Dijkstra-POJ-1502-MPI Maelstrom
2015-10-14 19:37
363 查看
MPI Maelstrom
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 7182 Accepted: 4375
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.
“How is Apollo’s port of the Message Passing Interface (MPI) working out?” Swigert asked.
“Is there anything you can do to fix that?”
“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
这真的是一道英语阅读题。。。无力吐槽。
直接求出单元最短路径,然后求出最远的距离输出就是所需时间。
再次用kb巨的优先队列模板做了一发。
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 7182 Accepted: 4375
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
5
50
30 5
100 20 50
10 x x 10
Sample Output
35
Source
East Central North America 1996
这真的是一道英语阅读题。。。无力吐槽。
直接求出单元最短路径,然后求出最远的距离输出就是所需时间。
// // main.cpp // 最短路练习-G-MPI maelstrom // // Created by 袁子涵 on 15/10/14. // Copyright © 2015年 袁子涵. All rights reserved. // // 32ms 812KB #include <iostream> #include <string.h> #include <stdio.h> #include <stdlib.h> #define minof(a,b) ((a)>(b)?(b):(a)) #define maxof(a,b) ((a)>(b)?(a):(b)) #define INF 999999999 using namespace std; int n; long long int map[105][105],dis[105],mins,out; bool book[105]; void dijkstra() { int next=0; for (int i=0; i<n; i++) { dis[i]=map[0][i]; } book[0]=1; for (int i=0; i<n; i++) { mins=INF; for (int j=0; j<n; j++) { if (mins>dis[j] && book[j]==0) { mins=dis[j]; next=j; } } book[next]=1; for (int j=0; j<n; j++) { if (mins+map[next][j]<dis[j]) { dis[j]=mins+map[next][j]; } } } } int main(int argc, const char * argv[]) { char in1[20]; long long int in2; memset(book, 0, sizeof(book)); cin >> n; for (int i=0; i<n; i++) { for (int j=0; j<i; j++) { cin >> in1; if (in1[0]=='x') { map[i][j]=INF; map[j][i]=INF; continue; } in2=atoi(in1); map[i][j]=in2; map[j][i]=in2; } map[i][i]=0; } dijkstra(); out=0; for (int i=0; i<n; i++) { out=maxof(out, dis[i]); } cout << out << endl; return 0; }
再次用kb巨的优先队列模板做了一发。
// // main.cpp // POJ-1502-MPI Maelstrom // // Created by 袁子涵 on 15/11/28. // Copyright © 2015年 袁子涵. All rights reserved. // // 16ms 804KB #include <iostream> #include <string.h> #include <stdio.h> #include <math.h> #include <algorithm> #include <stdlib.h> #include <vector> #include <queue> #define MAXN 105 #define INF 0x3f3f3f3f using namespace std; int n; struct qnode { int v,c; qnode(int _v=0,int _c=0):v(_v),c(_c){} bool operator <(const qnode &r) const { return c>r.c; } }; struct Edge { int v,cost; Edge(int _v=0,int _cost=0):v(_v),cost(_cost){} }; vector<Edge>E[MAXN]; bool vis[MAXN]; int dist[MAXN]; void Dijkstra(int start) { memset(vis, 0, sizeof(vis)); for (int i=1; i<=n; i++) { dist[i]=INF; } priority_queue<qnode>que; while (!que.empty()) { que.pop(); } dist[start]=0; que.push(qnode(start,0)); qnode tmp; while (!que.empty()) { tmp=que.top(); que.pop(); int u=tmp.v; if (vis[u]) { continue; } vis[u]=1; for (int i=0; i<E[u].size(); i++) { int v=E[tmp.v][i].v; int cost=E[u][i].cost; if (!vis[v]&&dist[v]>dist[u]+cost) { dist[v]=dist[u]+cost; que.push(qnode(v,dist[v])); } } } } void addedge(int u,int v,int w) { E[u].push_back(Edge(v,w)); } int main(int argc, const char * argv[]) { char x[20]; cin >> n; for (int i=1; i<=n; i++) { E[i].clear(); } for (int i=1; i<=n; i++) { for (int j=1; j<i; j++) { scanf("%s",x); if (x[0]=='x') { continue; } else { int temp=atoi(x); addedge(i,j,temp); addedge(j,i,temp); } } } Dijkstra(1); int out=0; for (int i=2; i<=n; i++) { out=max(out,dist[i]); } cout << out << endl; return 0; }
相关文章推荐
- codeforces 585B Phillip and Trains(bfs)
- lintcode 容易题:Count and Say 报数
- 信息项目管理师-整体管理知识点
- scheduleAtFixedRate 与 scheduleWithFixedDelay 区别
- 10个必需的iOS开发工具和资源
- Android 设计模式模式适配器
- CMMI(软件能力成熟度模型集成)名词术语整理
- SQL查询语句
- 1077. Kuchiguse
- 【hihocoder】1237 : Farthest Point ->微软2016校招在线笔试题
- iOS开发-14款状态栏(StatusBar)开源软件
- 笔试题——char能否存中文汉字
- 信息系统项目管理师:论项目的质量管理
- load和initialize的区分
- 1027. Colors in Mars
- 关于与运算和取余之间的关系
- 面试用友的前后;笔试跟面试
- The first blog:不称职的软件工程本科生到码农研究生的转变。
- NIM 博弈游戏
- 1015. Reversible Primes