Poj1011 Sticks 小木棍dfs搜索

Sticks

 

Description

George took sticks of the same length and cut them randomly until all parts became at most 50 units long. Now he wants to return sticks to the original state, but he forgot how many sticks he had originally and how long they were originally. Please help him and design a program which computes the smallest possible original length of those sticks. All lengths expressed in units are integers greater than zero.

Input

The input contains blocks of 2 lines. The first line contains the number of sticks parts after cutting, there are at most 64 sticks. The second line contains the lengths of those parts separated by the space. The last line of the file contains zero.

Output

The output should contains the smallest possible length of original sticks, one per line.

Sample Input

9
5 2 1 5 2 1 5 2 1
4
1 2 3 4
0

Sample Output

6
5

Source

 

中文题意

乔治有一些同样长的小木棍，他把这些木棍随意砍成几段，直到每段的长都不超过50。

现在，他想把小木棍拼接成原来的样子，但是却忘记了自己开始时有多少根木棍和它们的长度。

给出每段小木棍的长度，编程帮他找出原始木棍的最小可能长度

输出

仅一行，表示要求的原始木棍的最小可能长度。

 

思路

首先把小木棍长度从大到小排序， 从最长的木棍aim开始搜， 如果这些木棍的总长度能够整除该木棍aim的长度，就开始递归搜。

能够拼成的根数为sum/aim， 如果成功则输出当前的长度。如果能匹配成功retrun true,否则return false。

同时标记搜过的木棍， 如果不可以则取消标记。

如果当前木棍不可以， 那么和改木棍相同长度的木棍也不可用 （剪枝）。

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio>

using namespace std;
const int MAXN = 70;
int n, sum, aim, num, a[MAXN];
bool used[MAXN];
bool cmp(int a, int b){//从大到小排序
    return a > b;
}
bool dfs(int Stick, int len, int pos) {//Stick 当前组好的数目  len 组好的长度 pos 搜索到第几根
    bool sign = (len == 0 ? true:false);
    if(Stick == num) return true;//全部完成搜索， 退出函数
    for(int  i = pos + 1; i < n; i++) {
        if(used[i]) continue;//当前的棍子不能再用了
        if(len + a[i] == aim) {//正好可以和当前的棍子组成目标长度
            used[i] = true;
            if(dfs(Stick+1, 0, -1)) return true;//进入下一层搜索
            used[i] = false;//回溯
            return false;
        } else if(len + a[i] < aim) {//长度不够
            used[i] = true;
            if(dfs(Stick, len + a[i], i)) return true;//进入下一层搜索， 棍子长度变成 len  + a[i]
            used[i]  = false;
        //关键一步， 没有选择任何木棍， 跳出循环
            if(sign) return false;
            while(a[i] == a[i+1]) i++;//剪枝
        }
    }
    return false;
}

int main() {
    scanf("%d", &n);
        sum = 0;
        for(int i = 0; i < n; i++) {
            scanf("%d", &a[i]);
            sum += a[i];//求出所有木棍的总长度
        }
        sort(a, a+n, cmp);//排序
        for(aim = a[0]; aim <= sum; aim++) {
            if(sum % aim == 0) {
                num = sum/aim;
                  memset(used, false, sizeof(used));//初始化
                if(dfs(1, 0, -1)) {
                    printf("%d
", aim);//直接输出答案
                    break;
                }
            }

        }
    return 0;
}

还是比较不错滴