POJ 3268 Silver Cow Party(SPFA)

POJ 3268 Silver Cow Party(SPFA)

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 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?

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: 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.

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.

兩次SPFA水過。

題意：N個農場，每個農場的都有一個奶牛去參加派對，M條單向路。

求對於所有奶牛來說花費在路上的最長時間（PS：奶牛都是選擇最短的路）

1：以X為源點，各頂點的出邊構成鄰接表，求源點到各頂點的最短路。

2：以X為源點，各頂點的入邊做出邊，構成鄰接表，求源點到各頂電的最短路。

將兩次時間加起來最大的即為答案。

#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));
    queueq;
    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