Tree(并查集)

Tree

Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1771    Accepted Submission(s): 517

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
把满足题意的两点及最小花费。
#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;#define M 1000010int notprime[M];int s[1000010];int f[M]; struct node{int a;int b;int c;}p[1000010];void fun(){int i,j;memset(notprime,0,sizeof(notprime));notprime[1]=1;for(i=2;2*i<M;i++){if(!notprime[i]){for(j=2*i;j<M;j+=i)notprime[j]=1;}}}bool cmp(node x,node y){return x.c<y.c;}int find(int x){return x==f[x]?x:f[x]=find(f[x]);}int bin(int x1,int x2,int x3){int t1,t2,x4=0;t1=find(x1);t2=find(x2);if(t1!=t2){f[t1]=t2;x4+=x3;}return x4;}int main(){int T;fun();scanf("%d",&T);while(T--){int n,m;int i,j,k=0;memset(p,0,sizeof(p));scanf("%d",&n);for(i=1;i<=n;i++){f[i]=i;scanf("%d",&s[i]);}for(i=1;i<=n;i++){   // f[i]=i;//scanf("%d",&s[i]);//不能这样写。会超时的 for(j=1;j<i;j++){if(!notprime[s[i]]||!notprime[s[j]]||!notprime[s[i]+s[j]]){p[k].a=i;p[k].b=j;//f[i]=j;p[k++].c=min(min(s[i],s[j]),abs(s[i]-s[j]));}}}sort(p,p+k,cmp);//for(i=0;i<k;i++){//printf("%d %d %d\n",p[i].a,p[i].b,p[i].c);//}int sum=0;for(i=0;i<k;i++){ if(find(p[i].a)!=find(p[i].b)) sum+=bin(p[i].a,p[i].b,p[i].c);    }    int l=0;   for(i=1;i<=n;i++){   if(f[i]==i)   l++;   }   if(l>1)   printf("-1\n");   else   printf("%d\n",sum);} }