hdoj 2682 Tree 【最小生成树】

Tree

Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2099    Accepted Submission(s): 605

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

打表

代码：

#include<stdio.h>#include<string.h>#include<stdlib.h>#include<algorithm>#include<math.h>#define M 650#define INF 0x3f3f3fusing namespace std;int a[M];int judge[1000010];int vis[M];int low[M];int map[M][M];int n;void dabiao()//´ò±í£» {int i,j;memset(judge,0,sizeof(judge));for(i=2;i<1000010;i++){if(!judge[i]){for(j=i*2;j<1000010;j+=i){judge[j]=1;}}}judge[1]=1;}void init(){int i,j;for(i=1;i<=n;i++){for(j=1;j<=n;j++){if(i==j)map[i][j]=0;elsemap[i][j]=INF;}}}void prim(){int min,i,j,next,sum=0;memset(vis,0,sizeof(vis));for(i=1;i<=n;i++){low[i]=map[1][i];}vis[1]=1;for(i=1;i<n;i++){min=INF;next=1;for(j=1;j<=n;j++){if(!vis[j]&&min>low[j]){min=low[j];next=j;}}if(min==INF){printf("-1\n");return ;}sum+=min;vis[next]=1;for(j=1;j<=n;j++){if(!vis[j]&&low[j]>map[next][j]){low[j]=map[next][j];}}}printf("%d\n",sum);}int main(){int T;int i,j;dabiao();scanf("%d",&T);while(T--){scanf("%d",&n);init();for(i=1;i<=n;i++){scanf("%d",&a[i]);}for(i=1;i<=n;i++){    for(j=i+1;j<=n;j++)    {if(!judge[a[i]]||!judge[a[j]]||!judge[a[i]+a[j]]){map[i][j]=map[j][i]=min(min(a[i],a[j]),abs(a[i]-a[j]));}}}prim();}return 0;}