Tree HDOJ--2682

Tree

Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1533    Accepted Submission(s): 433

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

思路：由于题目对能相连的点有限制，必须将这些点处理，能相连的点合并到一个集合中，最后查看是否所有点都在一个集合里，若都在说明是一个连通图，存在最小生成树，否则图不连通，不存在最小花费。

AC代码：

 

 #include<stdio.h>
 #include<string.h>
 #include<stdlib.h>
 int CityHappy[],vis[];
 int isprime[],dist[];
 int map[][],n;
 int father[],depth[];
 void init_B()
 {
     int i;
     for(i = ;i <= n;i ++)
     {
         father[i] = i;
         depth[i] = ;
     }
 }

 int find(int x)
 {
     if(x == father[x])
         return x;
     return father[x] = find(father[x]);
 }

 void unit(int x,int y)
 {
     x = find(x);
     y = find(y);
     if(x == y)
         return ;
     if(depth[x] > depth[y])
         father[y] = x;
     else
     {
         if(depth[x] < depth[y])
             father[x] = y;
         else
         {
             father[x] = y;
             depth[y]++;
         }
     }
 }

 void prime()
 {
     int i,j;
     isprime[] = isprime[] = ;
     for(i = ;i <= 1e6;i ++)
     {
         if(!isprime[i])
         {
             for(j = i << ;j <= 1e6;j += i)
                 isprime[j] = ;
         }
     }
 }

 int judge(int a,int b)
 {
     if(!isprime[a] || !isprime[b])
         return ;
     if(!isprime[a+b])
         return ;
     return ;
 }

 int min(int a,int b)
 {
     return a < b?a:b;
 }

 void opration()
 {
     int i,j,a,b;
     init_B();
     for(i = ;i <= n;i ++)
     {
         for(j = ;j <= n;j ++)
         {
             if(i != j)
             {
                 a = CityHappy[i];
                 b = CityHappy[j];
                 if(judge(a,b))
                 {
                     map[i][j] = min(min(a,b),abs(a-b));
                     map[j][i] = map[i][j];
                     unit(i,j);
                 }
                 else
                     map[i][j] = map[j][i] =  << ;
             }
         }
     }
 }

 void init()
 {
     int i;
     memset(vis,,sizeof(vis));
     for(i = ;i <= n;i ++)
         dist[i] = map[][i];
 }

 int main()
 {
     int t,i,j,k,cnt,min,sum;
     scanf("%d",&t);
     prime();
     while(t--)
     {
         sum = cnt = ;
         scanf("%d",&n);
         for(i = ;i <= n;i ++)
             scanf("%d",&CityHappy[i]);
         opration();
         init();
         for(i = ;i <= n;i ++)
         {
             if(i == find(i))
                 cnt++;
             if(cnt == )
                 break;
         }
         if(cnt == )
         {
             printf("-1\n");
             continue ;
         }
         for(i = ;i < n;i ++)
         {
             min =  << ;
             for(j = ;j <= n;j ++)
             {
                 if(!vis[j] && min > dist[j])
                 {
                     min = dist[j];
                     k = j;
                 }
             }
             vis[k] = ;
             if(min !=  << )
                 sum += min;
             for(j = ;j <= n;j ++)
             {
                 if(!vis[j] && dist[j] > map[k][j])
                     dist[j] = map[k][j];
             }
         }
         printf("%d\n",sum);
     }
     return ;
 }