POJ 2377 Bad Cowtractors （kruskal求MST）

Bad Cowtractors

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 11136		Accepted: 4688

Description

Bessie has been hired to build a cheap internet network among Farmer John's N (2 <= N <= 1,000) barns that are conveniently numbered 1..N. FJ has already done some surveying, and found M (1 <= M <= 20,000) possible connection routes between pairs of barns. Each possible connection route has an associated cost C (1 <= C <= 100,000). Farmer John wants to spend the least amount on connecting the network; he doesn't even want to pay Bessie. 

Realizing Farmer John will not pay her, Bessie decides to do the worst job possible. She must decide on a set of connections to install so that (i) the total cost of these connections is as large as possible, (ii) all the barns are connected together (so that it is possible to reach any barn from any other barn via a path of installed connections), and (iii) so that there are no cycles among the connections (which Farmer John would easily be able to detect). Conditions (ii) and (iii) ensure that the final set of connections will look like a "tree".

Input

* Line 1: Two space-separated integers: N and M 

* Lines 2..M+1: Each line contains three space-separated integers A, B, and C that describe a connection route between barns A and B of cost C.

Output

* Line 1: A single integer, containing the price of the most expensive tree connecting all the barns. If it is not possible to connect all the barns, output -1.

Sample Input

5 8
1 2 3
1 3 7
2 3 10
2 4 4
2 5 8
3 4 6
3 5 2
4 5 17

Sample Output

42

求最小生成树 不能生成输出-1 判断 E = V - 1 是否满足即可

AC代码如下：

//
//  POJ 2377 Bad Cowtractors
//
//  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;
int par[1005], rank[1005];
struct Edge {
	int u, v, cost;
	bool operator<(const Edge& rhs) const {
		return cost < rhs.cost;
	}
} G[20005];

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 = 0, cnt = 0;
	sort(G + 1, G + 1 + m);
	init(n);
	for(int i = 1; i <= m; ++i) {
		Edge &e = G[i];
		if(!same(e.u, e.v)) {
			++cnt; ret += e.cost;
			unite(e.u, e.v);
		}
	}
	return ret = cnt == n - 1 ? -ret : -1;
}


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

	scanf("%d%d", &n, &m);
	for(int i = 1; i <= m; ++i) {
		int x, y, v; scanf("%d%d%d", &x, &y, &v);
		G[i] = (Edge) {x, y, -v};
	}
	printf("%d\n", kruskal());
	return 0;
}