HDU 2682 Tree

Tree

Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1704    Accepted Submission(s): 502

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

251234544444

 

Sample Output

4-1
题目大意：n个城市每个城市都有自己的幸福感，对于每两个城市A,B,对应的幸福感分别是VA,VB，如果满足VA或VB或VA+VB是素数，他们则是连通的，否则，嘿嘿，不连通。
如果连通，其连通的代价是VA,VB,|VA-VB|.
求将所有城市连通的最小值。
虽然他加了素数判断这一外壳，但是它依然遮盖不了他是一道水题的事实。。。
kruskal又来了。套模板的干活。
我就不多解释了，直接上代码
#include<cstdio>#include<algorithm>#include<cmath>#include<cstring>#include<iostream>using namespace std;const int L=1000005,inf=1<<30,maxn=1005;struct node{int s,y,w;}edge[L];int Fa[L],n,m,dot[L];bool is[L];void init();void unite(int x,int y);void getPrime();int Find(int x);int kruskal();bool cmp(node a,node b);bool same(int x,int y);bool judge(int x,int y);int main(){    int t;    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);m=0;        getPrime();        for(int i=1;i<=n;i++) scanf("%d",&dot[i]);        for(int i=1;i<=n;i++)            for(int j=i+1;j<=n;j++)            {                    if(judge(i,j))                    {                        edge[m].s=i;                        edge[m].y=j;                        edge[m++].w=min(min(dot[i],dot[j]),abs(dot[i]-dot[j]));                       // printf("s=%d,y=%d,w=%d\n",i,j,abs(dot[i]-dot[j]));                    }            }        int ans=kruskal();        if(ans==-1) printf("-1\n");        else printf("%d\n",ans);    }    return 0;}void init()//初始化并查集{    for(int i=0;i<=n;i++) Fa[i]=i;}int Find(int x)//查询属于哪个集合，并直接拜“祖宗”为师{    if(Fa[x]==x) return x;    else return Fa[x]=Find(Fa[x]);}void unite(int x,int y)//合并x,y两个元素{    x=Find(x);y=Find(y);    if(x==y) return ;    Fa[y]=x;}bool same(int x,int y)//【判断是否属于同个集合{    return Find(x)==Find(y);}bool cmp(node a,node b){    return a.w<b.w;}int kruskal(){    sort(edge,edge+m,cmp);    init();    int sum=0,cnt=0;    for(int i=0;i<m;i++)    {        if(cnt==n-1) break;        if(!same(edge[i].s,edge[i].y))        {            unite(edge[i].s,edge[i].y);            sum+=edge[i].w;            cnt++;        }    }    if(cnt!=n-1) return -1;    return sum;}void getPrime(){    fill(is,is+L,1);    is[1]=0;   // int np=0;    for(int i=2;i<L;i++)        if(is[i])        {           // prime[++np]=i;            for(int j=2*i;j<L;j+=i) is[j]=0;        }}bool judge(int x,int y){    if(is[dot[x]]||is[dot[y]]||is[dot[x]+dot[y]]) return 1;    return 0;}