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POJ 1502 MPI Maelstrom (Dijkstra算法+輸入處理)

來源:程序員人生   發布時間:2015-01-21 08:39:52 閱讀次數:3534次

MPI Maelstrom
Time Limit: 1000MS   Memory Limit: 10000K
Total Submissions: 5712   Accepted: 3553

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 n⑴ processors, they just do a sequence of n⑴ 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

Source

East Central North America 1996



題意:信息傳輸,總共有n個傳輸機,先要從1號傳輸機向其余n⑴個傳輸機傳輸數據,傳輸需要時間,給出1個嚴格的下3角(其實就是對角線之下的不包括對角線的部份)時間矩陣,a[i][j]代表從i向j傳輸數據需要的時間,并規定數據傳輸之間并沒有影響,即第1個傳輸機可以同時向其余傳輸機傳輸數據。求所有傳輸任務所需的最短時間。


解析:1個很裸的單源最短路,并且按題意可知邊不可能為負值,那末直接用Dijkstra便可。求編號為1的傳輸機到所有傳輸機的最短傳輸時間以后,那末所有最短時間中的最大值即為完成傳輸任務的最短時間。

PS:這題的讀入還是很惡心的,由于有x的存在,所以我們就需要把數字當做字符串去讀,然后再將其轉成int。

切記:無窮大不能開太大啊了,開大了如果超過數據類型的范圍的話,就會算出來負數,結果固然也就不對了。



AC代碼:

#include <cstdio> #include <cstring> #include <algorithm> using namespace std; #define INF 123456789 #define MAXN 100 + 2 int V, E; int w[MAXN][MAXN]; int vis[MAXN], dis[MAXN]; int input(){ //手寫讀入函數 char str[10]; scanf("%s", &str); if(!strcmp(str, "x")) return INF; //'x'代表兩點無路徑,用INF表示 else{ int ans = 0; int len = strlen(str); for(int i=0; i<len; i++){ ans = ans*10 + str[i] - '0'; } return ans; } } void dijkstra(){ //dijkstra函數 memset(vis, 0, sizeof(vis)); //預處理 for(int i=1; i<=V; i++) dis[i] = (i == 1 ? 0 : INF); for(int i=1; i<=V; i++){ //dijkstra int x, m = INF; for(int y=1; y<=V; y++) if(!vis[y] && dis[y] <= m){ x = y; m = dis[x]; } vis[x] = 1; for(int y=1; y<=V ;y++) dis[y] = min(dis[y], dis[x] + w[x][y]); //松弛操作 } } int main(){ #ifdef sxk freopen("in.txt", "r", stdin); //此舉為調試語句,可疏忽 #endif // sxk while(scanf("%d", &V)!=EOF){ for(int i=1; i<=V; i++) //讀入數據 for(int j=1; j<=i; j++){ if(i == j) w[i][j] = 0; else w[j][i] = w[i][j] = input(); } dijkstra(); //處理 int ans = -INF; //找其中的最大跳躍高度 for(int i=1; i<=V; i++){ if(ans < dis[i]) ans = dis[i]; } printf("%d ", ans); } return 0; }




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