UVALive-2957 Bring them there (maximum flow graph theory modeling)

The main problem: There are n planets, your task is to use the shortest time to the K supercomputer from Planet S to the planet T. Each supercomputer needs a spaceship to transport it, there are m two-way tunnels between the planets, each tunnel takes a day to pass, and no two ships can use the same tunnel at the same time, and the tunnel does not connect two identical planets, and there is at most one tunnel between each pair of planets.

The idea of solving problems: according to Great White books, the idea is to dismantle
For example, the delivery time is T, then each point u split into T + 1, respectively, u0, U1 ... uT, respectively, corresponds to the first day of the U Planet
For each given edge (U,V), assuming that I day, even the edge ui–>vi+1 and vi–>ui+1, two sides of the capacity of 1
You also have to add a point ui->ui+1, which is INF, which means that the ship is not moving forward.
The calculation of the process, the gradual expansion of the graph, day-by-day enumeration, the day-to-day expansion until the flow of K
Note here, at some point, a–>bi+1 and bi–>ai+1 at the same time the flow is not allowed, here, if there is a flow in two directions, then we are equivalent to the two spacecraft to stay on planet A and Planet B each day, so it will not be contradictory

#include <cstdio>#include <cstring>#include <algorithm>#include <queue>using namespace STD;#define M 1000010#define N 10010#define INF 0x3f3f3f3fstructedge{intU, V, cap, flow, next;} E[M];structdinic{intHead[n], d[n];intTot, sink, source;voidInit () {memset(Head,-1,sizeof(head)); tot =0; }inline voidAddedge (intUintVintCAP) {e[tot].u = u; E[TOT].V = v; E[tot].cap = cap; E[tot].flow =0; E[tot].next = Head[u];        Head[u] = tot++; u = u ^ v; v = u ^ v;        u = u ^ v; e[tot].u = u; E[TOT].V = v; E[tot].cap =0; E[tot].flow =0; E[tot].next = Head[u];    Head[u] = tot++; }inline BOOLBFsints) {intU, v;memset(d,0,sizeof(d)); Queue<int>Q;        Q.push (s); D[s] =1; while(! Q.empty ()) {u = Q.front (); Q.pop ();if(U = = sink)return true; for(inti = Head[u]; ~i; i = e[i].next) {v = e[i].v;if(!d[v] && e[i].cap-e[i].flow >0) {D[v] = D[u] +1;                Q.push (v); }            }        }return false; }intDfsintXintA) {if(x = = Sink | | a = =0)returnAintF, Flow =0; for(inti = head[x]; ~i; i = e[i].next) {intv = e[i].v;if(D[v] = = D[x] +1&& E[i].cap-e[i].flow >0) {f = Dfs (V, min (A, e[i].cap-e[i].flow));                E[i].flow + = f; e[i^1].flow-= f;                Flow + + F; A-= f;if(!a) Break; }        }if(Flow = =0) D[x] =0;returnFlow }intMaxflow (intSourceintSinkintNeed) {intFlow =0; This->sink = sink; while(BFS (source)) {Flow + = DFS (source, need-flow);if(Flow >= need)returnFlow }returnFlow }};D inic dinic;structNode {intx, y;} Node[n];intN, M, K, S, t;intPos[n];BOOLVis[n];voidPrintintCNT) { for(inti =1; I <= K; i++) Pos[i] = s;printf("%d\n", CNT); for(inti =1; I <= CNT; i++) { vector<int>U, v;memset(Vis,0,sizeof(VIS)); for(intj =0; J <4* m; J + =4) {intp = j + (I-1) *4* m + n * I *2;//positive flow, reverse flow, i.e. U-->v            if(E[p].flow &&!) e[p+2].flow) {U.push_back (e[p ^1].V-(I-1) * n);            V.push_back (E[P].V-I * n); }//Reverse flow, positive no flow, that is, v-->u, so that the two sides have to exclude the flow of the situation, the equivalent of two points of exchange, that is, both ships stay in situ            Else if(! E[p].flow && E[p +2].flow) {u.push_back (e[(p +2) ^1].V-(I-1) * n); V.push_back (E[p +2].v-i * N); }        }printf("%d", U.size ()); for(intp =0; P < u.size (); p++) for(intj =1; J <= K; J + +) {if(!vis[j] && pos[j] = = U[p]) {Vis[j] =1;printf("%d%d", J, V[p]); POS[J] = v[p]; Break; }            }printf("\ n"); }}voidSolve () { for(inti =0; I < m; i++)scanf("%d%d", &node[i].x, &AMP;NODE[I].Y);intCNT =0, Maxflow =0, sink = t; Dinic.init (); while(Maxflow < k) {++cnt;//stay where you are         for(inti =1; I <= N; i++) Dinic. Addedge (i + (CNT-1) * N, i + CNT * n, INF);//Side         for(inti =0; I < m; i++) {dinic. Addedge (node[i].x + (CNT-1) * N, NODE[I].Y + cnt * N,1); Dinic. Addedge (Node[i].y + (CNT-1) * N, node[i].x + cnt * N,1);        } sink + = n; Maxflow + = Dinic.    Maxflow (S, sink, k-maxflow); } print (CNT);}intMain () { while(scanf("%d%d%d%d%d", &n, &m, &k, &s, &t)! = EOF) solve ();return 0;}

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UVALive-2957 Bring them there (maximum flow graph theory modeling)