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2013
11-13

POJ 3670 Eating Together [解题报告] Java

Eating Together

问题描述 :

The cows are so very silly about their dinner partners. They have organized themselves into three groups (conveniently numbered 1, 2, and 3) that insist upon dining together. The trouble starts when they line up at the barn to enter the feeding area.

Each cow i carries with her a small card upon which is engraved Di (1 ≤ Di ≤ 3) indicating her dining group membership. The entire set of N (1 ≤ N ≤ 30,000) cows has lined up for dinner but it’s easy for anyone to see that they are not grouped by their dinner-partner cards.

FJ’s job is not so difficult. He just walks down the line of cows changing their dinner partner assignment by marking out the old number and writing in a new one. By doing so, he creates groups of cows like 111222333 or 333222111 where the cows’ dining groups are sorted in either ascending or descending order by their dinner cards.

FJ is just as lazy as the next fellow. He’s curious: what is the absolute mminimum number of cards he must change to create a proper grouping of dining partners? He must only change card numbers and must not rearrange the cows standing in line.

输入:

* Line 1: A single integer: N
* Lines 2..N+1: Line i describes the i-th cow’s current dining group with a single integer: Di

输出:

* Line 1: A single integer representing the minimum number of changes that must be made so that the final sequence of cows is sorted in either ascending or descending order

样例输入:

5
1
3
2
1
1

样例输出:

1

解题代码:

/* @author: */
import java.util.Scanner;
import java.util.Arrays;
public class Main{
  static int mintwo(int a,int b){ return a< b?a:b; }
  static int minthird(int a,int b,int c){
    int temp=mintwo(a,b);
    return mintwo(temp,c);   
} 

 public static void main(String args[]){
   Scanner sc=new Scanner(System.in);
 
   int f[][]=new int[30010][3];
   int g[][]=new int[30010][3];
   int v[]=new int[30010];
   int n=sc.nextInt();
    
  for(int i=1;i<=n;i++){
   v[i]=sc.nextInt();  
   f[i][0]=f[i-1][0];
   if(v[i]!=1) f[i][0]++;
   f[i][1]=mintwo(f[i-1][0],f[i-1][1]);
   if(v[i]!=2) f[i][1]++;
   f[i][2]=minthird(f[i-1][0],f[i-1][1],f[i-1][2]);
   if(v[i]!=3) f[i][2]++;                 
  }
     for(int i=1;i<=n;i++){
        g[i][2]=g[i-1][2];
        if(v[i]!=3) g[i][2]++;
        g[i][1]=mintwo(g[i-1][2],g[i-1][1]);
        if(v[i]!=2) g[i][1]++;
        g[i][0]=minthird(g[i-1][2],g[i-1][1],g[i-1][0]);
        if(v[i]!=1) g[i][0]++;        
     }
     int mins=100000000;
     for(int i=0;i< 3;i++){
          if(mins>f[n][i]) mins=f[n][i];
          if(mins>g[n][i]) mins=g[n][i];       
     }
     System.out.printf("%d\n",mins);
   }
}

  1. 有两个重复的话结果是正确的,但解法不够严谨,后面重复的覆盖掉前面的,由于题目数据限制也比较严,所以能提交通过。已更新算法

  2. 给你一组数据吧:29 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 1000。此时的数据量还是很小的,耗时却不短。这种方法确实可以,当然或许还有其他的优化方案,但是优化只能针对某些数据,不太可能在所有情况下都能在可接受的时间内求解出答案。

  3. 您没有考虑 树的根节点是负数的情况, 若树的根节点是个很大的负数,那么就要考虑过不过另外一边子树了