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2013
11-09

POJ 1160 Post Office [解题报告] Java

Post Office

问题描述 :

There is a straight highway with villages alongside the highway. The highway is represented as an integer axis, and the position of each village is identified with a single integer coordinate. There are no two villages in the same position. The distance between two positions is the absolute value of the difference of their integer coordinates.

Post offices will be built in some, but not necessarily all of the villages. A village and the post office in it have the same position. For building the post offices, their positions should be chosen so that the total sum of all distances between each village and its nearest post office is minimum.

You are to write a program which, given the positions of the villages and the number of post offices, computes the least possible sum of all distances between each village and its nearest post office.

输入:

Your program is to read from standard input. The first line contains two integers: the first is the number of villages V, 1 <= V <= 300, and the second is the number of post offices P, 1 <= P <= 30, P <= V. The second line contains V integers in increasing order. These V integers are the positions of the villages. For each position X it holds that 1 <= X <= 10000.

输出:

The first line contains one integer S, which is the sum of all distances between each village and its nearest post office.

样例输入:

10 5
1 2 3 6 7 9 11 22 44 50

样例输出:

9

解题代码:

/* @author: */
import java.util.Scanner;
import java.util.Arrays;
public class Main{
 static int num[]=new int[302];
 static int s[][]=new int[302][302];

 static void cost(int n){
     int i,j,k;
     for(i=1;i< n;i++){
         for(k=i+1; k <= n; k ++){
            int mid=(i+k)/2;
            for(j=i;j< mid;j++) s[i][k]+=num[mid]-num[j];
            for(j=mid+1;j<=k;j++) s[i][k]+=num[j]-num[mid];
         }
      }
 }

 public static void main(String args[])
{
 Scanner sc=new Scanner(System.in);
     int f[][]=new int[302][32];
     int n,k;

     int i,j,m;
     for(i=0;i< s.length;i++)
          Arrays.fill(s[i],0);
     for(i=0;i< f.length;i++)
          Arrays.fill(f[i],1000000);
     n=sc.nextInt();
     m=sc.nextInt();
    
     for(i=1;i<=n;i++)
        num[i]=sc.nextInt();
     cost(n);
     for(i=1;i<=n;i++) f[i][1]=s[1][i];
     for(k=2;k<=m;k++){
         for(i=1;i<=n;i++){
            for(j=1;j< i;j++)
                 if(f[j][k-1]+s[j+1][i]< f[i][k]) f[i][k]=f[j][k-1]+s[j+1][i];        
         }                  
     }
     System.out.printf("%d\n",f[n][m]);
    }
}

  1. 。。苹果那个可不是后门,就是漏洞而已。基本原理就是黑客用穷举法列出来的。苹果应该在所有输密码的地方对于连续尝试错误密码的这种情况下行锁定啊或是别的举动,的确,99%的地方都有处理,但其中find my iphone的一个不太常用的地方被发现漏了这种处理,

  2. 5.1处,反了;“上一个操作符的优先级比操作符ch的优先级大,或栈是空的就入栈。”如代码所述,应为“上一个操作符的优先级比操作符ch的优先级小,或栈是空的就入栈。”