生成树专题

1、POJ 1251 Jungle Roads

　　题意：给出个村庄之间的道路及其维修费；从中选取一些路使道路的维修费最小且保证各村庄之间道路畅通。

　　思路：最小生成树模板。

①Kruskal

 1 #include<iostream>
 2 #include<vector>
 3 #include<algorithm>
 4 using namespace std;
 5 int n;
 6 struct side
 7 {
 8     int v1;
 9     int v2;
10     int len;
11     side(int vv1=0,int vv2=0,int ll=0):v1(vv1),v2(vv2),len(ll){}
12     friend bool operator <(const side&a,const side&b)
13     {
14         return a.len > b.len;
15     }
16 };
17 vector<side>minHeap;
18 const int maxn = 30;
19 int pre[maxn];
20 int Find(int x)
21 {
22     int r = x;
23     while (r != pre[r])
24     {
25         r = pre[r];
26     }
27     int c = x, p;
28     while (c != r)
29     {
30         p = pre[c];
31         pre[c] = r;
32         c = p;
33     }
34     return r;
35 }
36 void Join(int x, int y)
37 {
38     int f1 = Find(x), f2 = Find(y);
39     if (f1 != f2) pre[f1] = f2;
40 }
41 int Kruskal()
42 {
43     side tmp;
44     int cnt = 1;
45     for (int i = 0; i <= n; i++)pre[i] = i;
46     int ans = 0;
47     while (cnt < n)
48     {
49         pop_heap(minHeap.begin(), minHeap.end());
50         tmp = minHeap.back();
51         minHeap.pop_back();
52         int u = Find(tmp.v1);
53         int v = Find(tmp.v2);
54         if (u != v)
55         {
56             Join(tmp.v1, tmp.v2);
57             ans += tmp.len;
58             cnt++;
59         }
60     }
61     return ans;
62 }
63 
64 int main()
65 {
66     while (~scanf("%d", &n))
67     {
68         if (n == 0)break;
69         minHeap.clear();
70         char v1, v2;
71         int len;
72         int k;
73         for (int i = 0; i < n - 1; i++)
74         {
75 
76             cin >> v1;
77             scanf("%d", &k);
78             for (int j = 0; j < k; j++)
79             {
80                 cin >> v2;
81                 scanf("%d", &len);
82                 minHeap.push_back(side(v1 - 'A', v2 - 'A', len));
83             }
84         }
85         make_heap(minHeap.begin(), minHeap.end());
86         int ans=Kruskal();
87         printf("%d\n", ans);
88     }
89     return 0;
90 }

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