Hdu1539 Shredding Company dfs搜索
来源:互联网 发布:自学日语教材知乎 编辑:程序博客网 时间:2024/04/28 01:21
Shredding Company
Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 679 Accepted Submission(s): 222
Problem 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.
The shredder takes as input a target number and a sheet of paper with a number written on it.
It shreds (or cuts) the sheet into pieces each of which has one or more digits on it.
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 52 (= 12 + 34 + 6) is greater than the target number of 50.
There are also three special rules:
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.
If it 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.
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 shedder should "cut up" the second number.
The shredder takes as input a target number and a sheet of paper with a number written on it.
It shreds (or cuts) the sheet into pieces each of which has one or more digits on it.
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 52 (= 12 + 34 + 6) is greater than the target number of 50.
There are also three special rules:
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.
If it 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.
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 shedder should "cut up" the second number.
Input
The input consists of several test cases, each on one line, as follows:
t1 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 bother integers are at most 6 digits in length. A line consisting of two zeros signals the end of the input.
t1 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 bother 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:
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.
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.
sum part1 part2 . . .
rejected
error
In the first type, partj and sum have the following meaning:
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.
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 12346376 144139927438 92743818 33129 314225 1299111 33333103 8621506 11040 0
Sample Output
43 1 2 34 6283 144 139927438 92743818 3 3 12error21 1 2 9 9rejected103 86 2 15 0rejected
Source
Asia 2002, Kanazawa (Japan)
这题可能poj的数据比较水,之前贴的poj的代码在hdu是会wa的,但是hdu的数据也有问题,超出了题目的范围(6位),看讨论版也算坑吧。于是就重写了一遍搜索,原先贴poj上的那题wa到死,至今没有找出来是什么问题。
感觉自己写dfs还是不行,要想清楚后再下手
下面代码:
#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>#include <cmath>#include <vector>#include <sstream>#include <queue>#include <map>using namespace std;const int maxn = 100010;int roof,len,maxNow;char str[100];vector<int> nowWay;vector<int> finalWay;map<int, int> vis;int charToInt(char *s,int l,int r){ int sum = 0,i,j=0; for(i=r;i>=l;i--){ sum += (s[i]-'0')*pow(10, j++); } return sum;}void dfs(int pos,int sum){ int now; if(sum>roof){ return ; } if(pos==len){ vis[sum]++; if(sum>maxNow&&sum<=roof){ maxNow = sum; finalWay.clear(); for(int i = 0;i<nowWay.size();i++){ finalWay.push_back(nowWay[i]); } } return ; } int i = pos; for(int j=i;j<len;j++){ now = charToInt(str, i, j); nowWay.push_back(now); dfs(j+1,sum+now) ; nowWay.pop_back(); }}int main(){ while(~scanf("%d %s",&roof,str)){ if(str[0]=='0'&&roof==0){ return 0; } vis.clear(); len = strlen(str); maxNow = -1; nowWay.clear(); dfs(0, 0); if(maxNow==-1){ printf("error\n"); continue; } if(vis[maxNow]==1){ printf("%d",maxNow); for(int i = 0;i<finalWay.size();i++){ printf(" %d",finalWay[i]); } printf("\n"); }else{ printf("rejected\n"); continue; } } return 0;}
0 0
- Hdu1539 Shredding Company dfs搜索
- HDU1539 Shredding Company
- poj1461 Shredding Company(DFS)
- Shredding Company--DFS
- Shredding Company--(DFS)
- Shredding Company (dfs)
- Shredding Company (dfs)
- poj1416--Shredding Company(搜索练习4,按位dfs)
- poj-1416-Shredding Company-dfs
- poj 1416 Shredding Company dfs
- poj 1416 Shredding Company (dfs)
- Shredding Company (hdu 1539 dfs)
- poj -1416-Shredding Company-DFS
- poj-1416 Shredding Company DFS
- hdoj 1539 Shredding Company【DFS】
- [POJ 1416]Shredding Company[DFS]
- POJ-1416 Shredding Company(DFS)
- POJ.1416 Shredding Company (DFS)
- 自己作为一个程序员的心路历程
- Qt中local and expressions无法查看当前栈变量
- Android响应式编程开发RxAndroid(3):map
- HDU5961 传递
- HDU-Monkey and Banana
- Hdu1539 Shredding Company dfs搜索
- GPL协议的相关条款
- 1023. 组个最小数
- Ubuntu更改Mysql新建的数据库存放位置
- 为学Linux,我看了这些书
- C/C++ 中的0长数组(柔性数组)
- liunx 查看网卡驱动信息
- 胶囊体阴影
- String字符串用法集合