POJ 3268 Silver Cow Party (Shortest way)

POJ 3268 Silver Cow Party

Description
One cow from each of N farms (1≤n≤1000) conveniently numbered 1..N are going to attend the big Cow party to being held at Farm #X (1≤x≤n). A total of M (1≤m≤100,000) unidirectional (one-way roads connects pairs of farms; Road I requires Ti (1≤ti≤100) UN Its of time to traverse.

Each of the cow must walk to the "party" and "when the" is "over" return to her farm. Each cow is a lazy and thus picks an optimal route with the shortest time. A Cow ' s return route might is different from her original route to the party since roads is one-way.

Of all the cows, what's the longest amount of time a cow must spend walking to the party and back?

Input
Line 1:three space-separated integers, respectively:n, M, and X
Lines 2.. M+1:line I+1 describes road I with three space-separated Integers:ai, Bi, and Ti. The described road runs from farm Ai to farm Bi, requiring Ti time units to traverse.
Output
Line 1:one integer:the maximum of the time any one cow must walk.
Sample Input
4 8 2
1 2 4
1 3 2
1 4 7
2 1 1
2 3 5
3 1 2
3 4 4
4 2 3
Sample Output
10

The main topic: There are N farms, M Road, and on the X Farm Open party. Now in addition to the other farms on the X farm, there are cattle to the X farm open Party,party after the end of each home, of course, the shortest way to go. This n-1 the largest short-circuit of the return in a cow. Problem-solving ideas: first to ask for the X as the starting point of the shortest, and then the side of the reverse read again, and then to the X as the starting point of the shortest. Add up to two d arrays and find the maximum value, which is the answer.

#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <cstdlib>#include <queue>using namespace STD;typedef Long LongllConst intINF =0x3f3f3f3f;Const intN =1005;Const intM =500005;intN, M, s;structEdge {intFrom, to, Dist; }; vector<Edge>edges, edgess; vector<int>G[M], g2[m];intVis[n], D[n], rec[n];intL[n];voidInit () { for(inti =0; I <= N; i++) {L[i] =0; Vis[i] =0; } for(inti =0; I < (n/2) * N; i++) g[i].clear (); for(inti =0; I < (n/2) * N;      i++) g2[i].clear ();      Edges.clear (); Edgess.clear ();}voidAddedge (intFromintTo,intDist) {Edges.push_back (Edge) {from, To, Dist}); Edgess.push_back (Edge) {To, from, Dist});intpos = Edges.size (); G2[to].push_back (POS-1); G[from].push_back (POS-1);}voidInput () {intU, V, Len; for(inti =0; I < m; i++) {scanf(" %d%d%d", &u, &v, &len);     Addedge (U, V, Len); }}voidSPFA ( vector<int>*gg, vector<Edge>E) {memset(Vis,0,sizeof(VIS)); for(inti =0; i < N; i++) D[i] = INF; for(inti =0; i < N;    i++) Rec[i] = INF; D[s] =0; Rec[s] =0; Vis[s] =1; Queue<int>Q; Q.push (s); while(! Q.empty ()) {intU = Q.front ();          Q.pop (); Vis[u] =0; for(inti =0; I < gg[u].size (); i++) {intv = e[gg[u][i]].to;if(D[v] > D[u] + e[gg[u][i]].dist)                  {D[v] = D[u] + e[gg[u][i]].dist; REC[V] = Rec[u] +1;if(!vis[v]) {Vis[v] =1;                  Q.push (v); }              }Else if(D[v] = = D[u] + e[gg[u][i]].dist) {if(Rec[v] > Rec[u] +1) {Rec[v] = Rec[u] +1;if(!vis[v]) {Vis[v] =1;                      Q.push (v); }                  }              }          }      }  }intMain () {intx, Max; while(scanf(" %d%d%d", &n, &m, &x) = =3) {Max =0;             Init ();        Input ();        s = x; SPFA (G, edges); for(inti =1; I <= N; i++) {if(i = = x)Continue;        L[i] + = D[i]; } SPFA (G2, edgess); for(inti =1; I <= N; i++) {if(i = = x)Continue;        L[i] + = D[i]; } for(inti =1; I <= N;        i++) {max = max (l[i], max); }printf("%d\n", Max); }return 0;}

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POJ 3268 Silver Cow Party (Shortest way)