POJ 2395 Out of Hay【最小生成树】

Out of Hay

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 14141		Accepted: 5489

Description

The cows have run out of hay, a horrible event that must be remedied immediately. Bessie intends to visit the other farms to survey their hay situation. There are N (2 <= N <= 2,000) farms (numbered 1..N); Bessie starts at Farm 1. She'll traverse some or all
of the M (1 <= M <= 10,000) two-way roads whose length does not exceed 1,000,000,000 that connect the farms. Some farms may be multiply connected with different length roads. All farms are connected one way or another to Farm 1. 

Bessie is trying to decide how large a waterskin she will need. She knows that she needs one ounce of water for each unit of length of a road. Since she can get more water at each farm, she's only concerned about the length of the longest road. Of course, she
plans her route between farms such that she minimizes the amount of water she must carry. 

Help Bessie know the largest amount of water she will ever have to carry: what is the length of longest road she'll have to travel between any two farms, presuming she chooses routes that minimize that number? This means, of course, that she might backtrack
over a road in order to minimize the length of the longest road she'll have to traverse.
Input

* Line 1: Two space-separated integers, N and M. 

* Lines 2..1+M: Line i+1 contains three space-separated integers, A_i, B_i, and L_i, describing a road from A_i to B_i of length L_i.
Output

* Line 1: A single integer that is the length of the longest road required to be traversed.
Sample Input

3 3
1 2 23
2 3 1000
1 3 43

Sample Output

43

Hint

OUTPUT DETAILS: 

In order to reach farm 2, Bessie travels along a road of length 23. To reach farm 3, Bessie travels along a road of length 43. With capacity 43, she can travel along these roads provided that she refills her tank to maximum capacity before she starts down a
road.
Source

USACO 2005 March Silver

题意：

给出若干个顶点，以及他们之间的某些道路的长度，规划一个方案使得所有点能相互到达，而且总的边权值最小，求其中所有的边中最长的边的权值

题解：

最小生成树的变形，只需要在生成树的时候进行边的最大值的更新就行了

/* http://blog.csdn.net/liuke19950717 */
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
const int maxn=2005;
int pre[maxn];
struct node
{
int a,b,len;
}edge[maxn*5];
void init(int n)
{
for(int i=1;i<=n;++i)
{
pre[i]=i;
}
}
int find(int x)
{
int r=x;
while(r!=pre[r])
{
r=pre[r];
}
int i=x,j;
while(i!=r)
{
j=pre[i];
pre[i]=r;
i=j;
}
return r;
}
int join(int x,int y)
{
int fx=find(x),fy=find(y);
if(fx!=fy)
{
pre[fx]=fy;
return 1;
}
return 0;
}
int cmp(node a,node b)
{
return a.len<b.len;
}
int kruscal(int n,int m)
{
sort(edge,edge+m,cmp);
int cnt=0,ans=0;
for(int i=0;i<m&&cnt<n-1;++i)
{
if(join(edge[i].a,edge[i].b))
{
ans=max(ans,edge[i].len);
++cnt;
}
}
return ans;
}
int main()
{
int n,m;
//freopen("shuju.txt","r",stdin);
while(~scanf("%d%d",&n,&m))
{
init(n);
for(int i=0;i<m;++i)
{
int a,b,c;
scanf("%d%d%d",&a,&b,&c);
node tp={a,b,c};
edge[i]=tp;
}
printf("%d\n",kruscal(n,m));
}
return 0;
}