hdu_2682_prime

Tree

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
mean
一个无向图中给你两点,a,b如果a 或者b 或者a+b是prime(素数) 那么a和b相连,距离为a,b,a+b中最小值
第一行给你T有t组案例,
第二行输入n 接下来有n个点的值
求最小生成树

#include<bits/stdc++.h>using namespace std;#define ll long longconst ll N=(ll)1e6*2+10;const ll inf=(ll)1e18;ll num[N]= {0,0,1,1},mp[666][666],vis[N],d[N],n;void get(){    ll i,j;    for(i=3; i<(ll)1e6+5; i++)    {        num[i++]=1;        num[i]=0;    }    for(i=3; i<(ll)1e6+5; i++)        for(j=i+i; j<(ll)1e6+5; j+=i)            num[j]=0;}ll prime(ll s){    ll i,j,ans=0;    for(i=0; i<n; i++)    {        d[i]=mp[s][i];        vis[i]=0;    }    d[s]=0,vis[s]=1;    for(i=0; i<n-1; i++)    {        ll mm=inf,k=0;        for(j=0; j<n; j++)        {            if(!vis[j]&&d[j]<mm)            {                mm=d[j];                k=j;            }        }        if(mm==inf) return -1;        ans+=mm;        vis[k]=1;        for(j=0; j<n; j++)        {            if(!vis[j]&&mp[k][j]<d[j])               d[j]=mp[k][j];        }    }    return ans;}int main(){    ll t,i,j,v[666];    get();    while(cin>>t)    {        while(t--)        {            scanf("%lld",&n);            for(i=0; i<n; i++)                scanf("%lld",v+i);            for(i=0; i<=n; i++)                for(j=0; j<=n; j++)                    mp[i][j]=inf;            for(i=0; i<n; i++)                for(j=0; j<n; j++)                {                    if(num[v[i]]||num[v[j]]||num[v[i]+v[j]])                        mp[i][j]=mp[j][i]=min(min(v[i],v[j]),abs(v[i]-v[j]));                }            ll tt=prime(0);            cout<<tt<<endl;        }    }    return 0;}