题目大意：
Nasa应邻居们的要求，决定用一个网络把大家链接在一起。给出v个点，e条可行路线，每条路线分别是x连接到y需要花费w。
1：如果不存在最小生成树，输出“No way”.
2：如果不存在次小生成树，输出“No second way”.
3：如果两者都存在，输出次小生成树的长度.
解题思路：
次小生成数+kruskal模板
1 #include <cmath> 2 #include <queue> 3 #include <string> 4 #include <cstdio> 5 #include <cstring> 6 #include <iostream> 7 #include <algorithm> 8 using namespace std; 9 10 const int maxn = 210; 11 const int INF = 0x3f3f3f3f; 12 const double Exp = 1e-10; 13 struct node 14 { 15 int x, y, w; 16 17 }; 18 bool cmp (node a, node b) 19 { 20 return a.w < b.w; 21 } 22 node stu[maxn]; 23 int v, e, father[maxn], path[maxn]; 24 25 void init () 26 { 27 for (int i=0; i<=v; i++) 28 father[i] = i; 29 } 30 31 int find (int x) 32 { 33 if (father[x] != x) 34 father[x] = find(father[x]); 35 return father[x]; 36 } 37 int kruskal () 38 { 39 int i, j, num = 0; 40 for (i=0,j=1; i<e; i++) 41 { 42 int px = find (stu[i].x); 43 int py = find (stu[i].y); 44 if (px != py) 45 { 46 num += stu[i].w; 47 path[j++] = i; 48 father[px] = py; 49 } 50 } 51 int ans = 0; 52 for (i=1; i<=v; i++) 53 if (father[i] == i) 54 ans ++; 55 if (ans == 1) 56 return num; 57 return INF; 58 } 59 int smst (int x) 60 { 61 int i, num = 0; 62 for (i=0; i<e; i++) 63 { 64 int px = find(stu[i].x); 65 int py = find(stu[i].y); 66 if (px != py && i != x) 67 { 68 num += stu[i].w; 69 father[px] = py; 70 } 71 } 72 int ans = 0; 73 for (i=1; i<=v; i++) 74 if (father[i] == i) 75 ans ++; 76 if (ans == 1) 77 return num; 78 return INF; 79 } 80 int main () 81 { 82 int t, l = 1; 83 scanf ("%d", &t); 84 while (t --) 85 { 86 scanf ("%d %d", &v, &e); 87 for (int i=0; i<e; i++) 88 scanf ("%d %d %d", &stu[i].x, &stu[i].y, &stu[i].w); 89 sort (stu, stu+e, cmp); 90 int num1 , num2; 91 num1 = num2 = INF; 92 init (); 93 num1 = kruskal(); 94 for (int i=1; i<v; i++) 95 { 96 init (); 97 num2 = min (num2, smst(path[i])); 98 } 99 if (num1 == INF) 100 printf ("Case #%d : No way ", l++); 101 else if (num2 == INF) 102 printf ("Case #%d : No second way ", l++); 103 else 104 printf ("Case #%d : %d ", l++, num2); 105 } 106 return 0; 107 }