HDU 1015 Safecracker【数值型DFS】

Safecracker

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3966    Accepted Submission(s): 2028

Problem Description

 

=== Op tech briefing, 2002/11/02 06:42 CST === 
"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."

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

"For example, given target 1 and letter set ABCDEFGHIJKL, one possible solution is FIECB, since 6 - 9^2 + 5^3 - 3^4 + 2^5 = 1. There are actually several solutions in this case, and the combination turns out to be LKEBA. Klein thought it was safe to encode the combination within the engraving, because it could take months of effort to try all the possibilities even if you knew the secret. But of course computers didn't exist then."

=== Op tech directive, computer division, 2002/11/02 12:30 CST ===

"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."

 

Sample Input

 

1 ABCDEFGHIJKL

11700519 ZAYEXIWOVU

3072997 SOUGHT

1234567

THEQUICKFROG

0 END

 

Sample Output

 

LKEBA

YOXUZ

GHOST

no solution

【参考】：http://blog.csdn.net/pengwill97/article/details/54882698

【题意】：在字符串中找五个字符可以满足上面的式子。存在多组，输出字典序最大的，没有的话，输出no solution。

【分析】：下面的for循环非常像dfs地图的四向搜索，但是len指的是数据中给定的字母序列的长度。那么就指，下一个搜索的目标要在所有的字母序列中找，哪些可以作为搜索目标呢？首先就是这个字母没有被选定过（!visit[i] ）并且现在解还没有找到（!judge）。 进入if后，首先把数组b中depth的位置赋值为a[i]，代表我数组b选定了你a中i这个位置的数字（或者说是字母）,并且在visit中置为选择过了，dfs(depth+1)继续寻找下一个位置的搜索目标。别忘了最后把visit[i]置为0（无后效性）。相比前边2题，此题的收获就在于：原先的地图四向搜索，也可以变成这样从几个字符，数字中寻找可行的解。活学活用，非常重要呀！不要忘记题目要求：找到一组字典序最大的解即可。首先是递归边界，如果找到了解（judge为真），停止递归；亦或是当depth为5（代表找到了5个数字）的时候，用check函数判断一下是否满足题目要求。若都不满足递归边界，继续搜索。 

【代码】：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
char a[], b[];
int visit[];
int n,num,len;
bool judge = false;

bool cmp(char a, char b)
{
    return a>b;
}
void init()
{
    len = strlen(a);
    judge = false;
    memset(visit,,sizeof(visit));
    sort(a,a+len,cmp);
    for(int i = ;i<len;++i){
        a[i] = a[i] -'A' + ;
    }
}
void lnit()
{
    for(int i = ;i<;++i)
        b[i] = b[i]+'A'-;
}
bool check()
{
    if(n == b[] - b[]*b[] + b[]*b[]*b[] - b[]*b[]*b[]*b[] + b[]*b[]*b[]*b[]*b[]){
        judge = true;
        return true;
    }else return false;
}
void dfs(int depth)
{
    //递归边界
    if(judge) return;
    if(depth == ) {check(); return;}
    for(int i = ;i<len;++i){
        if(!visit[i]&&!judge){
            b[depth] = a[i];
            visit[i] = ;
            dfs(depth+);
            visit[i] = ;
        }
    }
}
int main()
{

    //freopen("in.txt","r",stdin);
    while(scanf("%d %s",&n,a)&& !(n== && !strcmp(a,"END"))){
        init();
        dfs();
        lnit();
        if(judge)printf("%s\n",b);
        else printf("no solution\n");
    }
    return ;
}

数值型DFS