POJ 1502 MPI Maelstrom

Posted 2020-08-09 风中的簌雨

MPI Maelstrom

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 20000/10000K (Java/Other)

Total Submission(s) : 2   Accepted Submission(s) : 1

Problem Description

BIT has recently taken delivery of their new supercomputer, a 32 processor Apollo Odyssey distributed shared memory machine with a hierarchical communication subsystem. Valentine McKee‘s research advisor, Jack Swigert, has asked her to benchmark the new system.
``Since the Apollo is a distributed shared memory machine, memory access and communication times are not uniform,‘‘ Valentine told Swigert. ``Communication is fast between processors that share the same memory subsystem, but it is slower between processors that are not on the same subsystem. Communication between the Apollo and machines in our lab is slower yet.‘‘

``How is Apollo‘s port of the Message Passing Interface (MPI) working out?‘‘ Swigert asked.

``Not so well,‘‘ Valentine replied. ``To do a broadcast of a message from one processor to all the other n-1 processors, they just do a sequence of n-1 sends. That really serializes things and kills the performance.‘‘

``Is there anything you can do to fix that?‘‘

``Yes,‘‘ smiled Valentine. ``There is. Once the first processor has sent the message to another, those two can then send messages to two other hosts at the same time. Then there will be four hosts that can send, and so on.‘‘

``Ah, so you can do the broadcast as a binary tree!‘‘

``Not really a binary tree -- there are some particular features of our network that we should exploit. The interface cards we have allow each processor to simultaneously send messages to any number of the other processors connected to it. However, the messages don‘t necessarily arrive at the destinations at the same time -- there is a communication cost involved. In general, we need to take into account the communication costs for each link in our network topologies and plan accordingly to minimize the total time required to do a broadcast.‘‘

 

 

Input

The input will describe the topology of a network connecting n processors. The first line of the input will be n, the number of processors, such that 1 <= n <= 100. <br> <br>The rest of the input defines an adjacency matrix, A. The adjacency matrix is square and of size n x n. Each of its entries will be either an integer or the character x. The value of A(i,j) indicates the expense of sending a message directly from node i to node j. A value of x for A(i,j) indicates that a message cannot be sent directly from node i to node j. <br> <br>Note that for a node to send a message to itself does not require network communication, so A(i,i) = 0 for 1 <= i <= n. Also, you may assume that the network is undirected (messages can go in either direction with equal overhead), so that A(i,j) = A(j,i). Thus only the entries on the (strictly) lower triangular portion of A will be supplied. <br> <br>The input to your program will be the lower triangular section of A. That is, the second line of input will contain one entry, A(2,1). The next line will contain two entries, A(3,1) and A(3,2), and so on.

 

Output

Your program should output the minimum communication time required to broadcast a message from the first processor to all the other processors.

 

Sample Input

5

50

30 5

100 20 50

10 x x 10

 

Sample Output

35

题目大意：给你一个不完全的矩阵，数字表示权值，x表示两点间不可达
　　　　　由于自身到自身花费的时间为0，所以没有给出，由于i到j和j到i距离相同，互达时间相同
　　　　　所以只给出了一半的临界矩阵。
　　　　　根据给你的这个临界矩阵，让你来求从点1到其他点所花费最短时间集里面的的最大值。
　　　　　其实这是一个很直接的最短路

解题思路：特殊处理一下值为x的情况，建立邻接阵，运用单源最短路径算法求出第一个处理器到其他所有处理器的最短传递信息时间，

　　　　　然后取这些最短时间中的最大者即可。

AC代码：

 1 #include <stdio.h>
 2 #include <string.h>
 3 #define inf 9999999
 4 int p[110][110];
 5 int dis[110];
 6 int visit[110];
 7 int n;
 8 void dijkstra()  //模板
 9 {
10     int i,j,pos,minn;
11     for (i = 1; i <= n; i ++)
12     {
13         dis[i] = p[1][i];
14         visit[i] = 0;
15     }
16     dis[1] = 0;
17 
18     for (i = 1; i <= n; i ++)
19     {
20         minn = inf;
21         for(j = 1; j <= n; j ++)
22         {
23             if (!visit[j] && dis[j] < minn)
24             {
25                 minn = dis[j];
26                 pos = j;
27             }
28         }
29         visit[pos] = 1;
30         for (j = 1; j <= n; j ++)
31         {
32             if (!visit[j] && dis[j] > dis[pos]+p[pos][j])
33                 dis[j] = dis[pos]+p[pos][j];
34         }
35     }
36 }
37 int change(char str[110])  //处理输入数据
38 {
39     int i;
40     int ans = 0;
41     for (i = 0; str[i]; i ++)
42         ans = ans*10+str[i]-‘0‘;
43     return ans;
44 }
45 int main ()
46 {
47     int i,j;
48     char str[110];
49     while (~scanf("%d",&n))
50     {
51         for (i = 1; i <= n; i ++)
52             for (j = 1; j <= n; j ++)
53                 if (i == j)
54                     p[i][j] = 0;
55                 else
56                     p[i][j] = inf;
57 
58         for (i = 2; i <= n; i ++)
59         {
60             for (j = 1; j < i; j ++)
61             {
62                 scanf("%s",str);  //将数据全存为字符型
63                 if (str[0] != ‘x‘)
64                     p[i][j] = p[j][i] = change(str);
65             }
66         }
67 
68         dijkstra();
69         int maxx = 0;
70         for (i = 2; i <= n; i ++)
71         {
72             if (dis[i] > maxx)
73                 maxx = dis[i];
74         }
75         printf("%d\n",maxx);
76     }
77     return 0;
78 }

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