[poj 1011]sticks

題目

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

Sample Output

6

5

Source

Central Europe 1995

題解

我們可以從小到大枚舉原始木棒的長度 l e n len len，處理出木棒的總長度 s u m sum sum，那麼顯然有 l e n len len是 s u m sum sum的約數，是以你要拼成的木棒的個數 c n t = s u m / l e n cnt = sum / len cnt=sum/len。

于是我們可以 d f s dfs dfs一下， s t i c k stick stick表示目前拼接到了第幾個木棒， c a b cab cab表示目前所拼接的木棒的長度， l a s t last last為所拼接的上一根木棒。

但是這樣直接的 d f s dfs dfs會逾時，那麼我們就來思索一下如何剪枝。

• 假如有長度為 2 2 2， 4 4 4的兩根木棒，先選 2 2 2後選 4 4 4，和先選 4 4 4後選 2 2 2所得到的結果顯然是一樣的。
• 對于所有木棒我們先從大到小排序，然後在搜尋一定會降低複雜度。
• 如果目前拼接的第一根木棒就失敗那麼就判定改搜尋分支失敗，回溯
• 如果目前拼接完成但是下一個拼接失敗，就判定該搜尋分支失敗，回溯

code

#include <algorithm>
#include <cctype>
#include <cmath>
#include <complex>
#include <cstdio>
#include <cstring>
#include <deque>
#include <functional>
#include <list>
#include <map>
#include <iomanip>
#include <iostream>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>
#include <bitset>
//#define T 0
#define clean(x, y) memset(x, y, sizeof(x))
const int maxn = 105;
const int inf = 0x3f3f3f3f;
typedef long long LL;
using namespace std;

template <typename T>
inline void read(T &s) {
    s = 0;
    T w = 1, ch = getchar();
    while (!isdigit(ch)) { if (ch == '-') w = -1; ch = getchar(); }
    while (isdigit(ch))  { s = (s << 1) + (s << 3) + (ch ^ 48); ch = getchar(); }
    s *= w;
}

int n;
int val;
int sum;
int cnt;
int len;
int a[maxn];
int v[maxn];

inline bool cmp(int x, int y) { return x > y; }

bool dfs(int stick, int cab, int last) {
//	printf("%d ())(()())(())) \n", cnt);
	if (stick > cnt) return true;
	if (cab == len) return dfs(stick + 1, 0, 1);
	int fail = 0;
	for (int i = last; i <= n; ++i) {
		if (!v[i] && fail != a[i] && a[i] + cab <= len) {
			v[i] = 1;
			if (dfs(stick, a[i] + cab, i + 1)) return true;
			fail = a[i];
			v[i] = 0;
			if (cab == 0 || cab + a[i] == len) return false;
		}
	}
	return false;
}

int main() {
	while (1) {
		read(n);
		if (!n) exit(0);
		memset(a, 0, sizeof(a));
		sum = 0; val = 0;
		for (int i = 1; i <= n; ++i)
			read(a[i]), sum += a[i], val = max(val, a[i]);
		sort(a + 1, a + n + 1, cmp);
		for (len = val; len <= sum; ++len) {
			if (sum % len) continue;
			else {
				memset(v, 0, sizeof(v));
				cnt = sum / len;    
//				printf("%d ======= \n", cnt);
				if (dfs(1, 0, 1)) break;
			}
		}
		printf("%d\n", len);
	}
	return 0;
}