B - Out of Hay POJ - 2395

最新推荐文章于 2020-08-03 14:23:13 发布

原创最新推荐文章于 2020-08-03 14:23:13 发布 · 336 阅读

0 ·

CC 4.0 BY-SA版权

图论专栏收录该内容

5 篇文章

订阅专栏

帮助Bessie规划路线，使用Kruskal算法找出连接N个农场的最小生成树中最大边的长度，确保她携带的水量足够通过最长的道路。

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

Sample Output

题意，找到所有村庄中一条最长的路，用Kruskal算法，求最小生成树，刚开始的时候要把权值从小到大排序即可，然后求最大的一个权值，就是题中所要求的。

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
struct note
{
    int u;
    int v;
    int w;
} q[10010];
int f[2010];
int cmp(note a,note b)
{
    return a.w<b.w;
}
int find(int v)
{
    if(f[v]==v)return v;
    else return find(f[v]);
}
int merge(int x,int y)
{
    int t1,t2;
    t1=find(x);
    t2=find(y);
    if(t1!=t2)
    {
        f[t2]=t1;
        return 1;
    }
    return 0;
}
int main()
{
    int i,k,n,m;
    while(~scanf("%d%d",&n,&m))
    {
        for(i=0; i<m; i++)
            scanf("%d%d%d",&q[i].u,&q[i].v,&q[i].w);
        sort(q,q+m,cmp);   //按照权值从小到大排序
        for(i=1; i<=n; i++)
            f[i]=i;
        int count=0;
        for(i=0; i<m; i++)
        {
            if(merge(q[i].u,q[i].v))//并查集看是否被连通
            {
                count++;
                if(count==n-1)
                {
                    k=i;    //最长的一条路的下标
                    break;
                }
            }
        }
        printf("%d\n",q[k].w);
    }
}