Prim && Kruskal

Electrification Plan

Prim

 1 #include<iostream>
 2 #include<cstring>
 3 using namespace std;
 4 const int INF = 0x3f3f3f3f;
 5 const int N = 105;
 6 int n, k;
 7 int dis[N], ct[N][N], vis[N];
 8 
 9 int Prim()
10 {
11     int ans = 0;
12     memset(vis, 0, sizeof(vis));
13     /*
14         for(int i = 1; i <= n; i++)
15             dis[i] = ct[1][i];
16         vis[1] = 1;
17     */
18     while(1)
19     {
20         int v = -1;
21         for(int i = 1; i <= n; i++)
22         {
23             if(!vis[i] && (v == -1||dis[v]>dis[i])) v = i;
24         }
25         if(v == -1) break;
26         ans += dis[v];
27         vis[v] = 1;
28         for(int i = 1; i <= n; i++)
29             if(!vis[i])
30                 dis[i] = min(dis[i], ct[v][i]);
31     }
32     return ans;
33 }
34 int main()
35 {
36     ios::sync_with_stdio(false);
37     cin.tie(0);
38     cout.tie(0);
39     cin >> n >> k;
40     memset(dis, INF, sizeof(dis));
41     int tmp;
42     while(k--)
43     {
44         cin >> tmp;
45         dis[tmp] = 0;
46     }
47     for(int i = 1; i <= n; i++)
48         for(int j = 1; j <= n; j++)
49             cin >> ct[i][j];
50     cout << Prim() << endl;
51     return 0;
52 }

 

Kruskal

 1 #include<iostream>
 2 #include<algorithm>
 3 #include<cstring>
 4 using namespace std;
 5 const int N = 105;
 6 struct node
 7 {
 8     int l, r, c;
 9     bool operator < (const node & x)
10     {
11         return c < x.c;
12     }
13 }A[N*N];
14 int pre[N];
15 int Find(int x)
16 {
17     if(x == pre[x]) return x;
18     return pre[x] = Find(pre[x]);
19 }
20 int main()
21 {
22     ios::sync_with_stdio(false);
23     cin.tie(0);
24     cout.tie(0);
25     int n, m;
26     cin >> n >> m;
27     for(int i = 1; i <= n; i++)
28         pre[i] = i;
29     int t, tmp;
30     cin >> t;
31     m--;
32     while(m--)
33     {
34         cin >> tmp;
35         t = Find(t);
36         tmp = Find(tmp);
37         pre[tmp] = t;
38     }
39     int cnt = 0;
40     for(int i = 1; i <= n; i++)
41     {
42         for(int j = 1; j <= n; j++)
43         {
44             cin >> tmp;
45             if(i <= j) continue;
46             A[cnt].l = i, A[cnt].r = j, A[cnt++].c = tmp;
47         }
48     }
49     int ans = 0;
50     sort(A, A+cnt);
51     for(int i = 0; i < cnt; i++)
52     {
53         int x =A[i].l, y = A[i].r, ct = A[i].c;
54         x = Find(x), y = Find(y);
55         if(x == y) continue;
56         pre[y] = x;
57         ans += ct;
58     }
59     cout << ans << endl;
60     return 0;
61 }

 

posted @ 2018-02-04 13:43  Schenker  阅读(167)  评论(0)  编辑  收藏  举报