Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 27173 | Accepted: 12421 |
Description
Flatopian towns are numbered from 1 to N. Each highway connects exactly two towns. All highways follow straight lines. All highways can be used in both directions. Highways can freely cross each other, but a driver can only switch between highways at a town that is located at the end of both highways.
The Flatopian government wants to minimize the length of the longest highway to be built. However, they want to guarantee that every town is highway-reachable from every other town.
Input
The first line of each case is an integer N (3 <= N <= 500), which is the number of villages. Then come N lines, the i-th of which contains N integers, and the j-th of these N integers is the distance (the distance should be an integer within [1, 65536]) between village i and village j. There is an empty line after each test case.
Output
Sample Input
1 3 0 990 692 990 0 179 692 179 0
Sample Output
692
Hint
題意:求出最小生成樹中,最長的那一段,還是比較裸的Prim
代碼 如下:
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#define MAXN 700
using namespace std;
int sum,n;
int dis[MAXN][MAXN];
int cost[MAXN];
bool vis[MAXN];
const int INF=0x3f3f3f;
void Prim(int start)
{
memset(vis,0,sizeof(vis));
for(int i=1;i<=n;i++)
cost[i]=dis[start][i];
sum=0;
vis[start]=1;
for(int i=2;i<=n;i++)
{
int nm=0x3f3f3f,k=0;
for(int j=1;j<=n;j++)
{
if(!vis[j] && cost[j]<nm)
{
nm=cost[j];
k=j;
}
}
vis[k]=true;
sum+=nm;
for(int j=1;j<=n;j++)
{
if(!vis[j] && dis[k][j]<cost[j])
cost[j]=dis[k][j];
}
}
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int ans=-1;
memset(dis,0x3f3f3f,sizeof(dis));
scanf("%d",&n);
for(int i=1;i<=n;i++)
{
for(int j=1;j<=n;j++)
{
scanf("%d",&dis[i][j]);
}
}
for(int i=1;i<=n;i++)
Prim(i);
for(int i=1;i<=n;i++)
ans=max(ans,cost[i]);
printf("%d\n",ans);
}
return 0;
}