hdu 2682 Tree 最小生成树~~~~水题一枚，，用到了筛法求素数，我竟然在格式上面PE了两次！！

最新推荐文章于 2022-02-19 22:33:39 发布

原创最新推荐文章于 2022-02-19 22:33:39 发布 · 1.3k 阅读

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#hdu 2682 #Tree #最小生成树 #筛法求素数 #Prim

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本文讨论了如何在给定的城市集合中，通过连接质数幸福值的城市以达到最小化总连接成本的目的，提供了求解算法及实例分析。

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Tree

Time Limit: 6000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1754 Accepted Submission(s): 509

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

Author

Teddy

转载请申明： http://blog.youkuaiyun.com/lionel_d

题目给的输出样例很具有迷惑性啊！！竟然只有一个换行，o(╯□╰)o

好啊，是我笨

看代码：

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#define MAX 650
#define INF 1000000000

int graph[MAX][MAX] , value[MAX];
bool NotPrime[2001000] ;
int prim(int n)
{
	int lowCost[MAX];
	bool visited[MAX] ;
	memset(visited,false,sizeof(visited)) ;
	int sum = 0 ;
	for(int i = 0 ; i < n ; ++i)
	{
		lowCost[i] = graph[0][i] ;
	}
	visited[0] = true ;
	for(int i = 0 ; i < n-1 ; ++i)
	{
		int index = -1 , min = INF ;
		for(int j = 0 ; j < n ; ++j)
		{
			if(!visited[j] && lowCost[j]<min)
			{
				index = j ;
				min = lowCost[j] ;
			}
		}
		if(index == -1)
		{
			if(i < n-1)
			{
				return INF ;
			}
			break ;
		} 
		sum += min ;
		visited[index] = true ;
		for(int j = 0 ; j < n ; ++j)
		{
			if(!visited[j] && lowCost[j] > graph[index][j])
			{
				lowCost[j] = graph[index][j] ;
			}
		}
	}
	return sum ;
}

int min(int a , int b)
{
	return a>b?b:a ;
}

int main()
{
	int t ;
	NotPrime[0] = NotPrime[1] = true ;
	for(int i = 2 ; i < 2001000/2 ; ++i)
	{
		for(int j = 2 ; j*i < 2001000 ; ++j)
		{
			NotPrime[j*i] = true ;
		}
	}
	scanf("%d",&t);
	while(t--)
	{
		int n ;
		scanf("%d",&n);
		for(int i = 0 ; i < n ; ++i)
		{
			scanf("%d",&value[i]) ;
		}
		for(int i = 0 ; i < n ; ++i)
		{
			for(int j = 0 ; j < i ; ++j)
			{
				if(!NotPrime[value[i]] || !NotPrime[value[j]] || !NotPrime[abs(value[i]+value[j])])
				{
					graph[i][j] = graph[j][i] = min(min(value[i],value[j]),abs(value[i]-value[j]))  ;
				}
				else
				{
					graph[i][j] = graph[j][i] = INF ;
				}
			}
			graph[i][i] = 0 ;
		}
		int sum = prim(n) ;
		if(sum == INF)
		{
			puts("-1");
		}
		else
		{
			printf("%d\n",sum) ;
		}
	}
	return 0 ;
}

与君共勉