poj 3268 Silver Cow Party

Silver Cow Party
Time Limit: 2000MS
 
Memory Limit: 65536K
Total Submissions: 9222
 
Accepted: 4156
Description
One cow from each of N farms (1 ≤ N ≤ 1000) conveniently numbered 1..N is going to attend the big cow party to be 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) 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
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 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

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.
Source
USACO 2007 February Silver

 
 

#include <iostream>

#include <stdio.h>

#include <queue>

#include <stack>

#include <set>

#include <vector>

#include <math.h>

#include <string.h>

#include <algorithm>

using namespace std;

#define N 100002

#define Max 1000000002

struct node

{

    int to;

    int next;

    int value;

}Eg[N];

struct Record

{

   int x,y;

   int value;

}rc[N];



int dis1[1002],dis2[1002];

int v[1002];

bool b[1002];

int n,m,x;

void SFPA(int dis[])

{

   memset(b,0,sizeof(b));

   queue<int> Q;

   Q.push(x);

   b[x]=1;

   int e,i,k;

   while(!Q.empty())

   {

         i=Q.front(); Q.pop();

         b[i]=0;

       for(e=v[i];e!=-1;e=Eg[e].next)

       {

           k=Eg[e].to;

           if(dis[k]>dis[i]+Eg[e].value)

           {

               if(!b[k])  {

                    Q.push(k);

                    b[k]=1;

                    }

               dis[k]=dis[i]+Eg[e].value;

           }

       }

   }



}

int main()

{

   int i;

   while(scanf("%d %d %d",&n,&m,&x)!=EOF)

   {

     for(i=1;i<=n;i++)

         v[i]=-1;

     for(i=0;i<m;i++)

     {

        scanf("%d %d %d",&rc[i].x,&rc[i].y,&rc[i].value);

        Eg[i].value=rc[i].value;

        Eg[i].to=rc[i].y;

        Eg[i].next=v[rc[i].x];

        v[rc[i].x]=i;

     }

     for(i=1;i<=n;i++)

         dis1[i]=Max;

    dis1[x]=0;

     SFPA(dis1);

     __int64 ans=0;

     for(i=1;i<=n;i++)

         v[i]=-1;

     for(i=0;i<m;i++)

     {

        Eg[i].value=rc[i].value;

        Eg[i].to=rc[i].x;

        Eg[i].next=v[rc[i].y];

        v[rc[i].y]=i;

     }

     for(i=1;i<=n;i++)

         dis2[i]=Max;

     dis2[x]=0;

     SFPA(dis2);

     for(i=1;i<=n;i++)

        {

            dis1[i]+=dis2[i];

            ans=ans>dis1[i]?ans:dis1[i];

        }

     printf("%I64d
",ans);

   }

   return 0;

}