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

  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\n", l++);
101         else if (num2 == INF)
102             printf ("Case #%d : No second way\n", l++);
103         else
104             printf ("Case #%d : %d\n", l++, num2);
105     }
106     return 0;
107 }

 

posted @ 2015-06-09 15:27  罗茜  阅读(613)  评论(0)  编辑  收藏  举报