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最新推荐文章于 2017-06-27 09:32:00 发布

原创最新推荐文章于 2017-06-27 09:32:00 发布 · 549 阅读

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最小、最大、次小生成树专栏收录该内容

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本文探讨了最小生成树算法的应用场景，并详细解释了如何使用打表法优化算法执行效率，以解决特定问题。通过实例演示，读者能够理解最小生成树的概念及其在实际问题中的应用。

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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

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+1
using 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+50
int 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;
}