UVa 11733 - Airports

題目：在城市之間建立公路和飛機場，是的每個城市都能到飛機場，求最小花費。

分析：圖論，最小生成樹。kruskal算法，每次選擇加一條變便或者建一個飛機場。

            如果邊長不小於飛機場的花費選擇建立新的飛機場；

說明：花費一樣時，選擇最所的飛機場數量。

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

using namespace std;

typedef struct _enode
{
int point1;
int point2;
int weight;
}enode;
enode edge[100001];

//union_set
int union_sets[10001];
int union_rank[10001];

void union_inital(int a, int b)
{
for (int i = a; i <= b; ++ i) {
union_rank[i] = 0;
union_sets[i] = i;
}
}

int union_find(int a)
{
if (a != union_sets[a]) {
union_sets[a] = union_find(union_sets[a]);
}
return union_sets[a];
}

void union_union(int a, int b)
{
if (union_rank[a] < union_rank[b]) {
union_sets[a] = b;
}else {
if (union_rank[a] == union_rank[b]) {
union_rank[a] ++;
}
union_sets[b] = a;
}
}
//end_union_set

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

int kruskal(int n, int m, int cost)
{
sort(edge, edge+m, cmp_e);
union_inital(1, n);
int sum = 0;
for (int i = 0; i < m; ++ i) {
if (edge[i].weight >= cost) {
continue;
}
int A = union_find(edge[i].point1);
int B = union_find(edge[i].point2);
if (A != B) {
union_union(A, B);
sum += edge[i].weight;
}
}
int count = 0;
for (int i = 1; i <= n; ++ i) {
if (i == union_sets[i]) {
count ++;
}
}
printf("%d %d\n",sum+cost*count,count);
}

int main()
{
int T, n, m, a, b, c, cost;
while (~scanf("%d",&T)) {
for (int t = 1; t <= T; ++ t) {
scanf("%d%d%d",&n,&m,&cost);
for (int i = 0; i < m; ++ i) {
scanf("%d%d%d",&a,&b,&c);
edge[i].point1 = a;
edge[i].point2 = b;
edge[i].weight = c;
}

printf("Case #%d: ",t);
kruskal(n, m, cost);
}
}
return 0;
}