hdu1233 还是畅通工程(kruskal+并查集)
作者:互联网
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 71430 Accepted Submission(s): 32276
当N为0时,输入结束,该用例不被处理。 Output 对每个测试用例,在1行里输出最小的公路总长度。 Sample Input 3 1 2 1 1 3 2 2 3 4 4 1 2 1 1 3 4 1 4 1 2 3 3 2 4 2 3 4 5 0 Sample Output 3 5
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 int n,m; 5 const int MAX=5005; 6 struct Edge{ 7 int from,to,val; 8 }edge[MAX]; 9 int fa[MAX]; 10 11 bool cmp(Edge x,Edge y){ 12 return x.val<y.val; 13 } 14 15 int find(int x){ 16 int k=x; 17 while(fa[k]!=k)k=fa[k]; 18 return k; 19 } 20 21 int kruskal(){ 22 int res=0; 23 for(int i=1;i<=n;i++)fa[i]=i; 24 sort(edge,edge+m,cmp); 25 for(int i=0,k=0;i<m&&k<n-1;i++){ 26 int fx=find(edge[i].from),fy=find(edge[i].to); 27 if(fx!=fy){ 28 fa[fx]=fy; 29 k++; 30 res+=edge[i].val; 31 } 32 } 33 return res; 34 } 35 36 int main(){ 37 while(scanf("%d",&n),n){ 38 m=n*(n-1)/2; 39 for(int i=0;i<m;i++){ 40 scanf("%d%d%d",&edge[i].from,&edge[i].to,&edge[i].val); 41 } 42 printf("%d\n",kruskal()); 43 } 44 }
标签:int,hdu1233,kruskal,查集,测试用例,村庄,MAX,公路,Edge 来源: https://www.cnblogs.com/ChangeG1824/p/11679535.html