hdoj 2682 Tree

Tree

Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1760    Accepted Submission(s): 514

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

 

 

最小生成树，附上两种代码。  提醒此题需要用打表法，要不会超时。

 

kruskal：

#include<stdio.h>#include<string.h>#include<stdlib.h> #include<math.h>#include<algorithm>#define max 360000+10#define M 1000000+1using namespace std;int set[1000],prime[M];struct line{    int start;    int end;    int cost;}num[max];bool cmp(line a,line b){    return a.cost<b.cost;}int find(int p){    int child=p;    int t;    while(p!=set[p])    p=set[p];    while(child!=p)    {        t=set[child];        set[child]=p;        child=t;    }    return p;}void merge(int x,int y){    int fx=find(x);    int fy=find(y);    if(fx!=fy)    set[fx]=fy;}void dabiao(){    int i,j;    memset(prime,0,sizeof(prime));//素数为0     for(i=2;i<M;i++)    {    if(!prime[i])    {    for(j=2*i;j<M;j+=i)    {    prime[j]=1;    }    }    }    prime[1]=1;}int min(int x,int y){    if(x>y)    return y;    else    return x;}int main(){    int t,city;    __int64 need;    int exist;    int n[1000];    int i,j,x,y,c,k;    dabiao();    scanf("%d",&t);    while(t--)    {        scanf("%d",&city);        for(i=1;i<=city;i++)        {            set[i]=i;            scanf("%d",&n[i]);        }        k=0;        for(i=1;i<=city;i++)        {            for(j=i+1;j<=city;j++)            {                if(!prime[n[i]]||!prime[n[j]]||!prime[n[i]+n[j]])//可以连通  不能连通不记录                 {                    num[k].start=i;                    num[k].end=j;                    num[k].cost=min(min(n[i],n[j]),abs(n[i]-n[j]));                    k++;                }                            }        }        sort(num,num+k,cmp);        need=0;        for(i=0;i<k;i++)        {            if(find(num[i].start)!=find(num[i].end))            {                merge(num[i].start,num[i].end);                need+=num[i].cost;            }        }        exist=0;        for(i=1;i<=city;i++)        {            if(set[i]==i)            exist++;            if(exist>1)            break;        }        if(exist>1)        printf("-1\n");        else        printf("%I64d\n",need);    }    return 0;}

 

prime：

 

#include<stdio.h>#include<math.h>#include<stdlib.h>#include<string.h>#define INF 0x3f3f3f#define max 600+50int map[max][max],visit[max],low[max];int judge[1000000+1];int city;void prime(){    int i,j,next;    int min,mincost=0;    for(i=1;i<=city;i++)    {        visit[i]=0;        low[i]=map[1][i];    }    visit[1]=1;    for(i=2;i<=city;i++)    {        min=INF;        next=1;        for(j=1;j<=city;j++)        {            if(!visit[j]&&min>low[j])            {                min=low[j];                next=j;            }        }        if(min==INF)        {            printf("-1\n");            return ;        }        visit[next]=1;        mincost+=min;        for(j=1;j<=city;j++)        {            if(!visit[j]&&map[next][j]<low[j])            low[j]=map[next][j];        }    }    printf("%d\n",mincost);}void dabiao(){    int i,j;    memset(judge,0,sizeof(judge));    for(i=2;i<1000000+1;i++)    {    if(!judge[i])    {    for(j=2*i;j<1000000+1;j+=i)    judge[j]=1;    }    }    judge[1]=1;}int M(int x,int y){    if(x>y)    return y;    else    return x;}int main(){    int t,i,j,x,y,c;    int n[max];    dabiao();    scanf("%d",&t);    while(t--)    {        scanf("%d",&city);        for(i=1;i<=city;i++)        {            scanf("%d",&n[i]);            for(j=1;j<=city;j++)            {                if(i==j)                map[i][j]=0;                else                map[i][j]=INF;            }        }        for(i=1;i<=city;i++)        {            for(j=i+1;j<=city;j++)            {                if(!judge[n[i]]||!judge[n[j]]||!judge[n[i]+n[j]])                {                    map[i][j]=map[j][i]=M(M(n[i],n[j]),abs(n[i]-n[j]));                }            }        }        prime();    }    return 0;}