HDU 2682 Tree

Tree

Time Limit : 6000/2000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 15   Accepted Submission(s) : 2

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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

Author

Teddy

Source

2009浙江大学计算机研考复试（机试部分）——全真模拟

 

思路：Prim，这题数据有问题

 

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
using namespace std;
int prime[2000040];
int number[1020];
int map[1020][1020];
int pre[1020];
const int MAXN = 100000000;
bool p[1020];
int sum;
int dist[1020];
int the_last_flag;
int t,n;
void Prim(int n,int dist[1020],int map[1020][1020],int pre[1020])
{
    int minn;
    for(int i = 2;i <= n;i ++)
    {
        p[i] = false;
        dist[i] = map[1][i];
        pre[i] = 1;
    }
    dist[1] = 0;
    p[1] = true;
    for(int i = 1;i <= n - 1;i ++)
    {
        minn = MAXN;
        the_last_flag = 0;
        for(int j = 1;j <= n;j ++)
        {
            if(!p[j] && minn > dist[j])
            {
                minn = dist[j];
                the_last_flag = j;
            }
        }
        if(the_last_flag == 0)
           return ;
        p[the_last_flag] = true;
        for(int j = 1;j <= n;j ++)
        {
            if(!p[j] && map[the_last_flag][j] != MAXN && dist[j] > map[the_last_flag][j])
            {
                dist[j] = map[the_last_flag][j];
                pre[j] = the_last_flag;
            }
        }

    }
}
int main()
{
    memset(prime,0,sizeof(prime));
    prime[1] = 1;
    prime[0] = 1;
    for(int i = 2;i <= 2000020;i ++)
       for(int j = 2;i * j <= 2000020;j ++)
           prime[i * j] = 1;
    scanf("%d",&t);
    while(t --)
    {
        scanf("%d",&n);
        for(int i = 1;i <= n;i ++)
           scanf("%d",&number[i]);
        for(int i = 1;i <= n;i ++)
          for(int j = 1;j <= n;j ++)
          {
              if(i == j)
                 map[i][j] = 0;
              else
              {
                  if(prime[number[i]] == 0 || prime[number[j]] == 0
                      || prime[number[i] + number[j]] == 0)
                  {
                      //printf("%d %d    ",number[i],number[j]);
                      int x = min(number[i],number[j]);
                      int y = abs(number[i] - number[j]);
                      map[i][j] = min(x,y);
                      //printf("%d %d %d\n",i,j,map[i][j]);
                  }
                  else
                  {
                      map[i][j] = MAXN;map[j][i] = MAXN;
                  }
              }
          }
        /*for(int i = 1;i <= n;i ++)
        {
            for(int j = 1;j <= n;j ++)
                 printf("%d ",map[i][j]);
            printf("\n");
        }*/
        the_last_flag = 0;
        sum = 0;
        Prim(n,dist,map,pre);
        if(the_last_flag == 0)
            printf("-1\n");
        else
        {
            for(int i = 1;i <= n;i ++)
               sum += dist[i];
            printf("%d\n",sum);
        }
    }
    return 0;
}