MPI Maelstrom(POJ-1502)

Problem 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

题意:给出一 n 个点的无向图的邻接矩阵的下三角部分,输出从第 0 个点到其他所有点的距离的最大值

思路:补全邻接矩阵,Dijkstra 模版直接处理即可

Source Program

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#define PI acos(-1.0)
#define E 1e-6
#define MOD 16007
#define INF 0x3f3f3f3f
#define N 1001
#define LL long long
using namespace std;
int n,m;
struct Edge{//边
    int from;//下一条边的编号
    int to;//边到达的点
    int dis;//边的长度
    Edge(int f,int t,int d){//构造函数
        from=f;
        to=t;
        dis=d;
    }
};

struct HeapNode{//Dijkstra用到的优先队列的结点
    int dis;//点到起点距离
    int u;//点的序号
    HeapNode(int a,int b){
        dis=a;
        u=b;
    }
    bool operator < (const HeapNode &rhs) const  {
        return dis > rhs.dis;
    }
};

struct Dijkstra{
    int n,m;//边数、点数,均从0开始
    vector edges;//边列表
    vector G[N];//每个结点出发的边的编号
    bool vis[N];//是否走过
    int dis[N];//起点s到各点的距离
    int p[N];//从起点s到i的最短路中的最后一条边的编号

    void init(int n) {//初始化
        this->n = n;
        for(int i=0;i Q;//优先队列
        Q.push(HeapNode(0,s));
        while(!Q.empty()) {
            HeapNode x=Q.top();
            Q.pop();

            int u=x.u;
            if(vis[u])//若已被访问
                continue;

            vis[u]=true;//标记为已访问
            for(int i=0;i dis[u]+e.dis) {//更新距离
                    dis[e.to] = dis[u]+e.dis;
                    Q.push(HeapNode(dis[e.to],e.to));
                }
            }
        }
    }
}DJ;

int main()
{
    while(scanf("%d",&n)!=EOF&&n)
    {
        DJ.init(n);//初始化
        for(int i=1;i

 

你可能感兴趣的:(#,POJ,#,图论——最短路)