(简单) POJ 1502 MPI Maelstrom，Dijkstra。

　　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.''

 

　　就是求最短路里面最大的那个。。。

 

代码如下：

#include<iostream>
#include<cstring>
#include<cstdio>

using namespace std;

const int MaxN=110;
const int INF=10e8;

bool vis[MaxN];

void Dijkstra(int cost[][MaxN],int lowcost[],int n,int start)
{
    for(int i=1;i<=n;++i)
    {
        vis[i]=0;
        lowcost[i]=INF;
    }
    lowcost[start]=0;

    for(int j=1;j<=n;++j)
    {
        int k=-1;
        int minn=INF;

        for(int i=1;i<=n;++i)
            if(!vis[i] && lowcost[i]<minn)
            {
                minn=lowcost[i];
                k=i;
            }

        if(k==-1)
            break;

        vis[k]=1;

        for(int i=1;i<=n;++i)
            if(!vis[i])
                lowcost[i]=min(lowcost[k]+cost[k][i],lowcost[i]);
    }
}

int ans[MaxN];
int map1[MaxN][MaxN];

int main()
{
    int n;
    int maxn;
    char c[20];

    while(~scanf("%d",&n))
    {
        for(int i=1;i<=n;++i)
            for(int j=1;j<=i;++j)
                if(i==j)
                    map1[i][j]=0;
                else
                {
                    scanf("%s",c);

                    if(c[0]!='x')
                        sscanf(c,"%d",&map1[i][j]);
                    else
                        map1[i][j]=INF;

                    map1[j][i]=map1[i][j];
                }

        Dijkstra(map1,ans,n,1);

        maxn=-1;

        for(int i=1;i<=n;++i)
            if(ans[i]>maxn)
                maxn=ans[i];

        cout<<maxn<<endl;
    }

    return 0;
}