POJ 1502 MPI Maelstrom Dijkstra算法+输入处理

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

题意：信息传输，总共有n个传输机，先要从1号传输机向其余n⑴个传输机传输数据，传输需要时间，给出1个严格的下3角（其实就是对角线之下的不包括对角线的部份）时间矩阵，a[i][j]代表从i向j传输数据需要的时间，并规定数据传输之间并没有影响，即第1个传输机可以同时向其余传输机传输数据。求所有传输任务所需的最短时间。

解析：1个很裸的单源最短路，并且按题意可知边不可能为负值，那末直接用Dijkstra便可。求编号为1的传输机到所有传输机的最短传输时间以后，那末所有最短时间中的最大值即为完成传输任务的最短时间。

PS：这题的读入还是很恶心的，由于有x的存在，所以我们就需要把数字当做字符串去读，然后再将其转成int。

切记：无穷大不能开太大啊了，开大了如果超过数据类型的范围的话，就会算出来负数，结果固然也就不对了。

AC代码：

#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(){ //手写读入函数 char str[10]; scanf(%s,&str); if(!strcmp(str,x)) return INF; //'x'代表两点无路径，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函数 memset(vis,sizeof(vis)); //预处理 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); //此举为调试语句，可疏忽 #endif // sxk while(scanf(%d,&V)!=EOF){ for(int i=1; i<=V; i++) //读入数据 for(int j=1; j<=i; j++){ if(i == j) w[i][j] = 0; else w[j][i] = w[i][j] = input(); } dijkstra(); //处理 int ans = -INF; //找其中的最大跳跃高度 for(int i=1; i<=V; i++){ if(ans < dis[i]) ans = dis[i]; } printf(%d ,ans); } return 0; }