大意：有n個牧場，編號1到n，每個牧場有一頭奶牛。現在所有奶牛要到編號為x的牧場聚會，路徑是單向的，奶牛都很聰明，只走最短路徑，問哪頭奶牛來回走的路徑之和最大，輸出這個最大值。

思路：建立兩個鄰接表（一個出邊表，一個入邊表），然後分別對兩個鄰接表使用一次SPFA，得到的路徑長度分別存到dist_o[]和dist_i[]。dist_o + dist_i的最大值就是問題結果。

程式碼

PS：第一次寫SPFA，做題時建鄰接表有個指標搞錯了找了半天bug，操蛋。。程式碼也寫的很糟糕。以後要多寫多練

#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
#include<algorithm>
using namespace std;

#define N 1000
#define INF 1000000000

int vis_i[N + 5], dist_i[N + 5], vis_o[N + 5], dist_o[N + 5];
int n, m;

struct node
{
    int to, val;
    node * next;
};

node * e_i[N + 5];
node * e_o[N + 5];

void INIT(int v)
{
    for (int i = 1; i <= n; i++)
    {
        if (v == i)
        {
            dist_i[i] = dist_o[i] = 0;
        }
        else
        {
            dist_i[i] = dist_o[i] = INF;
        }
    }
}

void SPFA_i(int v)
{
    queue<int> q;
    q.push(v);
    while (!q.empty())
    {
        int a = q.front();
        q.pop();
        vis_i[a] = 0;
        node * p = e_i[a];
        //cout << a;
        while (p)
        {
            //cout << "  " << p->to;
            if (dist_i[a] < INF && dist_i[a] + p->val < dist_i[p->to])
            {
                dist_i[p->to] = dist_i[a] + p->val;
                if (!vis_i[p->to])
                {
                    vis_i[p->to] = 1;
                    q.push(p->to);
                }
            }
            p = p->next;
        }
        //cout << endl;
    }
}

void SPFA_o(int v)
{
    queue<int> q;
    q.push(v);
    while (!q.empty())
    {
        int a = q.front();
        q.pop();
        vis_o[a] = 0;
        node * p = e_o[a];
        //cout << a;
        while (p)
        {
            //cout << "  " << p->to;
            if (dist_o[a] < INF && dist_o[a] + p->val < dist_o[p->to])
            {
                dist_o[p->to] = dist_o[a] + p->val;
                if (!vis_o[p->to])
                {
                    vis_o[p->to] = 1;
                    q.push(p->to);
                }
            }
            p = p->next;
        }
        //cout << endl;
    }
}

int main()
{
    int x;
    while (cin >> n >> m >> x)
    {
        for (int i = 0; i <= N; i++)
        {
            e_i[i] = e_o[i] = NULL;
        }
        node * p;
        for (int i = 1; i <= m; i++)
        {
            int a, b, t;
            scanf("%d %d %d",&a, &b, &t);
            p = new node;
            p->to = a;
            p->val = t;
            p->next = NULL;
            if (e_i[b] == NULL)
                e_i[b] = p;
            else
            {
                p->next = e_i[b];
                e_i[b] = p;
            }
            p = new node;
            p->to = b;
            p->val = t;
            p->next = NULL;
            if (e_o[a] == NULL)
                e_o[a] = p;
            else
            {
                p->next = e_o[a];
                e_o[a] = p;
            }
        }
        INIT(x);
        SPFA_i(x);
        SPFA_o(x);
        int ans = 0;
        for (int i = 1; i <=n; i++)
        {
            dist_i[i] += dist_o[i];
        }
        sort(dist_i + 1, dist_i + n + 1);
        cout << dist_i[n] << endl;
    }
    return 0;
}