POJ 3268 Silver Cow Party（Dijkstra算法求解来回最短路问题）-优快云博客

本文详细解析了如何使用反向存储和Dijkstra算法解决POJ-3268题目，即计算N个农场中每头牛参加聚会并返回的最长总时间。通过实例输入输出，展示了算法的具体实现过程。

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题目链接：

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

Sample Output

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.

 1 /*
 2     关键是反向存储，求最短路的思维转换 
 3 */
 4 #include<stdio.h>
 5 #include<string.h>
 6 #include<algorithm>
 7 using namespace std;
 8 void Dijkstra(int s,int e[][1010]);
 9 #define inf 99999999
10 int dis[1010],book[1010],e[1010][1010],f[1010][1010],d[1010],n,m;
11 int main()
12 {
13     int i,j,k,x,a,b,c,maxn,temp;
14     while(scanf("%d%d%d",&n,&m,&x)!=EOF)
15     {
16         memset(d,0,sizeof(d));
17         for(i=1;i<=n;i++)
18             for(j=1;j<=n;j++)
19                 if(i==j)
20                 {
21                     e[i][j]=0;
22                     f[i][j]=0;
23                 }
24                 else
25                 {
26                     e[i][j]=inf;
27                     f[i][j]=inf;
28                 }
29                     
30         for(i=1;i<=m;i++)
31         {
32             scanf("%d%d%d",&a,&b,&c);
33             if(c<e[a][b])
34             {
35                 e[a][b]=c;
36                 f[b][a]=c;
37             }
38         }
39         Dijkstra(x,e);//回 
40         Dijkstra(x,f);//去 
41         maxn=-1;
42         for(i=1;i<=n;i++)
43         {
44             if(d[i]>maxn)    
45                 maxn=d[i];
46         }    
47         printf("%d\n",maxn);
48     }
49     return 0;
50 }
51 void Dijkstra(int s,int e[][1010])
52 {
53     int i,j,k,min,u;
54     for(i=1;i<=n;i++)
55         dis[i]=e[s][i];
56     memset(book,0,sizeof(book));
57     book[s]=1;
58     for(k=1;k<n;k++)
59     {
60         min=inf;
61         for(i=1;i<=n;i++)
62             if(book[i]==0&&dis[i]<min)
63             {
64                 min=dis[i];
65                 u=i;
66             }
67         book[u]=1;
68         for(i=1;i<=n;i++)
69             if(book[i]==0&&dis[i]>dis[u]+e[u][i])
70                 dis[i]=dis[u]+e[u][i];
71     }    
72     for(i=1;i<=n;i++)
73         d[i]+=dis[i];
74 }

转载于:https://www.cnblogs.com/wenzhixin/p/9429531.html