http://acm.hdu.edu.cn/showproblem.php?pid=2682
Tree
Time Limit: 6000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3581 Accepted Submission(s): 1118
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
2
5
1
2
3
4
5
4
4
4
4
4Sample Output
4
-1Author
Teddy
Source
Recommend
lcy
思路
题目大意：有N个城市，每个城市有一个幸福值，如果两个城市A、B的幸福值分别为VA、VB，如果VA是素数，
或者VB是素数，又或者VA+VB是素数，则城市A和B就能连接一条路，建路的所用花费为Min(Min(VA , VB),|VA-VB|)。
问：现在想要建几条路，使得能够连接所有的城市，所需要建设的最少路程和是多少？
思路：就是求最小生成树，先用素数筛选法将素数打表，然后根据题意建边。最后就是用Kruskal求最小生成树就行了。
这个大佬AC了，思路一致：点击链接
没AC Code （样例都过了啊，就是WA!求大佬赐教！）
//Min(Min(VA , VB),|VA-VB|)。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int nmax=610;
const int mmax=360000+10;
const int pmax=1000000+10;
int prime[pmax];//素数数组
int mark[pmax];//全体数的标记数组
int s[nmax];
int n,m;//点数、边数
struct Edge{
int u,v;
int val;//起点、终点、边权
}edge[mmax];
bool cmp(Edge a,Edge b){
return a.val<b.val; //按照边权从小到大排序，求最小生成树
}
int father[nmax];
int findFather(int u){
if(u==father[u]) return u;
else{
int f=findFather(father[u]);
father[u]=f;
return f;
}
}
void init(int n){
for(int i=1;i<=n;i++){
father[i]=i;
}
}
void Kruskal(){//返回最小生成树的边权和
init(n);
int cnt=0;//有效合并次数
int ans=0;//最小边权和
for(int i=0;i<m;i++){//遍历m条边
int fu=findFather(edge[i].u);
int fv=findFather(edge[i].v);
if(fu!=fv){
father[fu]=fv;
ans+=edge[i].val;
cnt++;
}
if(cnt==n-1)//合并了N-1条边，已经找到了最小生成树
break;
}
if(cnt==n-1)//找到最小生成树
printf("%d\n",ans);
else //图不连通
printf("-1\n");
}
int main(int argc, char** argv) {
int T;
while(scanf("%d",&T)!=EOF){
while(T--){
scanf("%d",&n);
memset(father,0,sizeof(father));
memset(prime,0,sizeof(prime));
memset(mark,0,sizeof(mark));
memset(s,0,sizeof(s));
memset(edge,0,sizeof(edge));
init(n);
prime[0]=1;
prime[1]=1;
prime[2]=0;
for(int i=2;i<=1000000;i++){
if(!prime[i]){
for(int j=i*2;j<=1000000;j+=i){
prime[j]=1;
}
}
}
for(int i=1;i<=n;i++){
scanf("%d",&s[i]);
}
int id=0;
for(int i=1;i<=n;i++){
for(int j=i+1;j<=n;j++){
int tmp=abs(s[i]-s[j]);
//if(!mark[s[i]]||!mark[s[j]]||!(mark[s[i]]+mark[s[j]])){
if(prime[s[i]]==0||prime[s[j]]==0||(prime[s[i]]+prime[s[j]])==0){
edge[id].u=i;
edge[id].v=j;
edge[id].val=min(tmp,min(s[j],s[i]));
id++;
}
}
}
m=id;
sort(edge,edge+m,cmp);
if(m>=n-1){
Kruskal();
}
else{
printf("-1\n");
}
}
}
return 0;
}