POJ1502_MPI Maelstrom_最短路::朴素的dijkstra

MPI Maelstrom

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 9173 Accepted: 5613

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. 

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. 

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. 

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

55030 5100 20 5010 x x 10

Sample Output

35

大致题意：

从1号节点出发，向其他所有节点传递消息，同时可以向无限个节点传送消息。并且，接收到消息的点也可以立即向无数个没有接收到消息的点传递消息。问消息传遍整个网络的最小时间。

大体思路：

迪杰斯特算法 求起点到所有点的最短路。这些最短路中最大的那个就是答案。

#include<cstdio>#include<iostream>#include<cstdlib>using namespace std;#define _max 2147483647int Map [105][105];//地图记录int D [105];//起点到所有点的已知最短路bool Dis [105];//标记点是已是最短int N;int main (){scanf("%d",&N);char str [10];for(int i=2; i<=N; i++)for(int j=1; j<i; j++){scanf(" %s",str);//点的信息以字符串的形式输入if(*str != 'x') Map[j][i] = Map[i][j] = atoi(str);//双向边只输入一次//字符串转整数}/*******初始化D*******/for(int i=1; i<=N; i++)if(Map[1][i] != 0) D[i] = Map[1][i];else D[i] = _max;/*******迪杰斯特算法*******/for(int m=1; m<=N; m++){/****找到一个当前距起点最近的点 来利用****/int Min = _max, k;for(int i=1; i<=N; i++)if(! Dis[i] && D[i] < Min) Min = D[i], k = i;//如果dis为真，说明已经利用过了Dis[k] = 1;//能被利用，说明这个点已经达到了可能的最小值，标记为真for(int i=1; i<=N; i++)//遍历并标记if(Map[k][i] != 0 && D[i] > D[k] + Map[k][i])D[i] = D[k] + Map[k][i];}int Max = 0;for(int i=1; i<=N; i++)if(Max<D[i]) Max = D[i];printf("%d\n",Max);return 0;}