Silver Cow Party

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 12674		Accepted: 5651

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 T_i (1 ≤ T_i ≤ 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?

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: A_i, B_i, and T_i. The described road runs from farm A_i to farm B_i, requiring T_i time units to traverse.

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

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

10

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.

//Memory   Time

// 3521K    241MS

//by : Snarl_jsb

#include<algorithm>

#include<cstdio>

#include<cstring>

#include<cstdlib>

#include<iostream>

#include<vector>

#include<queue>

#include<stack>

#include<iomanip>

#include<string>

#include<climits>

#include<cmath>

#define MAXV 1010

#define MAXE 100010

#define LL long long

using namespace std;

namespace Adj

{

    struct Node

    {

        int to,next,val;

        bool flag;

    } edge[MAXE<<1];

    int top,head[MAXV];

    void init()

    {

        top=1;

        memset(head,0,sizeof(head));

    }

    void addEdge(int u,int v,int val)

    {

        edge[top].to=v;

        edge[top].val=val;

        edge[top].flag=1;

        edge[top].next=head[u];

        head[u]=top++;

        edge[top].to=u;

        edge[top].val=val;

        edge[top].flag=0;

        edge[top].next=head[v];

        head[v]=top++;

    }

}

using namespace Adj;

int n,m,x,ans;

bool vis[MAXV];

int dis[MAXV];

int dis1[MAXV];

void spfa(bool flag)

{

    memset(vis,0,sizeof(vis));

    for(int i=0;i<=n;i++)

        dis[i]=INT_MAX;

    queue<int>Q;

    Q.push(x);

    vis[x]=1;

    dis[x]=0;

    while(!Q.empty())

    {

        int now=Q.front();

        Q.pop();

        vis[now]=0;

        for(int i=head[now];i;i=edge[i].next)

        {

            if(flag==1)

            {

                if(edge[i].flag==0)

                    continue;

            }

            else

            {

                if(edge[i].flag==1)

                    continue;

            }

            int son=edge[i].to;

            int val=edge[i].val;

            if(dis[now]+val<dis[son])

            {

                dis[son]=dis[now]+val;

                if(!vis[son])

                {

                    vis[son]=1;

                    Q.push(son);

                }

            }

        }

    }

    if(flag==1)

    {

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

            dis1[i]=dis[i];

    }

    else

    {

        int Max=INT_MIN;

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

        {

            dis[i]+=dis1[i];

            if(dis[i]>Max)

                Max=dis[i];

        }

        ans=Max;

    }

}

int main()

{

//    freopen("cin.txt","r",stdin);

//    freopen("cout.txt","w",stdout);

    while(~scanf("%d %d %d",&n,&m,&x))

    {

        init();

        int u,v,w;

        while(m--)

        {

            scanf("%d %d %d",&u,&v,&w);

            addEdge(u,v,w);

        }

        spfa(1);

        spfa(0);

        printf("%d\n",ans);

    }

    return 0;

}