Tree（prime）

Tree

Time Limit : 6000/2000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 2   Accepted Submission(s) : 2

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

 题解：

就是给n个数，如果这个数跟另一个数的和或者这两个数有一个是素数就可以连接；权值为Min(Min(VA , VB),|VA-VB|).

注意要打素数表；

prime代码:

#include<stdio.h>
#include <string.h>
#include<math.h>
#define Min(x,y) (x<y?x:y)
const int INF=0x3f3f3f3f;
const int MAXN=610;
bool prim[1000010];
int N,answer;
int judge(int a,int b){
        if(!prim[a]||!prim[b]||!prim[a+b]){
            return Min(Min(a,b),fabs(a-b));
        }
        else return -1;
}
void pt(){
    memset(prim,false,sizeof(prim));
    prim[0]=prim[1]=true;
    for(int i=2;i<=1000;i++){
        if(!prim[i])for(int j=i*i;j<1000010;j+=i){
            prim[j]=true;
        }
    }
}
int map[MAXN][MAXN],vis[MAXN],low[MAXN];
int v[MAXN];
void prime(){
    int k;
    int temp,flot=1;
    answer=0;
    memset(vis,0,sizeof(vis));
    vis[0]=1;
    for(int i=0;i<N;i++)low[i]=map[0][i];
    for(int i=0;i<N;i++){
            temp=INF;
        for(int j=0;j<N;j++)
            if(!vis[j]&&temp>low[j])
                    temp=low[k=j];
    if(temp==INF){
        if(flot==N)printf("%d
",answer);
        else puts("-1");
        break;
    }
        answer+=temp;
        vis[k]=1;
        flot++;
        for(int j=0;j<N;j++)
            if(!vis[j]&&low[j]>map[k][j])
            low[j]=map[k][j];
    }
}
int main(){
        int T,t;
        scanf("%d",&T);
        while(T--){
                    pt();
                memset(map,INF,sizeof(map));
            scanf("%d",&N);
            for(int i=0;i<N;i++)scanf("%d",&v[i]);
            for(int i=0;i<N;i++)
            for(int j=i+1;j<N;j++){
                    t=judge(v[i],v[j]);
                if(t>=0){
                    if(t<map[i][j])map[i][j]=map[j][i]=t;
                }
            }
            prime();
        }
    return 0;
}