UVa307 - Sticks(DFS+剪枝)

题意:给出若干个一样长的sticks,剪断后有n个部分,问原来sticks的最小长度。

思路:首先要确定sticks的长度范围,范围应该在[max(n1,n2..., n), sum(n1,n2...,n)}之间。然后用深度优先算法,在深搜时注意剪枝

(1)如果当前木棒搜索没有成功,那么后面的木棒长度与当前木棒长度一样,就应该跳过。

(2)如果当前木棒是第一个是没有成功,后面的也不会成功

#include 
#include 
#include 
#include 
#include 

using namespace std;

const int N = 100;

bool vis[N];
int a[N];
int n;
int sum, Max;
int len;

bool input()
{
	scanf("%d", &n);
	if (n == 0) return false;
	
	Max = 0;
	sum = 0;
	for (int i = 0; i < n; i++) {
		scanf("%d", &a[i]);
		Max = max(Max, a[i]);
		sum += a[i];
	}
	return true;
}

bool dfs(int complete, int pos, int length)
{
		if (complete == n) return true;
		
		for (int i = pos; i < n; i++) {
			if (vis[i]) continue;
			if (a[i] + length < len) {
				vis[i] = true;
				if (dfs(complete + 1, i + 1, a[i] + length)) return true;
				vis[i] = false;
				while (a[i] == a[i + 1] && i + 1 < n) i++;
				if (length == 0) return false;
			} else if (a[i] + length == len) {
				vis[i] = true;
				if (dfs(complete + 1, 0, 0)) return true;
				vis[i] = false;
				return false;
			}
		}
		return false;
}

void solve()
{	
	sort(a, a + n, greater());
	for (len = Max; len <= sum / 2; len++) {
		if (sum % len == 0) {
			memset(vis, false, sizeof(vis));
			if (dfs(0, 0, 0)) {
				printf("%d\n", len);
				break;
			}
		}
	}
	
	if (len > sum / 2) printf("%d\n", sum);
}

int main()
{
	#ifndef ONLINE_JUDGE
		freopen("d:\\OJ\\uva_in.txt", "r", stdin);
	#endif
	
	while (input()) {
		solve();
	}
	return 0;
}


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