Time Limit: 4000/2000 MS
(Java/Others) Memory Limit: 65536/32768 K
(Java/Others)
Total Submission(s): 22101 Accepted
Submission(s): 9871
1 #include<stdio.h> 2 #include<string.h> 3 const int inf=0x3f3f3f3f; 4 int vis[101],lowc[101]; 5 int sta[101][101]; 6 int prim(int cost[][101], int n) 7 { 8 int i,j,p; 9 int minc,res=0; 10 memset(vis , 0 , sizeof(vis)); 11 vis[0] = 1; 12 for(i=1;i<n;i++) 13 { 14 lowc[i] = cost[0][i]; 15 } 16 for(i=1 ; i<n ; i++) 17 { 18 minc=inf ; //<初始化一个较大的数> 19 p=-1; 20 for(j=0 ; j<n ; j++) 21 { 22 if(0==vis[j] && minc > lowc[j] ) 23 { 24 minc = lowc[j]; 25 p = j; 26 } 27 } 28 if(inf == minc) return -1; //原图不联通 29 res += minc ; 30 vis[p] = 1; 31 for( j=0 ; j<n ; j++) 32 { 33 if(0==vis[j] && lowc[j] > cost[p][j]) 34 lowc[j] = cost[p][j]; 35 } 36 } 37 return res ; 38 } 39 int main( void ) 40 { 41 int n,i,a,b,c,j; 42 while(scanf("%d",&n),n) 43 { 44 for(i=0;i<n;i++) 45 { 46 for(j=0;j<n;j++) 47 { 48 sta[i][j]=inf; 49 } 50 } 51 for(i=0 ; i<n*(n-1)/2 ; i++) 52 { 53 scanf("%d%d%d",&a,&b,&c); 54 sta[a-1][b-1]=sta[b-1][a-1]=c; 55 } 56 printf("%d\n",prim(sta,n)); 57 } 58 return 0; 59 }
原文:http://www.cnblogs.com/gongxijun/p/3566553.html