Highways
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 27493 | Accepted: 12554 |
Description
The island nation of Flatopia is perfectly flat. Unfortunately, Flatopia has no public highways. So the traffic is difficult in Flatopia. The Flatopian government is aware of this problem. They‘re planning to build some highways so that it will be possible
to drive between any pair of towns without leaving the highway system.
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 input is an integer T, which tells how many test cases followed.
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
For each test case, you should output a line contains an integer, which is the length of the longest road to be built such that all the villages are connected, and this value is minimum.
Sample Input
Sample Output
1
3
0 990 692
990 0 179
692 179 0
692
Hint
Huge input,scanf is recommended.
Source
POJ Contest,Author:[email protected]
题目大意：要想连接所有城市并建造高速公路，使得花费最小的情况下，输出消耗最大的那条高速公路需要的花费。
思路：克鲁斯卡尔贪心求最大边，代码实现很简单。
AC代码：
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
struct path
{
int x,y,w;
}a[500*500+50];
int f[5000];
int cmp(path a,path b)
{
return a.w<b.w;
}
int find(int a)
{
int r=a;
while(f[r]!=r)
r=f[r];
int i=a;
int j;
while(i!=r)
{
j=f[i];
f[i]=r;
i=j;
}
return r;
}
void merge(int a,int b)
{
int A,B;
A=find(a);
B=find(b);
if(A!=B)
f[B]=A;
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int n;
int cont=0;
scanf("%d",&n);
for(int i=0;i<n;i++)
{
f[i]=i;
for(int j=0;j<n;j++)
{
int k;
scanf("%d",&k);
a[cont].x=i;
a[cont].y=j;
a[cont++].w=k;
}
}
int output;
int tmp=0;
sort(a,a+cont,cmp);
for(int i=0;i<cont;i++)
{
if(find(a[i].x)!=find(a[i].y))
{
merge(a[i].x,a[i].y);
tmp++;
if(tmp==n-1)//如果是最后一条入边，那么这条边一定是权值最大边
{
output=a[i].w;
}
}
}
printf("%d\n",output);
}
}