poj 1125:Stockbroker Grapevine
2014-06-23 12:07
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解题思路:
floyd算法,求解最短路径
(1)二维矩阵每一对顶点的最短距离。
(2)依次扫描每一个点,以其为基点再遍历所有每一对顶点之间距离的值,检查是否可通过该基点让这对顶点间的距离更小。
(3)注意算法中循环嵌套的顺序。
#include<iostream>
#include<cmath>
using namespace std;
int dis[110][110];
const int inf = 20;
int n,m;
int maxTime;
int startMan;
void floyd()
{
for(int k=1;k<=n;k++)
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
if(i!=j)
dis[i][j] = min(dis[i][j],dis[i][k]+dis[k][j]);
for(int i=1;i<=n;i++)
{
int maxPath = 0;
for(int j=1;j<=n;j++)
if(i!=j)
maxPath = max(maxPath,dis[i][j]);
if(maxPath < maxTime)
{
maxTime = maxPath;
startMan = i;
}
}
if(maxTime==inf)
cout<<"disjoint"<<endl;
else
cout<<startMan<<" "<<maxTime<<endl;
return;
}
int main()
{
while(true)
{
cin>>n;
if(n==0)
break;
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
dis[i][j] = inf;
for(int i=1;i<=n;i++)
dis[i][i] = 0;
for(int i=1;i<=n;i++)
{
cin>>m;
for(int j=1;j<=m;j++)
{
int p,q;
cin>>p>>q;
dis[i][p] = q;
}
}
maxTime = inf;
startMan = 0;
floyd();
}
return 0;
}
floyd算法,求解最短路径
(1)二维矩阵每一对顶点的最短距离。
(2)依次扫描每一个点,以其为基点再遍历所有每一对顶点之间距离的值,检查是否可通过该基点让这对顶点间的距离更小。
(3)注意算法中循环嵌套的顺序。
#include<iostream>
#include<cmath>
using namespace std;
int dis[110][110];
const int inf = 20;
int n,m;
int maxTime;
int startMan;
void floyd()
{
for(int k=1;k<=n;k++)
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
if(i!=j)
dis[i][j] = min(dis[i][j],dis[i][k]+dis[k][j]);
for(int i=1;i<=n;i++)
{
int maxPath = 0;
for(int j=1;j<=n;j++)
if(i!=j)
maxPath = max(maxPath,dis[i][j]);
if(maxPath < maxTime)
{
maxTime = maxPath;
startMan = i;
}
}
if(maxTime==inf)
cout<<"disjoint"<<endl;
else
cout<<startMan<<" "<<maxTime<<endl;
return;
}
int main()
{
while(true)
{
cin>>n;
if(n==0)
break;
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
dis[i][j] = inf;
for(int i=1;i<=n;i++)
dis[i][i] = 0;
for(int i=1;i<=n;i++)
{
cin>>m;
for(int j=1;j<=m;j++)
{
int p,q;
cin>>p>>q;
dis[i][p] = q;
}
}
maxTime = inf;
startMan = 0;
floyd();
}
return 0;
}
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