POJ 2395 Out of Hay（最小生成树+克鲁斯卡尔）

系统管理员 2022-06-11 06:41 143阅读 0赞

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

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.

题解：

很水的一题求最小生成树里面最大的边值。。不知道为什么用prim算法写蜜汁无限RE。。还RE了8次，吐血，改成克鲁斯卡尔一次就过了

代码：

#include<algorithm>
    #include<iostream>
    #include<cstring>
    #include<stdio.h>
    #include<math.h>
    #include<string>
    #include<stdio.h>
    #include<queue>
    #include<stack>
    #include<map>
    #include<deque>
    using namespace std;
    const int INF=100861111;
    int pre[2005];
    int find(int x)//并查集
    {
        if(pre[x]!=x)
            pre[x]=find(pre[x]);
        return pre[x];
    }
    struct edge//存边
    {
        int f,t;
        int v;
    }a[10005];
    int cmp(edge x,edge y)
    {
        return x.v<y.v;
    }
    int main()
    {
        int i,j,n,m,d1,d2,x,y,z,ans=0,tot=0,minn;
        scanf("%d%d",&n,&m);
        for(i=1;i<=n;i++)
            pre[i]=i;//初始化
        for(i=0;i<m;i++)
        {
            scanf("%d%d%d",&x,&y,&z);
            a[i].f=x;
            a[i].t=y;
            a[i].v=z;
        }
        sort(a,a+m,cmp);//值从小到大排序
        minn=0;
        for(i=0;i<m;i++)
        {
            d1=find(a[i].f);
            d2=find(a[i].t);
            if(d1!=d2)//如果不在一个集合里，加入集合，更新最大值
            {
                pre[d2]=d1;
                ans++;
                if(a[i].v>minn)
                    minn=a[i].v;
            }
        }
        printf("%d\n",minn);
        return 0;
    }