最小生成树 prime算法 求权值最大的边

Out of HayDescriptionThe 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 allof 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, sheplans 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 backtrackover 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

算法思想：最小生成树 prime算法  求最大权值的边

代码：

#include <iostream>#include<stdio.h>#include<algorithm>#include<string.h>#include<math.h>using namespace std;const int maxn=1000000009;int map[2010][2010];int visited[2010];int lowst[2010];int n,m;int prim(){int pos,minn,res=-1;memset(visited,0,sizeof(visited));visited[1]=1;pos=1;for(int i=1;i<=n;i++){if(pos!=i) lowst[i]=map[pos][i];}for(int i=1;i<n;i++){minn=maxn;for(int j=1;j<=n;j++){if(!visited[j]&&minn>lowst[j]){minn=lowst[j];pos=j;}}if(res<minn) res=minn;visited[pos]=1;for(int j=1;j<=n;j++){if(!visited[j]&&lowst[j]>map[pos][j])lowst[j]=map[pos][j];}}return res;}int main(){scanf("%d%d",&n,&m);for(int i=1;i<=n;i++){for(int j=1;j<=n;j++)map[i][j]=maxn;}for(int i=0;i<m;i++){int a,b,c;scanf("%d%d%d",&a,&b,&c);if(map[a][b]>c)map[a][b]=map[b][a]=c;}printf("%d\n",prim());return 0;}