Description
The islandnation of Flatopia is perfectly flat. Unfortunately, Flatopia has no publichighways. So the traffic is difficult in Flatopia. The Flatopian government isaware of this problem. They're planning to build some highways so that it willbe possible to drive between any pair of towns without leaving the highwaysystem.
Flatopian towns are numbered from 1 to N. Each highway connects exactly twotowns. All highways follow straight lines. All highways can be used in bothdirections. Highways can freely cross each other, but a driver can only switchbetween 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 tobe built. However, they want to guarantee that every town is highway-reachablefrom every other town.
Input
The firstline 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 thenumber 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 aninteger within [1, 65536]) between village i and village j. There is an emptyline after each test case.
Output
For eachtest case, you should output a line contains an integer, which is the length ofthe longest road to be built such that all the villages are connected, and thisvalue is minimum.
Sample Input
1
3
0 990 692
990 0 179
692 179 0
Sample Output
692
分析：
最小生成树问题，但不是求边的权值之和，而是求最小生成树中最长的那条边的长度。用Prim算法和Kruskal算法均可以。由于题目的测试数据是以邻接矩阵的形式给出的，因此可以考虑优先使用Prim算法。如果使用Kruskal算法，还需要把邻接矩阵中的顶点和边分离出来。
prim代码：
#include<stdio.h>
#include<iostream>
using namespace std;
int n;
int tu[505][505];
int dis[505],vis[505];
void prim()
{
for(int i=1;i<=n;i++)//初始化，所有点的距离以及是否访问过
{
dis[i]=tu[1][i];
vis[i]=0;
}
vis[1]=1;//标记1号顶点已经访问过
int Min;
int k;
int ans=0;//保存最小生成树里面的最大边
//int sun=0;//最小生成树的值
for(int i=1;i<n;i++)
{
Min=0x3f3f3f3f;
for(int j=1;j<=n;j++)
{
if(vis[j]==0&&Min>dis[j])
{
Min=dis[j];
k=j;
}
}
if(ans<dis[k])
ans=dis[k];
vis[k]=1;//标记k这个点已经访问过
//sum+=dis[k];
for(int j=1;j<=n;j++)
{
if(vis[j]==0&&dis[j]>tu[k][j])
dis[j]=tu[k][j];
}
}
printf("%d
",ans);
//printf("%d
",sum);
}
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
scanf("%d",&n);
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
scanf("%d",&tu[i][j]);
prim();
}
}
Kruskal算法：
#include<stdio.h>
#include<iostream>
#include<algorithm>
using namespace std;
int n;
int Count;
int tu[505][505];
int Tree[505];//并查集的数组
struct Node
{
int a,b,w;
} node[505*505];
bool cmp(Node A,Node B)
{
if(A.w!=B.w)
return A.w<B.w;
if(A.a!=B.a)
return A.a<B.a;
return A.b<B.b;
}
int findRoot(int x) //递归查找顶点x所在树的根
{
if (Tree[x] == x) return x;//若为x,则x的根就是自身
else //否则
{
int tmp =findRoot(Tree[x]);//递归查找x的父亲Tree[x]的根,tmp为最终的树根
Tree[x]= tmp; //查找过程中进行路径压缩:把x到根之间遇到的所有顶点的父亲设为tmp
return tmp; //返回树根
}
}
void Kruskal()
{
int ans=0;
int k=0;
int sum=0;
for(int i=0; i<Count; i++)//遍历排序好的每一条边
{
if(k==n-1) break;//如果当前的生成树中已经有了n-1条边，就不用在接着往下找了
int a=findRoot(node[i].a);
int b=findRoot(node[i].b);
if(a!=b)
{
k++;
Tree[a]=b;
//sum+=node[i].w;
if(ans<node[i].w)
ans=node[i].w;
}
}
printf("%d
",ans);
//printf("%d
",sum);
}
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
scanf("%d",&n);
for(int i=1; i<=n; i++)
for(int j=1; j<=n; j++)
{
scanf("%d",&tu[i][j]);
}
for(int i=1; i<=n; i++)
Tree[i]=i;//标记没一个点都是只属于自身的集合
Count=0;
for(int i=1; i<=n; i++)
for(int j=1; j<i; j++)
{
node[Count].a=i;
node[Count].b=j;
node[Count].w=tu[i][j];
Count++;
}
sort(node,node+Count,cmp);
Kruskal();
}
return 0;
}