = INF;
	queue<int>Q;
	Q.push(s);
	while(!Q.empty())
	{
		int u = Q.front();Q.pop();
		inq[u] = 0; int sz = G[u].size();
		for(int i=0;i<sz;i++)
		{
			Edge& e = edges[G[u][i]];
			if(e.cap > e.flow && d[e.to] > d[u] + e.cost)
			{
				d[e.to] = d[u] + e.cost;
				p[e.to] = G[u][i];
				a[e.to] = min(a[u], e.cap - e.flow);
				if(!inq[e.to]){Q.push(e.to);inq[e.to] = 1;}
			}
		}
	}
	if(d[t] == INF) return false;
	flow += a[t];
	cost += (LL) d[t] * (LL) a[t];
	for(int u=t;u!=s;u=edges[p[u]].from)
	{
		edges[p[u]].flow += a[t];
		edges[p[u]^1].flow -= a[t];
	}
	return true;
}
int MincostMaxflow(int s, int t, LL& cost)
{
	int flow = 0; cost = 0;
	while(SPFA(s,t,flow,cost));
	return flow;
}
int N, M;
int main()
{
	while(scanf("%d", &N)!=EOF && N )
	{
		scanf("%d", &M);
		int u, v, w;
		n = N + M + 2;
		init();
		int s = 0, t = n - 1;
		for(int i=1;i<=M;i++)
		{
			scanf("%d%d%d", &u, &v, &w);
			AddEdge(u, i + N, 1, w);
			AddEdge(i + N, v, 1, 0);
			AddEdge(v, i + N, 1, 0);
			AddEdge(i + N, u, 1, w);
		}
		AddEdge(s, 1, 2, 0);
		AddEdge(N, t, 2, 0);
		LL cost = 0;
		int ans = MincostMaxflow(s, t, cost);
		if(ans < 2) printf("Back to jail\n");
		else cout << cost << endl;
	}
	return 0;
}