poj 1125 Stockbroker Grapevine Floyd(基础题)

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d(k)[i][j]：i->j中间点序号不大于k时 =min(i-j中间节点序号小于k,i->j中间节点序号为k) 

d(k)[i][j]=min(d(k-1)[i][j],d(k-1)[i][k]+d(k-1)[k][j]))

#include <iostream>
#include <algorithm>
#include <cstring>
#include <vector>
#include <cstdio>
#include <queue>
using namespace std;
typedef pair<int,int> ii;
typedef long long ll;
const int N=2e2+20;
const ll inf=1e9;
int d

;
int n;
void Floyd()
{
//d(k)[i][j]：i->j中间点序号不大于k时 =min(i-j中间节点序号小于k,i->j中间节点序号为k)
//d(k)[i][j]=min(d(k-1)[i][j],d(k-1)[i][k]+d(k-1)[k][j]))
for(int k=1;k<=n;k++)
{
for(int i=1;i<=n;i++)
{
for(int j=1;j<=n;j++)
{
d[i][j]=min(d[i][j],d[i][k]+d[k][j]);
}
}
}
}
int main()
{
while(cin>>n&&n)
{
int a,b;
memset(d,0x3f3f3f,sizeof(d));
for(int i=1;i<=n;i++)
{
int x;
cin>>x;
for(int j=1;j<=x;j++)
{
cin>>a>>b;
d[i][a]=b;
}
}
Floyd();
int pos,ans=0,res=1e9;
int used
;
for(int i=1;i<=n;i++)//首选i时
{
ans=0;
for(int j=1;j<=n;j++)
{
if(i==j) continue;
ans=max(ans,d[i][j]);//从i中同时发送出去时
}
if(res>ans)
{
res=ans;
pos=i;
}
}
cout<<pos<<' '<<res<<endl;
}
return 0;
}