MPI Maelstrom
Time Limit:1000 MS |
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Memory Limit:10000 K |
Total Submissions:5538 |
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Accepted:3451 |
Question link: http://poj.org/problem? Id = 1502
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 special features of our network that we shoshould 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,. 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. that is, the second line of input will contain in one entry, A (2, 1 ). the next line will contain in two entries, A (3, 1) and A (3, 2), and so on.
Output
Your program shocould output the minimum communication time required to broadcast a message from the first processor to all the other processors.
Sample Input
55030 5100 20 5010x10
Sample Output
35
Solution:
The meaning is that n machines send information to each other, and the minimum time required for sending the remaining n-1 machines from the first machine is required.
Bare Dijkstra, which is stored in an adjacent matrix. First, pay attention to the initialization operation of the matrix. The value of the main diagonal is 0, because it is meaningless to send messages to yourself, and the other points are set to infinity. Next, what is amazing is its input, which is input according to the lower triangle matrix. g [I] [j] = g [j] [I], symmetric points have the same weight. If the input is x, it means that I to j has no edge, and it remains infinitely large.
In the template below, we finally traverse the obtained dis array to find the maximum value, which is the shortest time required. (This is a bit awkward. It can be understood that dis [I] represents the shortest time from src source point to I. Since n vertices need to be traversed, the time must be greater than or equal to the maximum value in dis)
Finally, after processing the input, we can see that some people directly use functions to simulate the conversion as few as I know about the function ····
Complete code:
# Include <functional> # include <algorithm> # include <iostream> # include <fstream> # include <sstream> # include <iomanip> # include <numeric> # include <cstring> # include <climits> # include <cassert> # include <complex> # include <cstdio> # include <string> # include <vector> # include <bitset> # include <queue> # include <stack> # include <cmath> # include <ctime> # include <list> # include <set> # include <map> using namespace std; # p Ragma comment (linker, "/STACK: 102400000,102400000") typedef long LL; typedef double DB; typedef unsigned uint; typedef unsigned long uLL;/** Constant List .. ** // {const int MOD = int (1e9) + 7; const int INF = 0x3f3f3f3f; const ll inff = 0x3f3f3f3f3f3fll; const db eps = 1e-9; const db oo = 1e20; const db pi = acos (-1.0); // M_PI; int n; string s; int g [201] [201]; int dis [201]; bool vis [201]; int res; Void init () {for (int I = 1; I <= n; I ++) {for (int j = 1; j <= n; j ++) {if (I = j) g [I] [j] = 0; else g [I] [j] = INF ;}} void dijkstra () {for (int I = 1; I <= n; I ++) {dis [I] = INF;} dis [1] = 0; memset (vis, 0, sizeof (vis); for (int I = 1; I <= n; I ++) {int mark =-1; int mindis = INF; for (int j = 1; j <= n; j ++) {if (! Vis [j] & dis [j] <mindis) {mindis = dis [j]; mark = j ;}} vis [mark] = 1; for (int j = 1; j <= n; j ++) {if (! Vis [j]) {dis [j] = min (dis [j], dis [mark] + g [mark] [j]) ;}} res =-INF; for (int k = 1; k <= n; k ++) {if (dis [k]> res) res = dis [k] ;}} int main () {# ifdef DoubleQ freopen ("in.txt", "r", stdin); # endif while (~ Scanf ("% d", & n) {init (); for (int I = 2; I <= n; I ++) {for (int j = 1; j <I; j ++) {cin> s; if (s = "x") continue; else {int sum = 0; int jin = 1; int len = s. length (); for (int k = len-1; k> = 0; k --) {if (k! = Len-1) jin * = 10; sum + = jin * (s [k]-'0');} g [I] [j] = sum; g [j] [I] = sum ;}} dijkstra (); cout <res <endl ;}}
POJ1502 (Dijkstra)