SRM DIV2 574 TheNumberGameDiv2

【题意】：给你两个数A、B，问从A得到B最少需要的步数,怎么走这样规定：可以A/10得到一个数，或者翻转A得到一个数。如果不能则输出-1；

【算法】

1.找出A中的B，计算步数。

2.找出A中B的逆串，计算并更新最优解。

3.输出最优解。

【编码调试BUG】

1.18行，A中可能不存在B，所以要考虑indexOf返回-1的情况。 

【Java代码】来自菜鸟 

import java.util.*;
import java.util.regex.*;
import java.text.*;
import java.math.*;


public class TheNumberGameDiv2
{
    public int minimumMoves(int A, int B)
    {
        String strA = ""+A;
        String strB = ""+B;
        int count1=Integer.MAX_VALUE,count2=Integer.MAX_VALUE;
        int left,right;
        
        //compute steps if B is in A
        left = strA.indexOf(strB);
        if(left!=-1){
            right = strA.length() - left - strB.length();
            
            if(left==0){
                count1 = right;
            }else{
                count1 = right+left+2;
            }
        }
        
        //compute steps if inverse B is in A
        StringBuilder sb = new StringBuilder(strB);
        sb.reverse();
        strB = sb.toString();
        
        left = strA.indexOf(strB);
        if(left!=-1){
            right = strA.length() - left - strB.length();
            
            if(left==0){
                count2 = right+1;
            }else{
                count2 = right+left+1;
            }
        }
        
        return Math.min(count1, count2)==Integer.MAX_VALUE?-1:Math.min(count1, count2);
    }
    

}
//Powered by KawigiEdit 2.1.4 (beta) modified by pivanof!

【改进版代码】

【分析】

1.步数只需要考虑两点：a，A比B长多少；b,是否需要翻转A。所以不需要初始版本中的left和right。

2.学习了大神对StringBuilder求逆串的使用方法。

import java.util.*;
import java.util.regex.*;
import java.text.*;
import java.math.*;


public class TheNumberGameDiv2
{
    public int minimumMoves(int A, int B)
    {
        String strA = ""+A;
        String strB = ""+B;
        int count1=Integer.MAX_VALUE,count2=Integer.MAX_VALUE;
        int index;
        
        //compute steps if B is in A
        index = strA.indexOf(strB);
        if(index!=-1){
            count1=strA.length() - strB.length();
            if(index!=0){
                count1+=2;
            }
        }
        
        //compute steps if inverse B is in A
        strB = new StringBuilder().append(strB).reverse().toString();        
        index = strA.indexOf(strB);
        if(index!=-1){
            count2 = strA.length() - strB.length()+1;
        }
        
        return Math.min(count1, count2)==Integer.MAX_VALUE?-1:Math.min(count1, count2);
    }
    
//
}
//Powered by KawigiEdit 2.1.4 (beta) modified by pivanof!

【Java代码】来自大神

【分析】算法不同，还没有看出来是什么算法。

import java.util.LinkedList; 
import java.util.HashSet; 

public class TheNumberGameDiv2 { 

  public int minimumMoves(int A, int B) { 
    HashSet<Integer> seen = new HashSet<Integer>(); 
     
     
    LinkedList<TheNumberGameDiv2.state> q = new LinkedList<TheNumberGameDiv2.state>(); 
    q.add(new state(A, 0)); 
    seen.add(A); 
    while (!q.isEmpty()){ 
      state s = q.poll(); 
      if (s.a == B)return s.time; 
      if (!seen.contains(s.a/10)){ 
        seen.add(s.a/10); 
        q.add(new state(s.a/10,s.time+1)); 
      } 
      int r = reverse(s.a); 
      if (!seen.contains(r)){ 
        seen.add(r); 
        q.add(new state(r,s.time+1)); 
      } 
       
    }   
     
    return -1; 
  } 
  public int reverse (int a){ 
    return Integer.valueOf(new StringBuilder().append(a).reverse().toString()); 
  } 
  public static class state{ 
     
    int a; 
    int time; 
    public state(int a, int time){ 
      this .a = a; 
      this.time = time; 
    } 
  } 
}

---

【C++代码】来自大神

【分析】算法不同，貌似用到了DP？

#include <iostream> 
#include <vector> 
#include <algorithm> 
#include <numeric> 
#include <climits> 
#include <bitset> 
#include <string> 
#include <iostream> 
#include <set> 
#include <map> 
#include <cstdio> 
#include <cstdlib> 
#include <list> 
#include <cstring> 
#define mod 1000000007 
using namespace std; 
map <long long, long long> m; 
class TheNumberGameDiv2 
{ 
    private: 
        long long a; 
        long long b; 
        long long f (long long x) 
        { 
            if (x == b) { 
                return 0; 
            } 
            if (m[x] != 0) { 
                return m[x]; 
            } 
            long long y = 0,mm; 
            mm = 100000000LL; 
            m[x] = mm; 
            mm = min (mm, 1 + f (x/10)); 
            long long xx = x; 
            while (x != 0) { 
                y = y*10 + (x%10); 
                x = x/10; 
            } 
            mm = min (mm, 1 + f(y)); 
             
            return m[xx]=mm; 
        } 

    public: 
        int minimumMoves(int A, int B) 
        { 
            this -> a = A; 
            this -> b = B; 
            long long x; 
            x = f (a); 
            if (x >= 100000000LL) { 
                return -1; 
            } 
            return x; 
        } 
};

---

【总结】

算法是多种多样的，大神站得高看得远，DP什么的信手拈来，还需多多学习。