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_backusing 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;}