POJ Out of Hay(2395 prim最小生成树)

Out of Hay

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 11814 Accepted: 4628

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 31 2 232 3 10001 3 43

Sample Output

43

#include<iostream>#include<cstring>#include<cstdio>using namespace std;const int maxn = 2001;int m,n;int map[maxn][maxn];bool vis[maxn];int dis[maxn];#define inf 0x7fffffffint prim(){    int i,j;    int ans;    memset(vis,0,sizeof(vis));    for(i=1; i<=m; i++)    {        dis[i] = inf;    }    ans = 0;    dis[1] = 0;    for(i=1; i<=m; i++)    {        int tmp = inf,k = 0;        for(j=1; j<=m; j++)        {            if(!vis[j] && dis[j] < tmp)            {                tmp = dis[j];                k = j;            }        }        if(tmp == inf) return 0;        vis[k] = true;        if(tmp > ans)        {            ans = tmp;        }        for(j=1; j<=m; j++) //更新新的权值，从k到j如果小于直接到j,那么就更新        {            if(!vis[j] && dis[j] > map[k][j])            {                dis[j] = map[k][j];            }        }    }    return ans;}int main(){    int i;    int pos1,pos2,num1;    freopen("111","r",stdin);    while(cin>>m>>n)    {        memset(map,100,sizeof(map));        memset(dis,0,sizeof(dis));        for(i=1; i<=n; i++)        {            cin>>pos1>>pos2>>num1;            if(map[pos1][pos2] > num1)                map[pos1][pos2] = map[pos2][pos1] = num1;            //cout<<map[pos1][pos2]<<endl;        }        cout<<prim()<<endl;    }    return 0;}