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Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2877    Accepted Submission(s): 883

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

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <queue>
#include <cstdlib>
#include <iomanip>
#include <cmath>
#include <ctime>
#include <map>
#include <set>
using namespace std;
#define lowbit(x) (x&(-x))
#define max(x,y) (x>y?x:y)
#define min(x,y) (x<y?x:y)
#define MAX 100000000000000000
#define MOD 1000000007
#define pi acos(-1.0)
#define ei exp(1)
#define PI 3.141592653589793238462
#define ios() ios::sync_with_stdio(false)
#define INF 0x3f3f3f3f
#define mem(a) (memset(a,0,sizeof(a)))
typedef long long ll;
int g[750][750];
int vis[750],n;
int dis[750],x,y,z;
int val[750],t;
int ans[2000009];
void get_prime()
{
    memset(ans,0,sizeof(ans));
    ans[1]=1;
    for(int i=2;i<2000009;i++)
    {
        if(ans[i]) continue;
        for(int j=2;j*i<2000009;j++)
        {
            ans[j*i]=1;
        }
    }
}
void init()
{
    for(int i=0;i<=n;i++)
    {
        for(int j=0;j<i;j++)
        {
            g[i][j]=g[j][i]=INF;
        }
        g[i][i]=0;
    }
}
int prime()
{
    for(int i=1;i<=n;i++)
    {
        dis[i]=g[1][i];
        vis[i]=0;
    }
    vis[1]=1;
    int minn,v=1,sum=0;
    for(int i=1;i<n;i++)
    {
        minn=INF;
        for(int j=1;j<=n;j++)
        {
            if(!vis[j] && minn>dis[j])
            {
                minn=dis[j];
                v=j;
            }
        }
        if(minn==INF) return -1;
        vis[v]=1;
        sum+=minn;
        for(int j=1;j<=n;j++)
        {
            if(!vis[j]) dis[j]=min(dis[j],g[v][j]);
        }
    }
    return sum;
}
int main()
{
    get_prime();
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        init();
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&val[i]);
            for(int j=1;j<=i;j++)
            {
                if(!ans[val[i]] || !ans[val[j]] || !ans[val[i]+val[j]])
                    g[i][j]=g[j][i]=min(g[i][j],min(min(val[i],val[j]),abs(val[i]-val[j])));
            }
        }
        printf("%d\n",prime());
    }
    return 0;
}

Sample Output

4
-1

必须保证va 或者 vb 或者 va+vb是素数