Bad Cowtractors POJ - 2377 (Kruskal算法求生成树)

àì夳堔傛蜴生んèń 2022-02-23 09:40 158阅读 0赞

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

解题思路:

Kruskal 模板题,套用模板即可,由于这里要给出最大生成树,在排序时要从大到小排序

#include <iostream>
    #include <algorithm>
    #include <vector>
    #include <set>
    using namespace std;
    const int MAXN = 50010;
    struct ksEdge {
    	int u, v, tcost;
    }ksEdges[MAXN];
    int node_nums;
    bool scmp(ksEdge a, ksEdge b) {
    	return a.tcost > b.tcost;   //将边从小到大排序
    }
    int sFather[MAXN];  //并查集
    int N,R;
    
    void sinit() {
    	for (int i = 1; i <= N; ++i) {
    		sFather[i] = i;   
        }
    }
    
    int findFather(int x) {
    	if (sFather[x] != x) {
    		sFather[x] = findFather(sFather[x]);  //开始查找
    	}
    	return sFather[x];
    }
    
    void unionSet(int a, int b) {  //将两个并查集合并
    	int fa = findFather(a);
    	int fb = findFather(b);
    	if (fa != fb) {   
    		sFather[fa] = fb;  //合并
    	}
    }
    
    //两个元素是否同属一个并查集f
    bool isSame(int a, int b) {
    	int fa = findFather(a);
    	int fb = findFather(b);
    	if (fa == fb) return true;
    	else return false;
    }
    
    int kruskal() {
    	sort(ksEdges, ksEdges + R, scmp);  //将边排序
    	sinit();  //初始化并查集
    	int res = 0;  //最后结果
    	node_nums = 1;
    	for (int i = 0; i < R; ++i) {
    		ksEdge e = ksEdges[i];   //遍历每一条边
    		if (!isSame(e.u, e.v)) {
    			unionSet(e.u, e.v);
    			res += e.tcost;
    			node_nums++;
    		}
    	}
    	return res;  //返回res
    }
    
    int main() {
    	scanf("%d %d", &N, &R);
    	for (int i = 0; i < R; ++i) {
    		scanf("%d %d %d", &ksEdges[i].u, &ksEdges[i].v, &ksEdges[i].tcost);
    	}
    	int tre = kruskal();
    	if (node_nums < N) {
    		printf("-1\n");
    	}
    	else {
    		printf("%d\n", tre);
    	}
    
    	system("PAUSE");
    	return 0;
    }