Electrification Plan(最小生成树)

http://acm.sdut.edu.cn:8080/vjudge/contest/view.action?cid=50#problem/D

最小生成树模板，注意的是这里有k个发电站，它们不再需要连接其他的发电站，所以任何两个发电站之间的权值是0；

 1 #include<stdio.h>

 2 #include<string.h>

 3 const int maxn = 110;

 4 const int INF = 0x3f3f3f3f;

 5 int map[maxn][maxn],power[maxn];

 6 int n,k;

 7 

 8 void prim()

 9 {

10     int dis[maxn],vis[maxn];

11     int ans = 0;

12     memset(vis,0,sizeof(vis));

13     for(int i = 1; i <= n; i++)

14         dis[i] = map[1][i];

15     vis[1] = 1;

16 

17     for(int i = 1; i <= n-1; i++)

18     {

19         int min = INF,pos;

20 

21         for(int j = 1; j <= n; j++)

22         {

23             if(!vis[j] && dis[j] < min)

24             {

25                 min = dis[j];

26                 pos = j;

27             }

28         }

29         vis[pos] = 1;

30         ans += min;

31         for(int j =1; j <= n; j++)

32         {

33             if(!vis[j] && dis[j] > map[pos][j])

34                 dis[j] = map[pos][j];

35         }

36     }

37     printf("%d\n",ans);

38 }

39 

40 int main()

41 {

42     scanf("%d %d",&n,&k);

43 

44     /*for(int i = 1; i <= n; i++)

45         for(int j = 1; j <= n; j++)

46         {

47             if(i != j)

48                 map[i][j] = INF;

49             else map[i][j] = 0;

50         }*/

51     memset(power,0,sizeof(power));

52     int x;

53 

54     for(int i = 0; i < k; i++)

55     {

56         scanf("%d",&x);

57         power[x] = 1;

58     }

59 

60     for(int i = 1; i <= n; i++)

61     {

62         for(int j = 1; j <= n; j++)

63             scanf("%d",&map[i][j]);

64     }

65 

66     for(int i = 1; i <= n; i++)

67     {

68         for(int j = 1; j <= n; j++)

69         {

70             if(power[i] && power[j])

71                 map[i][j] = 0;

72         }

73     }

74     prim();

75     return 0;

76 }

View Code