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. 題意：n個點，m條單向邊，每個點的牛都要到一個點k去並回來，要求來回路程的最大值。

剛開始看到這題，首先求出k點到每個點的最短距離，然後遍歷每一個點到k點的距離，求出兩段值的和的最大值。但這樣顯然會超時的。然後看到別人的部落格說將矩陣置換一下就好了，原因可以這樣理解：要求每個點到k點的最短距離，將路徑反過來再求k點到每個點的最短距離，那麼就變成了每個點到k點的最短距離了。

超時程式碼：

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<math.h>
using namespace std;
#define N 1000
#define inf 0x3fffffff
int m,n,k,dis1[N+1],dis[N+1],e[N+1][N+1];
bool vis[N+1];
void dijkstra(int x)
{
    memset(vis,false,sizeof(vis));
    for(int i=0;i<n;i++)
        dis[i]=e[x][i];
    for(int i=0;i<n-1;i++)
    {
        int tmp=inf,u=-1;
        for(int j=0;j<n;j++)
        {
            if(tmp>dis[j]&&!vis[j]&&dis[j])
            {
                tmp=dis[j];
                u=j;
            }
        }
        if(u==-1) break;
        vis[u]=true;
        for(int j=0;j<n;j++)
        {
            if(j==x) continue;
            if(!vis[j]&&e[u][j]&&(dis[j]>dis[u]+e[u][j]||!dis[j]))
                dis[j]=dis[u]+e[u][j];
        }
    }
}
int main()
{
    int a,b,c,ans=0;
    scanf("%d%d%d",&n,&m,&k);
    for(int i=0;i<m;i++)
    {
        scanf("%d%d%d",&a,&b,&c);
        e[a-1][b-1]=c;
    }
    for(int i=0;i<n;i++)
        dis1[i]=e[k-1][i];
    for(int i=0;i<n-1;i++)
    {
        int t=inf,uu=-1;
        for(int j=0;j<n;j++)
        {
            if(t>dis1[j]&&!vis[j]&&dis1[j])
            {
                t=dis1[j];
                uu=j;
            }
        }
        if(uu==-1) break;
        vis[uu]=true;
        for(int j=0;j<n;j++)
        {
            if(j==k-1) continue;
            if(!vis[j]&&e[uu][j]&&(dis1[j]>dis1[uu]+e[uu][j]||!dis1[j]))
                dis1[j]=dis1[uu]+e[uu][j];
        }
    }
//    for(int i=0;i<n;i++)
//        printf("%d ",dis1[i]);
    for(int i=0;i<n;i++)
    {
        if(i==k-1) continue;
        dijkstra(i);
        ans=max(ans,dis[k-1]+dis1[i]);
    }
    printf("%d\n",ans);
}

正解：

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<math.h>
using namespace std;
#define N 1000
#define inf 0x3fffffff
int m,n,k,dis1[N+1],dis[N+1],e[N+1][N+1];
bool vis[N+1];
void dijkstra(int x)
{
    memset(vis,false,sizeof(vis));
    for(int i=0;i<n;i++)
        dis[i]=e[x][i];
    for(int i=0;i<n-1;i++)
    {
        int tmp=inf,u=-1;
        for(int j=0;j<n;j++)
        {
            if(tmp>dis[j]&&!vis[j]&&dis[j])
            {
                tmp=dis[j];
                u=j;
            }
        }
        if(u==-1) break;
        vis[u]=true;
        for(int j=0;j<n;j++)
        {
            if(j==x) continue;
            if(!vis[j]&&e[u][j]&&(dis[j]>dis[u]+e[u][j]||!dis[j]))
                dis[j]=dis[u]+e[u][j];
        }
    }
}
int main()
{
    int a,b,c,ans=0;
    scanf("%d%d%d",&n,&m,&k);
    for(int i=0;i<m;i++)
    {
        scanf("%d%d%d",&a,&b,&c);
        e[a-1][b-1]=c;
    }
    dijkstra(k-1);
    for(int i=0;i<n;i++)
        dis1[i]=dis[i];
    for(int i=0;i<n;i++)
        for(int j=i+1;j<n;j++)
            swap(e[i][j],e[j][i]);
    dijkstra(k-1);
    for(int i=0;i<n;i++)
        ans=max(ans,dis1[i]+dis[i]);
    printf("%d\n",ans);
}