Uva 10806 来回最短路，不重复，MCMF

题目链接：https://uva.onlinejudge.org/external/108/10806.pdf

题意：无向图，从1到n来回的最短路，不走重复路。

分析：可以考虑为1到n的流量为2时的最小花费；

建图： 一个点到一个点的容量为1，费用为距离。

#include <cstring>
#include <cstdio>
#include <vector>
#include <queue>
#include <algorithm>
#include <iostream>
using namespace std;

const int maxn = 100 + 10, INF = 1000000000;


struct Edge
{
    int from, to, cap, flow, cost;
};

struct MCMF
{
    int n, m;
    vector<Edge> edges;
    vector<int> G[maxn];
    bool inq[maxn];         // 是否在队列中
    int d[maxn];           // Bellman-Ford
    int p[maxn];           // 上一条弧
    int a[maxn];           // 可改进量

    void init(int n)
    {
        this->n = n;
        for(int i = 0; i < n; 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
        });
        m = edges.size();
        G[from].push_back(m-2);
        G[to].push_back(m-1);
    }

    bool BellmanFord(int s, int t, int &flow, long long& cost)
    {
        memset(inq,0,sizeof(inq));
        for(int i=0;i<n;i++)
            d[i] = INF;
        d[s] = 0;
        inq[s] = true;
        p[s] = 0;
        a[s] = INF;

        queue<int> Q;
        Q.push(s);
        while(!Q.empty())
        {
            int u = Q.front();
            Q.pop();
            inq[u] = false;
            for(int i = 0; i < G[u].size(); 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] = true;
                    }
                }
            }
        }
        if(d[t] == INF) return false; //s-t 不连通，失败退出
        flow += a[t];
        cost += (long long)d[t] * (long long)a[t];
        int u = t;
        while(u != s)
        {
            edges[p[u]].flow += a[t];
            edges[p[u]^1].flow -= a[t];
            u = edges[p[u]].from;
        }
        return true;
    }

     pair<long long,int>Mincost(int s, int t)
    {
        long long cost = 0;
        int flow = 0;
        while(BellmanFord(s, t, flow, cost));
        return pair<long long ,int>{cost,flow};
    }
};

MCMF solver;
int n;


int main()
{
    while(true)
    {
        scanf("%d", &n);
        if(n == 0) break;
        solver.init(n + 1);
        int m, from, to, cost;
        scanf("%d", &m);
        for(int i = 0; i < m; i++)
        {
            scanf("%d%d%d", &from, &to, &cost);
            solver.AddEdge(from, to, 1, cost);
            solver.AddEdge(to, from, 1, cost);
        }
        solver.AddEdge(0, 1, 2, 0);

        pair<long long,int> ans = solver.Mincost(0, n);
        int flow = ans.second;

        if(flow != 2)
            puts("Back to jail");
        else
            printf("%lld
", ans.first);
    }
    return 0;
}