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[605],vis[605];
int isprime[1000005],dist[605];
int map[601][601],n;
int father[605],depth[605];
void init_B()
{
    int i;
    for(i = 1;i <= n;i ++)
    {
        father[i] = i;
        depth[i] = 0;
    }
}

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[0] = isprime[1] = 1;
    for(i = 2;i <= 1e6;i ++)
    {
        if(!isprime[i])
        {
            for(j = i << 1;j <= 1e6;j += i)
                isprime[j] = 1;
        }
    }
}

int judge(int a,int b)
{
    if(!isprime[a] || !isprime[b])
        return 1;
    if(!isprime[a+b])
        return 1;
    return 0;
}

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

void opration()
{
    int i,j,a,b;
    init_B();
    for(i = 1;i <= n;i ++)
    {
        for(j = 1;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] = 1 << 30;
            }
        }
    }
}

void init()
{
    int i;
    memset(vis,0,sizeof(vis));
    for(i = 1;i <= n;i ++)
        dist[i] = map[1][i];
}

int main()
{
    int t,i,j,k,cnt,min,sum;
    scanf("%d",&t);
    prime();
    while(t--)
    {
        sum = cnt = 0;
        scanf("%d",&n);
        for(i = 1;i <= n;i ++)
            scanf("%d",&CityHappy[i]);
        opration();
        init();
        for(i = 1;i <= n;i ++)
        {
            if(i == find(i))
                cnt++;
            if(cnt == 2)
                break;
        }
        if(cnt == 2)
        {
            printf("-1
");
            continue ;
        }
        for(i = 0;i < n;i ++)
        {
            min = 1 << 30;
            for(j = 1;j <= n;j ++)
            {
                if(!vis[j] && min > dist[j])
                {
                    min = dist[j];
                    k = j;
                }
            }
            vis[k] = 1;
            if(min != 1 << 30)
                sum += min;
            for(j = 1;j <= n;j ++)
            {
                if(!vis[j] && dist[j] > map[k][j])
                    dist[j] = map[k][j];
            }
        }
        printf("%d
",sum);
    }
    return 0;
}