UVa 11733 - Airports

来源：互联网发布：拼模特卡软件编辑：程序博客网时间：2024/05/01 14:54

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_setint 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_setint 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;}

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