UVa 10842 - Traffic Flow

題目:求最大生成樹的最小邊。

分析:圖論,最小生成樹,并查集。利用kruskal算法求解即可。

說明:╮(╯▽╰)╭。

#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstdio>

using namespace std;

typedef struct _enode
{
	int point1;
	int point2;
	int weight;
}enode;
enode edge[10001];

//union_set
int sets[101];
int rank[101];

void set_inital(int a, int b)
{
	for ( int i = a; i <= b; ++ i) {
		rank[i] = 0;
		sets[i] = i;
	}
}

int  set_find(int a)
{
	if (a != sets[a])
		sets[a] = set_find(sets[a]);
	return sets[a];
}

void set_union(int a, int b)
{
	if (rank[a] < rank[b])
		sets[a] = b;
	else {
		if (rank[a] == rank[b])
			rank[a] ++;
		sets[b] = a;
	}
}
//end_union_set

int cmp_e(enode a, enode b)
{
	return a.weight > b.weight;
}

int kruskal(int n, int m)
{
	sort(edge, edge+m, cmp_e);
	
	set_inital(0, n);
	int ans = 0;
	for (int i = 0; i < m; ++ i) {
		int A = set_find(edge[i].point1);
		int B = set_find(edge[i].point2);
		if (A != B) { 
			set_union(A, B);
			ans = edge[i].weight+0LL;
		}
	}
	return ans;
}

int main()
{
	int t, n, m;
	while (~scanf("%d",&t))
	for (int k = 1; k <= t; ++ k) {
		scanf("%d%d",&n,&m);
		for (int i = 0; i < m; ++ i)
			scanf("%d%d%d",&edge[i].point1,&edge[i].point2,&edge[i].weight);
	
		printf("Case #%d: %d\n",k,kruskal(n, m));
	}
	
	return 0;
}


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