hdoj 1015 Safecracker---盗窃贼

最新推荐文章于 2018-10-27 21:06:19 发布

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#combinations #numbers #integer #input #features #each

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本文详细解析了一道算法难题，涉及利用深搜技术解决与Klein安全箱组合有关的问题。通过实例演示，深入理解算法实现过程，并探讨了相关数据结构应用及排序技巧。同时，分享了在解决此类问题时的思考与经验总结。

AOJ 怎么了。。。还不好。。

English is very useful !深深为自己的英文感到惋惜。。。。。。

"The item is locked in a Klein safe behind a painting in the second-floor library. Klein safes are extremely rare; most of them, along with Klein and his factory, were destroyed in World War II. Fortunately old Brumbaugh from research knew Klein's secrets and wrote them down before he died. A Klein safe has two distinguishing features: a combination lock that uses letters instead of numbers, and an engraved quotation on the door. A Klein quotation always contains between five and twelve distinct uppercase letters, usually at the beginning of sentences, and mentions one or more numbers. Five of the uppercase letters form the combination that opens the safe. By combining the digits from all the numbers in the appropriate way you get a numeric target. (The details of constructing the target number are classified.) To find the combination you must select five letters v, w, x, y, and z that satisfy the following equation, where each letter is replaced by its ordinal position in the alphabet (A=1, B=2, ..., Z=26). The combination is then vwxyz. If there is more than one solution then the combination is the one that is lexicographically greatest, i.e., the one that would appear last in a dictionary."
"Develop a program to find Klein combinations in preparation for field deployment. Use standard test methodology as per departmental regulations. Input consists of one or more lines containing a positive integer target less than twelve million, a space, then at least five and at most twelve distinct uppercase letters. The last line will contain a target of zero and the letters END; this signals the end of the input. For each line output the Klein combination, break ties with lexicographic order, or 'no solution' if there is no correct combination. Use the exact format shown below."

v - w^2 + x^3 - y^4 + z^5 = target

Sample Input

1 ABCDEFGHIJKL
11700519 ZAYEXIWOVU
3072997 SOUGHT
1234567 THEQUICKFROG
0 END

Sample Output

LKEBA
YOXUZ
GHOST
no solution

英语看了半天。。诶呀呀。。。。谷哥、度娘终于告诉我这道题的意思！

果断深搜记得曾做过幂集原理一样啊！

一次就Accepted了。。。相当高兴。。。

今天加深了对栈保存数据的应用的印象本来想用C++的sort 的，额。。。好像不起作用。。学了快速排序其实也挺简单的。。只不过以前想的太多了。。。

记一下深搜的特点退出条件加搜索加设置相关变量恢复相关变量。。。

#include <iostream>
#include <algorithm>
#include <string.h>
#include <stdlib.h>
using namespace std;
typedef struct Node
{
  char ch;
  int num;
}Node;
int target;
char str[27];
Node a[27];
int len;
bool visited[27];
char stack[6];
int top=0;
int flag=0;
int cmp(const void *a,const void *b)
{
	Node *c=(Node *)a;
	Node *d=(Node *)b;
	return d->ch-c->ch;
}
int caculate(int dep,int n)
{
  switch(dep)
  {
  case 0: return n;
  case 1: return -n*n;
  case 2: return n*n*n;
  case 3: return -n*n*n*n;
  case 4: return n*n*n*n*n;
  }
  return 0;
}
void print()
{
	int i;
	for(i=0;i<top;i++)
	{
		cout<<stack[i];
	}
	cout<<endl;
}
void dfs(int dep,int ans)
{
  if(dep>5) return;
  if(dep==5&&ans==target)
  {
	  //cout<<"YES"<<endl;
	  flag=1;
	  print();
	  return;
  }
  int i;
  for(i=0;i<len;i++)
  {
      if(!visited[i])
	  {
                  visited[i]=true;
		stack[top++]=a[i].ch;
                  ans+=caculate(dep,a[i].num);
		dfs(dep+1,ans);
		if(flag==1)return;
		visited[i]=false;
		top--;
		ans-=caculate(dep,a[i].num);
	  }
  }
}
int main()
{
 while(cin>>target>>str,target!=0&&strcmp(str,"END")!=0)
 {
 len=strlen(str);
 int i;
 //将字符串转化为数字并保存到结点数组中
 for(i=0;i<len;i++)
 {
   a[i].ch=str[i];
   a[i].num=str[i]-'A'+1;
 }
 qsort(a,len,sizeof(a[0]),cmp);
 memset(visited,false,27);
 dfs(0,0);
 if(flag==0)
	 cout<<"no solution"<<endl;
 flag=0;
 top=0;

 }
 return 0;
}