poj2377 -- Bad Cowtractors(最大生成树)

最新推荐文章于 2019-05-25 13:40:15 发布

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最新推荐文章于 2019-05-25 13:40:15 发布

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分类专栏：图论最小生成树

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本文介绍了一个经典的算法问题：构建成本最高的生成树以连接所有节点，同时确保无环且构成一个树形结构。通过Kruskal算法实现这一目标，并提供了一段完整的C++代码示例。

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Problem J

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 131072/65536K (Java/Other)

Total Submission(s) : 5 Accepted Submission(s) : 3

Problem 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 <br> <br>* 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

Sample Output

42

也是一个裸的最（大）生成的 ，kurskal

#include <iostream>
#include <algorithm>
#include <string.h>
#include <stdio.h>
#include <cmath>
#include <vector>
#include <queue>
#include <utility>
using namespace std;
const int INF = 0x3fffffff;
struct Node {
    int u, v, w;
}node[100005];
int fa[1005];
int n, m;
int find(int x) {
    if (fa[x] == x)
        return x;
    fa[x] = find(fa[x]);
    return fa[x];
}
void merge(int x, int y) {
    int fx = find(x);
    int fy = find(y);
    if (fx != fy) {
        fa[fy] = fx;
    }
}
bool cmp(const Node& n1, const Node& n2) {
    return n1.w > n2.w;
}
int main() {
        //freopen("in.txt", "r", stdin);
    cin >> n >> m;
    for (int i = 1; i <= n; ++i) {
        fa[i] = i;
    }
    
    for (int i = 0; i < m; ++i) {
        cin >> node[i].u >> node[i].v >> node[i].w;
    }
    sort(node, node + m, cmp);
    
    int num = 0;
    int ans = 0;
    for (int i = 0; i < n; ++i) {
        for ( ; num < m; ++num) {
            int fu = find(node[num].u);
            int fv = find(node[num].v);
            if (fu != fv) {
                ans += node[num].w;
                merge(fv, fu);
                break;
            }
        }
    }
    int flag = 0;
    for (int i = 1; i <= n; ++i) {
        if (fa[i] == i)
            flag++;
    }
    cout << ((flag == 1) ? ans : -1) << endl;
}