POJ 3268 Silver Cow Party shortest circuit

Original title Link: http://poj.org/problem?id=3268

Silver Cow Party

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 15545		Accepted: 7053

Description

One cow from each of N Farms (1≤ n ≤1000) conveniently numbered 1.. N is 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 re Quires ti (1≤ ti ≤100) units 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 IWith three space-separated integers: Ai, Bi, and Ti. The described road runs from farm AiTo farm Bi, requiring TiTime units to traverse.

Output

Line 1:one integer:the maximum of the time any one cow must walk.

Sample Input

4 8 21 2 41 3 21 4 72 1 12 3 53 1 23 4 44 2 3

Sample Output

10

Hint

Cow 4 proceeds directly to the party (3 units) and returns via farms 1 and 3 (7 units), for a total of ten time units.

Source

Usaco February Silver Test instructions

There's a bull party, the other cows go, the cows take the shortest way, and the party ends up back at home. Ask which cow to walk the longest path, output the longest path.

Exercises

Running on both sides of the SPFA, is running two times against the opposite. And then it's done.

Code

#include <iostream>#include<queue>#include<cstring>#include<algorithm>#include<vector>#include<cstdio>#defineINF 1000006#defineMax_n 1003using namespacestd;structNode { Public:    intu, c; Node (intUuintcc): U (UU), C (cc) {} node () {}};structEdge { Public:    intto, cost; Edge (intTintc): to (t), cost (c) {} edge () {}};queue<node>que;intn,m,x;voidSPFA (intS,vector<edge> g[],intd[]) {Fill (d, D+ N +1, INF); Que.push (node (s),0)); D[s]=0;  while(Que.size ()) {node now=Que.front ();        Que.pop (); if(now.c! = d[now.u])Continue; intU =now.u;  for(inti =0; I < g[u].size (); i++) {            intv =g[u][i].to; intt = D[u] +G[u][i].cost; if(T <D[v]) {D[v]=T;            Que.push (Node (v, t)); }}}}vector<edge>G[max_n],rg[max_n];intD[max_n],rd[max_n];intMain () {scanf ("%d%d%d", &n, &m, &x);  for(inti =0; I < m; i++) {        intu, V, c; scanf ("%d%d%d", &u, &v, &c);        G[u].push_back (Edge (V, c));    Rg[v].push_back (Edge (U, c));    } SPFA (x, G, D);  while(Que.size ()) Que.pop ();    SPFA (x, RG, RD); intAns =0;  for(inti =1; I <= N; i++) ans = max (ans, d[i] +Rd[i]); cout<<ans<<Endl; return 0;}

POJ 3268 Silver Cow Party shortest circuit