POJ1502(最短路入门题)
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 7471 | Accepted: 4550 |
Description
``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 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
Sample Input
5
50
30 5
100 20 50
10 x x 10
Sample Output
35
题意:给出结点数,用邻接矩阵给出每个节点的距离,因为结点i到结点i的距离为0以及路径是双向的所以只给出邻接矩阵对角线的左下半块。
下面分别用求最短路径的方法实现
/*
dijkstra 1502 Accepted 428K 0MS G++
*/
#include"cstdio"
#include"cstring"
#include"algorithm"
using namespace std;
const int MAXN=;
const int INF=0x3fffffff;
int mp[MAXN][MAXN];
int V;
int dijkstra(int s)
{
int d[MAXN];
int vis[MAXN];
for(int i=;i<=V;i++)
{
vis[i]=;
d[i]=mp[s][i];
} int n=V;
while(n--)
{
int mincost,k;
mincost=INF;
for(int i=;i<=V;i++)
{
if(!vis[i]&&mincost>d[i])
{
mincost=d[i];
k=i;
}
} vis[k]=;
for(int i=;i<=V;i++)
{
if(!vis[i]&&d[i]>d[k]+mp[k][i])
{
d[i]=d[k]+mp[k][i];
}
}
} int ans=-;
for(int i=;i<=V;i++) ans=max(ans,d[i]);
return ans;
}
int main()
{
while(scanf("%d",&V)!=EOF)
{
for(int i=;i<=V;i++)
for(int j=;j<=i;j++)
if(i==j) mp[i][j]=;
else{
char x[];
scanf("%s",x);
if(x[]=='x') mp[i][j]=mp[j][i]=INF;
else mp[i][j]=mp[j][i]=atoi(x);
} printf("%d\n",dijkstra());
}
return ;
}
/*
堆优化dijkstra 1502 1502 Accepted 632K 0MS G++
*/
#include"cstdio"
#include"cstring"
#include"algorithm"
#include"vector"
#include"queue"
using namespace std;
const int MAXN=;
const int INF=0X3fffffff;
struct Edge{
int to,cost;
Edge(){}
Edge(int to,int cost)
{
this->to=to;
this->cost=cost;
}
friend bool operator<(const Edge &a,const Edge &b)
{
return a.cost < b.cost;
}
};
vector<Edge> G[MAXN];
int V;
int dijkstra(int s)
{
int d[MAXN];
for(int i=;i<=MAXN;i++) d[i]=INF;
d[s]=; priority_queue<Edge> que;
que.push(Edge(s,));
while(!que.empty())
{
Edge e=que.top();que.pop();
int v=e.to;
if(d[v]<e.cost) continue;
for(int i=;i<G[v].size();i++)
{
Edge ek=G[v][i];
if(d[ek.to]>d[v]+ek.cost)
{
d[ek.to]=d[v]+ek.cost;
que.push(Edge(ek.to,d[ek.to]));
}
}
}
int ans=-;
for(int i=;i<=V;i++)
if(d[i]<INF)
ans=max(ans,d[i]);
return ans;
}
int main()
{
while(scanf("%d",&V)!=EOF)
{
for(int i=;i<=V;i++)
G[i].clear();
for(int i=;i<=V;i++)
for(int j=;j<=i;j++)
if(i==j){
G[i].push_back(Edge(j,));
G[j].push_back(Edge(i,));
}
else{
char x[];
scanf("%s",x);
if(x[]=='x') ;
else{
G[j].push_back(Edge(i,atoi(x)));
G[i].push_back(Edge(j,atoi(x)));
}
}
printf("%d\n",dijkstra());
}
return ;
}
/*
ford 1502 Accepted 444K 0MS G++
*/
#include"cstdio"
#include"cstring"
#include"algorithm"
using namespace std;
const int MAXN=;
const int INF=0X3fffffff;
struct Edge{
int from,to,cost;
}es[MAXN];
int V,E;
int ford(int s)
{
int d[MAXN];
for(int i=;i<=V;i++) d[i]=INF;
d[s]=; while(true)
{
bool update=false;
for(int i=;i<E;i++)
{
Edge e=es[i];
if(d[e.from]!=INF&&d[e.to]>d[e.from]+e.cost)
{
d[e.to]=d[e.from]+e.cost;
update=true;
}
}
if(!update) break;
}
int ans=-;
for(int i=;i<=V;i++)
{
if(d[i]<INF) ans=max(ans,d[i]);
}
return ans;
}
int main()
{
while(scanf("%d",&V)!=EOF)
{
E=;
for(int i=;i<=V;i++)
for(int j=;j<i;j++)
{
char x[];
scanf("%s",x);
if(x[]!='x')
{
es[E].from=i,es[E].to=j,es[E++].cost=atoi(x);
es[E].from=j,es[E].to=i,es[E++].cost=atoi(x);
}
} printf("%d\n",ford());
} return ;
}
/*
poj1502 spfa Accepted 232K 16MS C++
*/
#include<cstdio>
#include<cstring>
#include<queue>
#include<algorithm>
#include<vector>
using namespace std;
const int MAXN=;
const int INF=0x3fffffff;
struct Edge{
int to,w;
Edge(){}
Edge(int cto,int cw):to(cto),w(cw){}
};
vector<Edge> mp[MAXN];
int n;
int d[MAXN],vis[MAXN];
void spfa(int s)
{
memset(vis,,sizeof(vis));
for(int i=;i<=n;i++)
d[i]=INF;
queue<int> que;
d[s]=,vis[s]=;
que.push(s);
while(!que.empty())
{
int now=que.front();que.pop();
vis[now]=;
for(int i=;i<mp[now].size();i++)
{
Edge e=mp[now][i];
if(d[e.to]>d[now]+e.w)
{
d[e.to]=d[now]+e.w;
if(!vis[e.to])
{
vis[e.to]=;
que.push(e.to);
}
}
}
}
int res=;
for(int i=;i<=n;i++)
res=max(d[i],res);
printf("%d\n",res);
}
int main()
{
while(scanf("%d",&n)!=EOF)
{ for(int i=;i<=n;i++)
mp[i].clear();
for(int i=;i<=n;i++)
for(int j=;j<i;j++)
{
char x[]="\0";
scanf("%s",x);
if(x[]=='x') continue;
else
{
int l=atoi(x);
mp[i].push_back(Edge(j,l));
mp[j].push_back(Edge(i,l));
}
}
spfa();
}
}
/*
floyd 1502 Accepted 428K 0MS G++
*/
#include"cstdio"
#include"algorithm"
using namespace std;
const int MAXN=;
const int INF=0X3fffffff;
int mp[MAXN][MAXN];
int V;
int main()
{
while(scanf("%d",&V)!=EOF)
{
for(int i=;i<=V;i++)
for(int j=;j<=i;j++)
{
if(i==j){
mp[i][j]=;
continue;
}
char x[];
scanf("%s",x);
if(x[]!='x') mp[i][j]=mp[j][i]=atoi(x);
else mp[i][j]=mp[j][i]=INF;
} for(int k=;k<=V;k++)
for(int i=;i<=V;i++)
for(int j=;j<=V;j++)
mp[i][j]=min(mp[i][j],mp[i][k]+mp[k][j]);
int ans=-;
for(int i=;i<=V;i++)
if(mp[][i]<INF)
ans=max(mp[][i],ans); printf("%d\n",ans);
} return ;
}
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