UVA 10806 Dijkstra, Dijkstra. (Minimum charge maximum flow)

UVA 10806 Dijkstra, Dijkstra. The main topic: you and your partner want to escape. Your partner goes to the scout first, and when your partner comes to the station, he will call you (the phone is hidden in the cake), and then you can run to the station where someone will pick you up. But because your partner was wearing a prison suit when he was running to the station, the streets he passed were under martial law and you had to run from the other streets. If you can get to the train station, please output the shortest time you and your partner will spend on the road, if not please "back to jail". Problem solving idea: minimum fee maximum flow. Set up a super source point to the prison (start 1), a capacity of 2 (two people), set a super meeting point, make the train station (end N) connect to him, the capacity is 2 (two people), the rest of the streets are non-direction (that is, forward to reverse all to consider), and capacity of 1 (each street can only Finally, the maximum flow, if the maximum flow is 2 of the output of the minimum fee, or go back to prison for the next jailbreak.

#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <cstdlib>#include <queue>using namespace STD;typedef Long LongllConst intN = the;Const intINF =0x3f3f3f3f;intN, M, S, t;intA[n], Pre[n], d[n], inq[n];structedge{intFrom, to, cap, flow; llCos;}; vector<Edge>Edges vector<int>g[3* N];voidInit () { for(inti =0; I <3N    i++) g[i].clear (); Edges.clear ();}voidAddedge (intFromintTo,intCapintFlow, LLCos{Edges.push_back (Edge) {from, to, Cap,0,Cos}); Edges.push_back (Edge) {To, from,0,0, -Cos});intm = Edges.size (); G[from].push_back (M-2); G[to].push_back (M-1);}voidInput () {intFrom, to; ll cost; for(inti =0; I < m; i++) {scanf("%d%d%lld", &from, &to, &cost); Addedge (from, to,1,0, cost); Addedge (To, from,1,0, cost); } Addedge (0,1,2,0,0); Addedge (n, n +1,2,0,0);}intBF (intSintTint& Flow, ll& cost) { Queue<int>Q;memset(INQ,0,sizeof(INQ));memsetA0,sizeof(a));memset(Pre,0,sizeof(pre)); for(inti =0; I <=2* n +1;    i++) D[i] = INF; D[s] =0;    A[s] = INF; Inq[s] =1;intFlag =1; Pre[s] =0; Q.push (s); while(! Q.empty ()) {intU = Q.front ();        Q.pop (); Inq[u] =0; for(inti =0; I < g[u].size (); i++) {Edge &e = edges[g[u][i]];if(E.cap > E.flow && d[e.to] > D[u] + E.Cos) {d[e.to] = D[u] + E.Cos;                A[e.to] = min (A[u], e.cap-e.flow); Pre[e.to] = G[u][i];if(!inq[e.to]) {Inq[e.to] =1;                Q.push (e.to); }}} flag =0; }if(D[t] = = INF)return 0;    Flow + = A[t]; Cost + = (LL) d[t] * (LL) a[t]; for(intu = t; U! = S;        U = edges[pre[u]].from) {Edges[pre[u]].flow + = a[t]; edges[pre[u]^1].flow-= a[t]; }return 1;}intMCMF (intSintT, ll& cost) {intFlow =0; Cost =0; while(BF (S, t, flow, cost));returnFlow;}intMain () { while(scanf("%d", &n)! = EOF) {if(n = =0) Break;scanf("%d", &m); s =0, T = n +1;        Init ();        Input (); ll cost =0;intAns = MCMF (s, t, cost);if(ans = =1)printf("Back to Jail\n");Else printf("%lld\n", cost); }return 0;}

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UVA 10806 Dijkstra, Dijkstra. (Minimum charge maximum flow)