ZJNU 1213 - water - Senior

I can play well in a village water costs spent Wi, water can also be connected to the water village

And because the village is not possible without a well drilling (ie, there is at least one village wells, connected to the rest of the village it)

In fact it can be understood as well regarded as the first N + 1 villages, the village needs to have connected with the N + 1 village, in order to ensure that all villages have water

I is connected to the village costs village N + 1, can be directly understood the cost of drilling as Wi

The remainder of the minimum spanning tree can be used Kruskal

 1 /*
 2 Written By StelaYuri
 3 */
 4 #include<cstdio>
 5 #include<algorithm>
 6 using namespace std;
 7 struct tube{
 8     int from,to,cost;
 9     bool operator < (const tube &a) const{
10         return cost<a.cost;
11     }
12 }p[90001];
13 int gp[301];
14 int find(int a){
15     return a==gp[a]?a:(gp[a]=find(gp[a]));
16 }
17 int main(){
18     int i,j,N,d,cnt=0,k=0,sum;
19     scanf("%d",&N);
20     for(i=1;i<=N;i++){
21         scanf("%d",&d);
22         p[cnt].from=i;
23         p[cnt].to=N+1;
24         p[cnt++].cost=d;
25         gp[i]=i;
26     }
27     for(i=1;i<=N;i++)
28         for(j=1;j<=N;j++){
29             scanf("%d",&d);
30             if(i==j)
31                 continue;
32             p[cnt].from=i;
33             p[cnt].to=j;
34             p[cnt++].cost=d;
35         }
36     sort(p,p+cnt);
37     for(sum=i=0;i<cnt;i++){
38         if(k==N)
39             break;
40         if(find(p[i].from)!=find(p[i].to)){
41             gp[find(p[i].to)]=find(p[i].from);
42             sum+=p[i].cost;
43             k++;
44         }
45     }
46     printf("%d\n",sum);
47     
48     return 0;
49 }