POJ 3268 Silver Cow Party 最短路—dijkstra算法的优化。

POJ 3268 Silver Cow Party

Description

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.

 

题目描述： 

一群牛分别从1~n号农场赶往x号农场参加聚会，农场与农场之间的路是单向的，在n个农场之间有m条路，给出 a ，b ， t表示从a号农场到b号农场需要t时间。 每头牛都会选择最短的路，问来回路上花费时间最长的牛花费的时间是多少。

思路描述：

题意并不难理解，即求农场 X 到其余农场的最短距离与其余农场到X农场的最短距离，农场 X 到其余农场的最短距离一次dijkstra便可得到，但是求其余农场到X农场的最短距离，需要n次dijkstra方可得到，观察数据量便可得知普通的dijkstra算法难以满足要求，因为普通的dijkstra算法时间复杂度为O（n^2），加之n+1次调用，根据数据量必定超时，所以下面代码便对普通的dijkstra算法进行了优化，使之时间复杂度变为O（n*n*log n），便可满足题目的要求。而主要的实现是通过优先队列对边进行排序，从而先去了每次寻找最小边所花费的时间，具体细节详见代码如下：

 

代码实现：

 1 #include <set>
 2 #include  ,int>,vectorint,int> >,greaterint,int> > >q;   //利用优先队列对里面的边从小到大进行排序；
33     for(int i = 1;i<=1199;i++) v[i].clear();   //vector 清空；
34     fill(dis,dis+1200,inf);                           //初始化为最大值；
35     N n2;
36     for(int i = 1;i<=m;i++)                         //将初始边加入到vector中；
37     {
38         n2.e = M[i].cost;
39         n2.to = M[i].v;
40         v[M[i].u].push_back(n2);
41     }
42     dis[s] = 0;
43     //优化与实现：
44     q.push(pair<int,int>(0,s));
45     while(!q.empty())
46     {
47         pair<int ,int> p = q.top();
48         q.pop();
49         int v1 = p.second;
50         if(dis[v1]continue;
51         for(int i = 0;i)
52         {
53             N n1 = v[v1][i];
54             if(dis[n1.to]>dis[v1]+n1.e)
55             {
56                 dis[n1.to]=dis[v1]+n1.e;
57                 q.push(pair<int,int>(dis[n1.to],n1.to));
58             }
59         }
60     }
61 }
62 int main()
63 {
64     int sum[1200];
65     while(~scanf("%d %d %d",&n,&m,&x))
66     {
67         memset(sum,0,sizeof(sum));
68         for(int i = 1;i<=m;i++) scanf("%d %d %d",&M[i].u,&M[i].v,&M[i].cost);
69         dijkstra(x);
70         for(int i = 1;i<=n;i++) sum[i]+=dis[i];
71         for(int i = 1;i<=n;i++)
72         {
73                 dijkstra(i);
74                 sum[i] += dis[x];
75         }
76         int max1 = 0;
77         for(int i = 1;i<=n;i++) if(sum[i]>max1)  max1 = sum[i];
78         printf("%d\n",max1);
79     }
80     return 0;
81 }

 

 

 

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转载于:https://www.cnblogs.com/LGJC1314/p/7231215.html