POJ 3268 Silver Cow Party (SPFA)

POJ 3268 Silver Cow Party (SPFA)

Description

One cow from eachNFarms (1 ≤N≤ 1000) conveniently numbered 1 ..NIs going to attend the big cow party to be held at farm #X(1 ≤X≤N). A totalM(1 ≤MLess than or equal to 100,000) unidirectional (one-way roads connects pairs of farms; roadIRequiresTi(1 ≤Ti≤ 100) units of time to traverse.

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

Of all the cows, what is 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 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 10 time units.

Two SPFA times.

N farms, each of which has a cow for a party, M one-way roads.

The longest time spent on the road for all cows (PS: the shortest path for the cows)

1: X is the source point, and the outdegree of each vertex forms an adjacent table, and finds the shortest short circuit from the source point to each vertex.

2: Use X as the source point, and the inbound edges of each vertex form an edge list to find the shortest path from the source point to each top point.

The greatest sum of the two times is the answer.

#include
 
  #include
  
   #include
   
    #include#include
    
     #include
     
      using namespace std;const int maxn=101000;const int INF=0x3f3f3f3f;int head[maxn],end[maxn];int next[maxn],cost[maxn];int u[maxn],v[maxn],w[maxn];int d1[maxn],d2[maxn],e;int visit[maxn],cnt[maxn];void init(){    e=0;    memset(head,-1,sizeof(head));}void add(int u,int v,int w){    end[e]=v;    cost[e]=w;    next[e]=head[u];    head[u]=e++;}void spfa(int x,int n){    memset(visit,0,sizeof(visit));    memset(cnt,0,sizeof(cnt));    memset(d1,INF,sizeof(d1));    queue
      
       q; q.push(x); visit[x]=1; cnt[x]++; d1[x]=0; while(!q.empty()) { int uu=q.front(); q.pop(); visit[uu]=0; for(int i=head[uu];i!=-1;i=next[i]) { int vv=end[i]; int ww=cost[i]; if(d1[vv]>d1[uu]+ww) { d1[vv]=d1[uu]+ww; if(!visit[vv]) { visit[vv]=1; q.push(vv); if(++cnt[vv]>n) return ; } } } }}int main(){ int n,m,x; while(~scanf("%d%d%d",&n,&m,&x)) { init(); for(int i=0;i