hdu 2682 Tree kruskal+并查集

Tree

Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2333    Accepted Submission(s): 710


Problem Description
There are N (2<=N<=600) cities,each has a value of happiness,we consider two cities A and B whose value of happiness are VA and VB,if VA is a prime number,or VB is a prime number or (VA+VB) is a prime number,then they can be connected.What's more,the cost to connecte two cities is Min(Min(VA , VB),|VA-VB|).
Now we want to connecte all the cities together,and make the cost minimal.
 

 

Input
The first will contain a integer t,followed by t cases.
Each case begin with a integer N,then N integer Vi(0<=Vi<=1000000).
 

 

Output
If the all cities can be connected together,output the minimal cost,otherwise output "-1";
 

 

Sample Input
2 5 1 2 3 4 5 4 4 4 4 4
 

 

Sample Output
4 -1
 
没什么好说的，最小生成树 kruskal+最小生成树，模板题
 
 1 #include<cstdio>
 2 #include<iostream>
 3 #include<algorithm>
 4 #include<queue>
 5 #include<cmath>
 6 #define MAX_N 1000005
 7 
 8 using namespace std;
 9 
10 int n,m;
11 int par[MAX_N];
12 int num[MAX_N];
13 
14 struct node
15 {
16     int a,b,va;
17 }edge[MAX_N];
18 bool cmp(node a,node b)
19 {
20     return a.va<b.va;
21 }
22 int su(int a)
23 {
24     int flag=1;
25     for(int i = 2; i <= sqrt(a)&&flag; i++)
26     {
27         if(a%i==0)
28             flag=0;
29     }
30     if(a==1)
31         flag=0;
32     return flag;
33 }
34 int ffind(int x)
35 {
36     if(x==par[x])
37         return x;
38     else
39         return par[x]=ffind(par[x]);
40 }
41 bool same(int x,int y)
42 {
43     return ffind(x)==ffind(y);
44 }
45 void mix(int x,int y)
46 {
47     x=ffind(x);
48     y=ffind(y);
49     if(x!=y)
50         par[x]=y;
51 }
52 int kruskal()
53 {
54     sort(edge,edge+m,cmp);
55     int ans=0,sum=0;
56     for(int i = 0; i < m; i++)
57     {
58         if(!same(edge[i].a,edge[i].b))
59         {
60             mix(edge[i].a,edge[i].b);
61             sum+=edge[i].va;
62             ans++;
63         }
64     }
65     if(ans==n-1)
66         return sum;
67     else
68         return -1;
69 
70 }
71 int main()
72 {
73     int t;
74     scanf("%d",&t);
75     while(t--)
76     {
77         scanf("%d",&n);
78         for(int i = 0; i <= n; i++)
79             par[i]=i;
80         for(int i = 1; i <= n; i++)
81             scanf("%d",&num[i]);
82         m=0;
83         for(int i = 1; i <= n; i++)
84         {
85             for(int j = i+1; j <= n; j++)
86             {
87                 if(su(num[i])||su(num[j])||su(num[i]+num[j]))
88                 {
89                     edge[m].a=i;
90                     edge[m].b=j;
91                     edge[m++].va=min(min(num[i],num[j]),abs(num[i]-num[j]));
92                 }
93             }
94         }
95         printf("%d\n",kruskal());
96     }
97     return 0;
98 }

 

posted @ 2016-11-22 20:03  Xycdada  阅读(239)  评论(0)  编辑  收藏  举报