UVA 10806 Dijkstra, Dijkstra.(费用流)

n个点的无向带权图，求1->n的最短往返路径，不走重复边。

这里涉及到一个知识点：求无向图上s->t的最短路，其实就是费用流。

而求1->n最短往返路径呢？增加源点s，由s到1加弧，容量为2（往返两次），费用为0；而对于原图中的边<u, v>，分别由u到v，由v到u增加容量为1（往返不能走重边），费用为边权的弧。然后跑费用流得到的最小费用便是答案。如果最后求得的最大流小于2，则说明无解。

 

#include<algorithm>

#include<iostream>

#include<cstring>

#include<cstdlib>

#include<fstream>

#include<sstream>

#include<bitset>

#include<vector>

#include<string>

#include<cstdio>

#include<cmath>

#include<stack>

#include<queue>

#include<stack>

#include<map>

#include<set>

#define FF(i, a, b) for(int i=a; i<b; i++)

#define FD(i, a, b) for(int i=a; i>=b; i--)

#define REP(i, n) for(int i=0; i<n; i++)

#define CLR(a, b) memset(a, b, sizeof(a))

#define debug puts("**debug**")

#define LL long long

#define PB push_back

using namespace std;



const int maxn = 111;

const int INF = 1e9;

int n, m, s, t, d[maxn], p[maxn], a[maxn], inq[maxn];

int flow, cost;

struct Edge

{

    int from, to, cap, flow, cost;

};

vector<Edge> edges;

vector<int> G[maxn];



inline void init()

{

    flow = cost = s = 0, t = n;

    REP(i, t+1) G[i].clear(); edges.clear();

}



void add(int from, int to, int cap, int cost)

{

    edges.PB((Edge){from, to, cap, 0, cost});

    edges.PB((Edge){to, from, 0, 0, -cost});

    int nc = edges.size();

    G[from].PB(nc-2);

    G[to].PB(nc-1);

}



bool spfa(int& flow, int& cost)

{

    REP(i, t+1) d[i] = INF;

    CLR(inq, 0);

    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 nc = G[u].size();

        REP(i, nc)

        {

            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 += d[t] * 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;

}



int main()

{

    while(scanf("%d", &n), n)

    {

        scanf("%d", &m);

        init();

        int a, b, c;

        add(s, 1, 2, 0);

        while(m--)

        {

            scanf("%d%d%d", &a, &b, &c);

            add(a, b, 1, c);

            add(b, a, 1, c);

        }

        while(spfa(flow, cost));

        if(flow < 2) puts("Back to jail");

        else printf("%d\n", cost);

    }

    return 0;

}