poj 3268 Silver Cow Party（最短路）

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.

思路
这题要求走去走回两次路得到的最短路的最大值，用spfa算法先求正走的最短路，将邻接表倒置，求反走的最短路，相加求它们的最大值

代码实现

#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
typedef long long ll;
const int N=1005;
typedef pair P;
int n,m,x;
ll dis[N],inf=ll(1e18), sl[N];
struct Edge
{
    int v;
    int cost;
};
vector edge1[N],edge2[N];
bool vis[N];
void spfa(vector edge[])
{
    queue point;
    memset(vis,false,sizeof(vis));
    for (int i = 1; i <=n; i++)
        dis[i] = inf;
    point.push(x);
    dis[x]=0;
    vis[x]=true;
    while(point.size())
    {
        int u=point.front();
        point.pop();
        vis[u]=false;
        int l=edge[u].size();
        for(int i=0; idis[u]+cost)
            {
                dis[v]=dis[u]+cost;
                if(!vis[v])
                {
                    point.push(v);
                    vis[v]=true;
                }
            }
        }
    }
}
int main()
{
    scanf("%d %d %d",&n,&m,&x);
    for(int i=1,u,v,w; i<=m; i++)
    {
        scanf("%d %d %d",&u,&v,&w);
        edge1[u].push_back({v,w});
        edge2[v].push_back({u,w});
    }
    spfa(edge1);
    for(int i=1; i<=n; i++)
        sl[i]+=dis[i];
    spfa(edge2);
    for(int i=1; i<=n; i++)
        sl[i]+=dis[i];
    ll ans=0;
    for(int i=1; i<=n; i++)
        ans=max(ans,sl[i]);
    printf("%lld\n",ans);
    return 0;
}