hdu 2485 Highways(最小生成数）

Highways

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu

Submit Status

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

130 990 692990 0 179692 179 0

Sample Output

692

题意：

每一次的第一行是一个整数N(3 < = N < = 500),这是村庄的数量。然后N行,其中包含N个整数,第i行和j 个整数为村庄i到j的的距离(距离应该是整数在[65536])村庄之间我和村j。在每个测试用例有一个空行。

求将村庄连接的最小公路中的最长一段

#include <iostream> #include <string.h> #include <stdio.h> #include <algorithm> using namespace std;const int maxn=125010;const int N=510; const int inf=0x7fffffff;int parent[N]; int t,n; struct Node{ int from,to,edge; }node[maxn];  void init(int n)  { for(int i=1;i<=n;i++) parent[i]=i; }  int find(int x) { int r=x;while(r!=parent[r])r=parent[r]; return r;}  bool cmp(Node a,Node b){ return a.edge<b.edge; }  int main() { scanf("%d",&t); while(t--) { scanf("%d",&n);int m=0; int len;  for(int i=1;i<=n;i++) for(int j=1;j<=n;j++){ scanf("%d",&len); if(i<j)  { int temp=m+1;  node[temp].from=i; //存储数据 node[temp].to=j;  node[temp].edge=len;m++; } } m=n*(n-1)/2; //数据总数init(n);  sort(node+1,node+1+m,cmp);len=-inf;for(int i=1;i<=m;i++) { int x=find(node[i].from);int y=find(node[i].to); if(x==y) continue; else { parent[x]=y; if(len<node[i].edge)len=node[i].edge; } } printf("%d\n",len);  } return 0;  }