POJ 1416-Shredding Company（DFS-分割数字）

素颜马尾好姑娘i 2022-07-16 14:07 155阅读 0赞

Shredding Company

<table> 
 <tbody> 
  <tr> 
   <td><strong>Time Limit:</strong>&nbsp;1000MS</td> 
   <td>&nbsp;</td> 
   <td><strong>Memory Limit:</strong>&nbsp;10000K</td> 
  </tr> 
  <tr> 
   <td><strong>Total Submissions:</strong>&nbsp;6035</td> 
   <td>&nbsp;</td> 
   <td><strong>Accepted:</strong>&nbsp;3299</td> 
  </tr> 
 </tbody> 
</table>

Description

You have just been put in charge of developing a new shredder for the Shredding Company Although a "normal" shredder would just shred sheets of paper into little pieces so that the contents would become unreadable, this new shredder needs to have the following unusual basic characteristics.  
  
1.The shredder takes as input a target number and a sheet of paper with a number written on it.  
  
2.It shreds (or cuts) the sheet into pieces each of which has one or more digits on it.  
  
3.The sum of the numbers written on each piece is the closest possible number to the target number, without going over it.  
  
For example, suppose that the target number is 50, and the sheet of paper has the number 12346. The shredder would cut the sheet into four pieces, where one piece has 1, another has 2, the third has 34, and the fourth has 6. This is because their sum 43 (= 1 + 2 + 34 + 6) is closest to the target number 50 of all possible combinations without going over 50. For example, a combination where the pieces are 1, 23, 4, and 6 is not valid, because the sum of this combination 34 (= 1 + 23 + 4 + 6) is less than the above combination's 43. The combination of 12, 34, and 6 is not valid either, because the sum 52 (= 12 + 34 + 6) is greater than the target number of 50.  
![1416_1.jpg][]  
Figure 1. Shredding a sheet of paper having the number 12346 when the target number is 50  
  
There are also three special rules :  
  
1.If the target number is the same as the number on the sheet of paper, then the paper is not cut.  
  
For example, if the target number is 100 and the number on the sheet of paper is also 100, then  
  
the paper is not cut.  
  
2.If it is not possible to make any combination whose sum is less than or equal to the target number, then error is printed on a display. For example, if the target number is 1 and the number on the sheet of paper is 123, it is not possible to make any valid combination, as the combination with the smallest possible sum is 1, 2, 3. The sum for this combination is 6, which is greater than the target number, and thus error is printed.  
  
3.If there is more than one possible combination where the sum is closest to the target number without going over it, then rejected is printed on a display. For example, if the target number is 15, and the number on the sheet of paper is 111, then there are two possible combinations with the highest possible sum of 12: (a) 1 and 11 and (b) 11 and 1; thus rejected is printed. In order to develop such a shredder, you have decided to first make a simple program that would simulate the above characteristics and rules. Given two numbers, where the first is the target number and the second is the number on the sheet of paper to be shredded, you need to figure out how the shredder should "cut up" the second number.

Input

The input consists of several test cases, each on one line, as follows :  
  
tl num1  
t2 num2  
...  
tn numn  
0 0  
  
Each test case consists of the following two positive integers, which are separated by one space : (1) the first integer (ti above) is the target number, (2) the second integer (numi above) is the number that is on the paper to be shredded.  
  
Neither integers may have a 0 as the first digit, e.g., 123 is allowed but 0123 is not. You may assume that both integers are at most 6 digits in length. A line consisting of two zeros signals the end of the input.

Output

For each test case in the input, the corresponding output takes one of the following three types :  
  
sum part1 part2 ...  
rejected  
error  
  
In the first type, partj and sum have the following meaning :  
  
1.Each partj is a number on one piece of shredded paper. The order of partj corresponds to the order of the original digits on the sheet of paper.  
  
2.sum is the sum of the numbers after being shredded, i.e., sum = part1 + part2 +...  
  
Each number should be separated by one space.  
The message error is printed if it is not possible to make any combination, and rejected if there is  
more than one possible combination.  
No extra characters including spaces are allowed at the beginning of each line, nor at the end of each line.

Sample Input

50 12346
    376 144139
    927438 927438
    18 3312
    9 3142
    25 1299
    111 33333
    103 862150
    6 1104
    0 0

Sample Output

43 1 2 34 6
    283 144 139
    927438 927438
    18 3 3 12
    error
    21 1 2 9 9
    rejected
    103 86 2 15 0
    rejected

Source

[Japan 2002 Kanazawa][]

## 题目意思： ##

给出一个目标数和一个要分割的数，纸条粉碎机能够将一个数分割成若干个小的数。

用粉碎机去分割要分割的数，使得其分成的若干个数的和是最大的且不大于目标数的分割方法。

如果无论怎么分割，和都大于目标数，则输出error；

否则如果存在不止一种方法能够分割出最大数，则输出rejected；

否则输出分割点的个数和从左往右依次分割出的各个数。

##    解题思路： ##

DFS，令void DFS(int p,int d,int sum)//碎的位置、碎了几刀、当前的和

搜索过程中剪枝：对求出的和大于目标值的情况直接返回不再计算。

对一串数字的各个分割点枚举，不断更新最大值。

#include<iostream>
    #include<cstdio>
    #include<iomanip>
    #include<cmath>
    #include<cstdlib>
    #include<cstring>
    #include<map>
    #include<algorithm>
    #include<vector>
    #include<queue>
    using namespace std;
    #define INF 0x3f3f3f3f
    #define MAXN 10
    int t,a[MAXN],n;//目标数、粉碎数
    int n,len;//目标值、目标值的位数
    int r[MAXN],res[MAXN];//切割位置
    int ans=0,num=0;//总和、刀数
    int flag=-1;//能否分割
    //-1 error 始终大于
    //0 rejected 多个情况
    //1 accept
    void DFS(int p,int d,int sum)//碎的位置、碎了几刀、当前的和
    {
        if(sum>t) return;//求出的和大于目标值，直接返回不再计算
        if(p>=len)//分割点到了末尾，即分割完毕
        {
            if(sum>ans)//更新最大值
            {
                ans=sum;
                num=d;//分割点个数
                for(int i=0; i<d; ++i)
                    res[i]=r[i];//记录各个分割点
                flag=1;//记录但不返回
            }
            else if(sum==ans)//多个情况
                flag=0;
        }
        int temp=0;
        for(int i=p; i<len; ++i)
        {
            temp*=10;//temp存储当前分隔段的值
            temp+=a[i];
            r[d]=temp;
            DFS(i+1,d+1,sum+temp);
        }
    }
    
    int main()
    {
    #ifdef ONLINE_JUDGE
    #else
        freopen("G:/cbx/read.txt","r",stdin);
        //freopen("F:/cb/out.txt","w",stdout);
    #endif
        ios::sync_with_stdio(false);
        cin.tie(0);
        char s[MAXN];
        while(cin>>t>>s)
        {
            n=atoi(s);//目标值
            if(t==0&&n==0) break;
            memset(r,0,sizeof(r));
            memset(res,0,sizeof(res));
            ans=num=0;
            flag=-1;
            len=strlen(s);
            for(int i=0; i<len; ++i)
                a[i]=s[i]-'0';
            DFS(0,0,0);
            if(flag==-1) cout<<"error"<<endl;
            else if(flag==0) cout<<"rejected"<<endl;
            else if(flag==1)
            {
                cout<<ans<<" ";
                for(int i=0; i<num-1; ++i)
                    cout<<res[i]<<" ";
                cout<<res[num-1]<<endl;
            }
        }
    }

[1416_1.jpg]: /images/20220709/5200c645224747cabe235240f4eb7ad0.png
[Japan 2002 Kanazawa]: http://poj.org/searchproblem?field=source&key=Japan+2002+Kanazawa