Silver Cow Party（2013.09.15）（最短路）

G. Silver Cow Party

Time Limit: 2000ms

Case Time Limit: 2000ms

Memory Limit: 65536KB

64-bit integer IO format:  %lld      Java class name:  Main

Submit  Status  PID: 3387

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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 irequires 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

这道题其实是一道最短路的变形。在建图的时候正反各建立一次，然后用spfa求得相应的距离。

这里要注意一下，以后建图的时候尽可能用邻接表来建立。（可以用数组，可以用vector、list）

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<queue>
#include<vector>
#define INF 10000000
using namespace std;
int n,m,x,dist_v1[1010]= {0},dist_v2[1010]= {0};
struct node
{
    int x,d;
};
vector<node> v1[1010];
vector<node> v2[1010];
void spfa_v1(int x)
{
    int i,j,visit[1010]= {0},t;
    queue<int> q;
    for (i=1; i<=n; i++)
        dist_v1[i]=INF;
    dist_v1[x]=0;
    visit[x]=1;
    q.push(x);
    int s,l;
    while(!q.empty())
    {
        t=q.front();
        q.pop();
        visit[t]=0;
        for (i=0; i<v1[t].size(); i++)
        {
            s=v1[t][i].x;
            l=v1[t][i].d;
            if (dist_v1[s]>dist_v1[t]+l)
            {
                dist_v1[s]=dist_v1[t]+l;
                if (!visit[s])
                {
                    visit[s]=1;
                    q.push(s);
                }
            }
        }
    }
}
void spfa_v2(int x)
{
    int i,j,visit[1010]= {0},t;
    queue<int> q;
    for (i=1; i<=n; i++)
        dist_v2[i]=INF;
    dist_v2[x]=0;
    visit[x]=1;
    q.push(x);
    int s,l;
    while(!q.empty())
    {
        t=q.front();
        q.pop();
        visit[t]=0;
        for (i=0; i<v2[t].size(); i++)
        {
            s=v2[t][i].x;
            l=v2[t][i].d;
            if (dist_v2[s]>dist_v2[t]+l)
            {
                dist_v2[s]=dist_v2[t]+l;
                if (!visit[s])
                {
                    visit[s]=1;
                    q.push(s);
                }
            }
        }
    }
}
int main ()
{
    int i,j;
    int a,b,t;
    scanf("%d%d%d",&n,&m,&x);
    node e;
    while(m--)
    {
        scanf("%d%d%d",&a,&b,&t);
        e.x=b;
        e.d=t;
        v1[a].push_back(e);
        e.x=a;
        v2[b].push_back(e);
    }
    spfa_v1(x);
    spfa_v2(x);
    int max=0,s1,s2;
    for (i=1; i<=n; i++)
    {
        if (max<dist_v1[i]+dist_v2[i])
            max=dist_v1[i]+dist_v2[i];
    }
    cout<<max<<endl;
    return 0;
}