題干：

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

5 50 30 5 100 20 50 10 x x 10

Sample Output

35

題目大意：

給你一個(gè)不完全的矩陣（下三角形），數(shù)字表示權(quán)值，x表示兩點(diǎn)間不可達(dá)
由于自身到自身花費(fèi)的時(shí)間為0，所以沒有給出，由于i到j(luò)和j到i距離相同，互達(dá)時(shí)間相同
所以只給出了一半的臨界矩陣。
根據(jù)給你的這個(gè)臨界矩陣，讓你來求從點(diǎn)1到其他點(diǎn)所花費(fèi)最短時(shí)間集里面的的最大值。
其實(shí)這是一個(gè)很直接的最短路

解題報(bào)告：

? ? ?用o(n^2)的方法，鄰接矩陣儲存圖，給一個(gè)下三角形矩陣，字符串讀入，用get函數(shù)處理一下成int型權(quán)值，然后跑一邊Dijkstra，再看從點(diǎn)1可以跑到那個(gè)點(diǎn)是最大值即可。

?

AC代碼：

#include<cstdio> #include<algorithm> #include<iostream> #include<cstring> using namespace std; const int MAX = 100000 + 5; const int INF = 0x3f3f3f3f ; int vis[MAX]; int maze[105][105]; int n,m; int dis[MAX];//表示從出發(fā)點(diǎn)開始到該點(diǎn)的最短距離。 void Dijkstra(int u,int v) {dis[u] = 0;int all = n,minw = INF,minv;//這里的初始化還是有講頭的。 for(int k = 1;k<= n;k++) {minw = INF;for(int i = 1; i<=n; i++) {if(vis[i] ) continue;if(dis[i] < minw) {minv = i; minw = dis[i];}}vis[minv] = 1; // if(minv == v) break;for(int i = 1; i<=n; i++) {if(vis[i]) continue;if(maze[minv][i] == 0) continue;dis[i] = min(dis[i], dis[minv] + maze[minv][i]);}} } int get(char *s) {if(s[0] == 'x') return INF;int len = strlen(s);int ans = 0;for(int i = 0; i<len; i++) {ans = ans*10 + s[i]-'0';} return ans; } int main() {int a,b;char w[20];while(~scanf("%d",&n) ) {memset(dis,0x3f3f3f3f,sizeof(dis) ) ;memset(maze,0,sizeof(maze) );memset(vis,0,sizeof(vis) ) ;for(int i = 2; i<=n; i++) {for(int j = 1; j<i; j++) {scanf("%s",w);maze[i][j] = maze[j][i] = get(w);}}Dijkstra(1,n); int ans = *max_element(dis+1,dis+n+1);printf("%d\n",ans);}return 0 ;}

?

?總結(jié)：

? ? ?只要改一下模板，不讓他在(minv == v) 的時(shí)候break掉就可以。并且此時(shí)傳入的v沒有任何卵用了。。。

? ? ?即 其實(shí)你傳入的v也只有在這個(gè)判斷的時(shí)候有用啊，其他時(shí)候你看模板里面都沒有v出現(xiàn)。所以針對這個(gè)題我們只需要不讓他在minv==v的時(shí)候停止，而是繼續(xù)跑程序，就可以了。最后輸出最大值。

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