坏的牛圈建筑
题目大意:就是现在农夫又要牛修建牛栏了,但是农夫想不给钱,于是牛就想设计一个最大的花费的牛圈给他,牛圈的修理费用主要是用在连接牛圈上
这一题很简单了,就是找最大生成树,把Kruskal算法改一下符号就好了,把边从大到小排列,然后最后再判断是否联通(只要找到他们的根节点是否相同就可以了!)
1 #include <iostream> 2 #include <algorithm> 3 #include <functional> 4 #define MAX_N 1005 5 #define MAX 20005 6 7 using namespace std; 8 9 typedef int Position; 10 typedef struct _edge 11 { 12 Position from; 13 Position to; 14 int cost; 15 }Edge_Set; 16 int fcomp(const void *a, const void *b) 17 { 18 return (*(Edge_Set *)b).cost - (*(Edge_Set *)a).cost; 19 } 20 21 static Edge_Set edge[MAX]; 22 static Position Set[MAX_N]; 23 24 Position Find(Position); 25 void Union(Position, Position); 26 void Kruskal(const int, const int); 27 bool Is_Connected(const int); 28 29 int main(void) 30 { 31 int Node_Sum, Path_Sum, cost; 32 Position tmp_from, tmp_to; 33 34 while (~scanf("%d%d", &Node_Sum, &Path_Sum)) 35 { 36 for (int i = 0; i < Path_Sum; i++)//读入边 37 { 38 scanf("%d%d%d", &tmp_from, &tmp_to, &cost); 39 edge[i].from = tmp_from; edge[i].to = tmp_to; edge[i].cost = cost; 40 } 41 qsort(edge, Path_Sum, sizeof(Edge_Set), fcomp); 42 Kruskal(Node_Sum, Path_Sum); 43 } 44 return 0; 45 } 46 47 Position Find(Position x) 48 { 49 if (Set[x] < 0) 50 return x; 51 else return Set[x] = Find(Set[x]); 52 } 53 54 void Union(Position px, Position py) 55 { 56 if (Set[px] < Set[py]) 57 { 58 Set[px] += Set[py]; 59 Set[py] = px; 60 } 61 else 62 { 63 Set[py] += Set[px]; 64 Set[px] = py; 65 } 66 } 67 68 bool Is_Connected(const int Node_Sum) 69 { 70 Position p_u, p_tmp; 71 p_u = Find(1); 72 for (int i = 2; i <= Node_Sum; i++) 73 { 74 p_tmp = Find(i); 75 if (p_u != p_tmp) 76 return false; 77 } 78 return true; 79 } 80 81 void Kruskal(const int Node_Sum, const int Path_Sum) 82 { 83 Position px, py, from, to; 84 long long ans = 0; 85 86 fill(Set, Set + Node_Sum + 1, -1); 87 for (int i = 0; i < Path_Sum; i++) 88 { 89 from = edge[i].from; to = edge[i].to; 90 px = Find(from); py = Find(to); 91 92 if (px != py) 93 { 94 ans += edge[i].cost; 95 Union(px, py); 96 } 97 } 98 if (Is_Connected(Node_Sum)) 99 printf("%lld\n", ans); 100 else 101 printf("-1\n"); 102 }