Out of Hay POJ - 2395 (Krusal最小生成树)

迈不过友情╰ 2022-02-23 09:54 150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

解题思路:

最小生成树模板题,保留树中最长的那条边即可

#define _CRT_SECURE_NO_WARNINGS
    #include <iostream>
    #include <algorithm>
    #include <vector>
    #include <string.h>
    using namespace std;
    const int MAXE = 10010;
    const int MAXV = 3010;
    struct nmEdge {
    	int u, v;
    	long long cost;
    }nmEdges[MAXE];
    
    bool nmCmp(nmEdge a, nmEdge b) {
    	return a.cost < b.cost;
    }
    int nmFather[MAXV]; 
    int N, M;
    
    void nmInit() {
    	for (int i = 1; i <= N; ++i) {
    		nmFather[i] = i;
    	}
    }
    
    int findFather(int x) {
    	if (nmFather[x] != x) {
    		nmFather[x] = findFather(nmFather[x]);
    	}
    	return nmFather[x];
    }
    
    void unintSet(int a, int b) {
    	int fa = findFather(a);
    	int fb = findFather(b);
    	if (fa != fb) {
    		nmFather[fa] = fb;  //把它们并到一起
    	}
    }
    
    bool bisSame(int a, int b) {
    	int fa = findFather(a);
    	int fb = findFather(b);
    	if (fa == fb) return true;
    	else return false;  
    }
    
    
    int main() {
    	scanf("%d %d", &N, &M);
    	for (int i = 0; i < M; ++i) {
    		scanf("%d %d %lld", &nmEdges[i].u, &nmEdges[i].v, &nmEdges[i].cost);
    	}
    	long long minRes = 0;
    	sort(nmEdges, nmEdges + M, nmCmp);
    	nmInit(); //初始化并查集
    	for (int i = 0; i < M; ++i) {
    		nmEdge e = nmEdges[i];
    		if (!bisSame(e.u, e.v)) {
    			unintSet(e.u, e.v);
    			if (e.cost > minRes) minRes = e.cost;
    		}
    	}
    
    	printf("%lld\n", minRes);
    	system("PAUSE");
    	return 0;
    }