UVA 10462 Is There A Second Way Left? (次小生成树+kruskal)

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

 #include <cmath>
 #include <queue>
 #include <string>
 #include <cstdio>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 using namespace std;

 const int maxn = ;
 const int INF = 0x3f3f3f3f;
 const double Exp = 1e-;
 struct node
 {
     int x, y, w;

 };
 bool cmp (node a, node b)
 {
     return a.w < b.w;
 }
 node stu[maxn];
 int v, e, father[maxn], path[maxn];

 void init ()
 {
     for (int i=; i<=v; i++)
         father[i] = i;
 }

 int find (int x)
 {
     if (father[x] != x)
         father[x] = find(father[x]);
     return father[x];
 }
 int kruskal ()
 {
     int i, j, num = ;
     for (i=,j=; i<e; i++)
     {
         int px = find (stu[i].x);
         int py = find (stu[i].y);
         if (px != py)
         {
             num += stu[i].w;
             path[j++] = i;
             father[px] = py;
         }
     }
     int ans = ;
     for (i=; i<=v; i++)
         if (father[i] == i)
             ans ++;
     if (ans == )
         return num;
     return INF;
 }
 int smst (int x)
 {
     int i, num = ;
     for (i=; i<e; i++)
     {
         int px = find(stu[i].x);
         int py = find(stu[i].y);
         if (px != py && i != x)
         {
             num += stu[i].w;
             father[px] = py;
         }
     }
     int ans = ;
     for (i=; i<=v; i++)
         if (father[i] == i)
             ans ++;
     if (ans == )
         return num;
     return INF;
 }
 int main ()
 {
     int t, l = ;
     scanf ("%d", &t);
     while (t --)
     {
         scanf ("%d %d", &v, &e);
         for (int i=; i<e; i++)
             scanf ("%d %d %d", &stu[i].x, &stu[i].y, &stu[i].w);
         sort (stu, stu+e, cmp);
         int num1 , num2;
         num1 = num2 = INF;
         init ();
         num1 = kruskal();
         for (int i=; i<v; i++)
         {
             init ();
             num2 = min (num2, smst(path[i]));
         }
         if (num1 == INF)
             printf ("Case #%d : No way\n", l++);
         else if (num2 == INF)
             printf ("Case #%d : No second way\n", l++);
         else
             printf ("Case #%d : %d\n", l++, num2);
     }
     return ;
 }