POJ 2395 Out of Hay （MST）

Out of Hay

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 11853		Accepted: 4642

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 3
1 2 23
2 3 1000
1 3 43

Sample Output

43

kruskal求最小生成森林 然后求出最大的那条边即可

AC代码如下：

//
//  POJ 2395 Out of Hay
//
//  Created by TaoSama on 2015-03-21
//  Copyright (c) 2015 TaoSama. All rights reserved.
//
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>
#define CLR(x,y) memset(x, y, sizeof(x))

using namespace std;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const int N = 1e5 + 10;

int n, m, par[2005], rank[2005];

struct Edge {
	int u, v, cost;
	bool operator < (const Edge& rhs) const {
		return cost < rhs.cost;
	}
};

vector<Edge> G;

void init(int n) {
	for(int i = 1; i <= n; ++i) {
		par[i] = i;
		rank[i] = 0;
	}
}

int find(int x) {
	if(par[x] == x) return x;
	return par[x] = find(par[x]);
}

void unite(int x, int y) {
	x = find(x); y = find(y);
	if(x == y) return;
	if(rank[x] < rank[y]) {
		par[x] = y;
	} else {
		par[y] = x;
		if(rank[x] == rank[y]) ++rank[x];
	}
}

bool same(int x, int y) {
	return find(x) == find(y);
}

int kruskal() {
	int ret = -INF;
	sort(G.begin(), G.end());
	init(n);
	for(int i = 0; i < G.size(); ++i) {
		Edge &e = G[i];
		if(!same(e.u, e.v)) {
			ret = max(ret, e.cost);
			unite(e.u, e.v);
		}
	}
	return ret;
}

int main() {
#ifdef LOCAL
	freopen("in.txt", "r", stdin);
//	freopen("out.txt","w",stdout);
#endif
	ios_base::sync_with_stdio(0);

	cin >> n >> m;
	for(int i = 1; i <= m; ++i) {
		int x, y, v; cin >> x >> y >> v;
		G.push_back((Edge) {x, y, v});
	}
	cout << kruskal() << endl;
	return 0;
}