hdu 2682

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

题意概括：给n个数Vi(0<=Vi<=1000000)，这n个数的编号是从1到n的，如果任意两个数如果他们中有一个数是素数或者他们相加的值是素数，那么可以认为这两个数可以连接，权值是Min(Min(VA , VB),|VA-VB|)。

解题思路：把符合情况的所有组合的编号权值存入结构体中，Kruskal算一遍。

错误分析：这一题输入的n个数Vi是有等于0的情况的，但是0不属于素数，但是这一题无论把0当成素数还是非素数可以AC。

代码：

#include<stdio.h>#include<math.h>#include<string.h>#include<algorithm>using namespace std;#define M 1500#define N 1000000int i,j,m,n,l;int f[N*2],prime[N+100];struct edge {int u,v,w;}e[M*M];void p(){memset(prime,0,sizeof(prime));for(i=2;i<N+100;i++){if(prime[i]==0){for(j=i*2;j<N+100;j+=i){prime[j]=1;}}}prime[1]=1;}int min(int a,int b){if(a>b)a=b;return a;}int abs(int a){if(a<0)a=-a;return a;}bool cmp(edge a,edge b){return a.w<b.w;}int getf(int u){if(f[u]==u){return u;}else{f[u]=getf(f[u]);return f[u];}}int merge(int u,int v){int t1=getf(u);int t2=getf(v);if(t1!=t2){f[t2]=t1;return 1;}return 0;}void Kruskal(){int num=0,sum=0;for(i=0;i<=1000100;i++){f[i]=i;}//printf("n=%d l=%d\n",n,l);for(int i=0;i<l;i++){if(merge(e[i].u,e[i].v)!=0){//printf("u=%d v=%d\n",e[i].u,e[i].v);num++;sum+=e[i].w;}//if(num==n-1)//break;}//printf("%d\n",num);if(num==0){printf("-1\n");}elseprintf("%d\n",sum);}int main(){int t,a[1000];scanf("%d",&t);while(t--){l=0;scanf("%d",&n);for(i=1;i<=n;i++){scanf("%d",&a[i]);}p();for(i=1;i<=n-1;i++){for(j=i+1;j<=n;j++){if(prime[a[i]]==0||prime[a[j]]==0||prime[a[i]+a[j]]==0){e[l].u=i;e[l].v=j;e[l].w=min(min(a[i],a[j]),abs(a[i]-a[j]));l++;}}}sort(e,e+l,cmp);Kruskal();}return 0;}