UVA 10600 ACM Contest and Blackout (次小生成树)

题目大意：

　　给n个节点，m条边，问最小生成树，次小生成树？

　　ps：以前做次小生成树的时候估计没有掌握牢固，这次wa的好辛苦哟。

 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 = 100;
11 const int INF = 0x3f3f3f3f;
12 const double Exp = 1e-10;
13 
14 int cost[maxn+10][maxn+10], lowc[maxn+10], vis[maxn+10];
15 int Max[maxn+10][maxn+10], used[maxn+10][maxn+10], pre[maxn+10];
16 int n;
17 void init ()
18 {
19     for (int i=1; i<=n; i++)
20         for (int j=1; j<=n; j++)
21             if (i == j)
22                 cost[i][j] = 0;
23             else
24                 cost[i][j] = INF;
25 }
26 int prim ()
27 {
28     int i, j, sum = 0;
29     memset (Max, 0, sizeof(Max));
30     memset (used, 0, sizeof(used));
31     memset (vis, 0, sizeof(vis));
32     vis[1] = 1;
33     for (i=1; i<=n; i++)
34     {
35         lowc[i] = cost[1][i];
36         pre[i] = 1;
37     }
38     for (i=1; i<n; i++)
39     {
40         int p, mini = INF;
41         for (j=1; j<=n; j++)
42             if (!vis[j] && mini > lowc[j])
43             {
44                 p = j;
45                 mini = lowc[j];
46             }
47         vis[p] = 1;
48         sum += mini;
49         used[pre[p]][p] = used[p][pre[p]] = 1;
50         for (j=1; j<=n; j++)
51         {//这一点的错误wa的好苦
52             if (vis[j] && j != p)//p!=j一定要有，否则Max[i][i]的距离就可能会不等于零，影响后面的计算过程
53                 Max[j][p] = Max[p][j] = max(Max[j][pre[p]], lowc[p]);
54             if (!vis[j] && lowc[j] > cost[p][j])
55             {
56                 lowc[j] = cost[p][j];
57                 pre[j] = p;
58             }
59         }
60     }
61     return sum;
62 }
63 int smst (int sum)
64 {
65     int i, j, mini = INF;
66     for (i=1; i<=n; i++)
67         for (j=i+1; j<=n; j++)
68             if (cost[i][j] != INF && !used[i][j])
69                 mini = min (mini, sum + cost[i][j] - Max[i][j]);
70     return mini;
71 }
72 int main ()
73 {
74     int t, m;
75     scanf ("%d", &t);
76     while (t --)
77     {
78         scanf ("%d %d", &n, &m);
79         init ();
80         while (m --)
81         {
82             int a, b, c;
83             scanf ("%d %d %d", &a, &b, &c);
84             cost[a][b] = cost[b][a] = c;
85         }
86         int num1, num2;
87         num1 = prim ();
88         num2 = smst (num1);
89         printf ("%d %d\n", num1, num2);
90     }
91     return 0;
92 }

 

posted @ 2015-06-03 16:40  罗茜  阅读(207)  评论(0)  编辑  收藏  举报