HDU2682 Tree 题解 【最小生成树】【图论】【Kruskal】

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
4
Sample Output
4
-1
解题报告
题意就是有N座城市，每个城市有一定的幸福值a[i]。对于任意两个城市i和j，如果a[i]，a[j]，a[i]+a[j]中任意一者的值为素数，那么他们的边权就是min(min(a[i],a[j]),abs(a[i]-a[j]))。问题就是这一幅图的最小生成树。
显然，边一旦建出来了，这就是一道裸题。

#include<cstdio>#include<cmath>#include<cstring>#include<algorithm>using namespace std;const int N=600;const int M=600*600;const int H=1000000*2;const int INF=522133279;int head[N+5],num,father[N+5];int n;int hap[N+5];bool prime[H*2+5];int T;struct edge{    int u,v;    int w;    int next;    edge(){next=-1;}    bool operator<(const edge& b)const    {return w<b.w;}}ed[M<<2];inline void build(int u,int v,int w){    ed[++num].u=u;    ed[num].v=v;    ed[num].w=w;    ed[num].next=head[u];    head[u]=num;}int getfather(int x)  {    return father[x]!=x?(father[x]=getfather(father[x])):x;  }  inline int unionn(int x,int y)  {    return ((x=getfather(x))!=(y=getfather(y)))&&(father[x]=y);  }void pri(){    prime[0]=1,prime[1]=1,prime[2]=0;    for(int i=2;i<=H;i++)    if(!prime[2])    {        for(int j=i+i;j<=H;j+=i)        prime[j]=1;    }}  inline int kruskal(){    int ans=0;    int tot=0;    for(int i=1;i<=n;++i)    father[i]=i;    sort(ed+1,ed+1+num);    for(int i=1;i<=num;++i)    {        int u=getfather(ed[i].v),v=getfather(ed[i].u);        if(u!=v)        {            ++tot,ans+=ed[i].w;            unionn(u,v);        }        if(tot==n-1)return ans;    }    return -1;}inline int w(int i,int j){    if(!prime[i]||!prime[j]||!prime[i+j])    return min(min(i,j),abs(i-j));}int main(){    pri();    for(scanf("%d",&T);T;T--)    {        num=0;        memset(head,-1,sizeof(head));        memset(hap,0,sizeof(hap));        scanf("%d",&n);        for(int i=1;i<=n;i++)        {            scanf("%d",&hap[i]);            for(int j=1;j<=i-1;j++)            {                if(!prime[hap[i]]||!prime[hap[j]]||!prime[hap[i]+hap[j]])                {                    build(i,j,w(hap[i],hap[j]));                    build(j,i,w(hap[i],hap[j]));                }            }        }        printf("%d\n",kruskal());    }    return 0;}