luoguP2329 [SCOI2005]栅栏

首先我们可以看出一个性质:取小的一定不比取大的劣.显然嘛

于是我们将约翰所需的木板从小到大排序,原材料也按从小到大排序,然后二分一个\(mid\)表示约翰取几个木板,用搜索\(check\)一下即可.

几个剪枝:

• 一.我们可以定义一个全局变量\(Waste\)表示割下木板后总共浪费的原材料长度.

  \(\therefore\) 如果\(Waste+\sum_{i=1}^{mid}b_i>sum\),直接退出,返回\(0\).

• 二.对于两个长度相等的所需木板,我们可以记一个\(last\)表示之前的所需木板找到了第\(last\)个,\(\therefore\)直接从\(last\)开始找即可.

• 三.我们在二分前可以先将总和大于\(sum\)的删去,即\(while (s[r] > sum)--r;\)

#include<bits/stdc++.h>
#define il inline
#define rg register
#define gi read<int>
using namespace std;
template<class TT>
il TT read() {
	TT o = 0,fl = 1; char ch = getchar();
	while (!isdigit(ch) && ch != '-') ch = getchar();
	if (ch == '-') fl = -1, ch = getchar();
	while (isdigit(ch)) o = o * 10 + ch - '0', ch = getchar();
	return fl * o;
}
int n, m, sum, l, r, mid, t, ans, a[55], b[1005], c[55], s[1005];
bool check(int x, int last) {
	if (!x) return true;
	if (sum < t + s[mid]) return false;
	for (int i = last; i <= m; ++i) {
		if (c[i] >= b[x]) {
			c[i] -= b[x];
			if (c[i] < b[1]) t += c[i];
			if (b[x] == b[x - 1]) {
				if (check(x - 1, last)) return true;
			}
			else if (check(x - 1, 1)) return true;
			if (c[i] < b[1]) t -= c[i];
			c[i] += b[x];
		}
	}
	return false;
}
int main() {
	m = gi();
	for (int i = 1; i <= m; ++i)
		a[i] = gi(), sum += a[i];
	sort(a + 1, a + m + 1);
	n = gi();
	for (int i = 1; i <= n; ++i) b[i] = gi();
	sort(b + 1, b + n + 1);
	for (int i = 1; i <= n; ++i) s[i] = s[i - 1] + b[i];
	l = 1, r = n;
	while (s[r] > sum) --r;
	while (l <= r) {
		for (int i = 1; i <= m; ++i) c[i] = a[i];
		mid = l + r >> 1, t = 0;
		if (check(mid, 1)) ans = mid, l = mid + 1;
		else r = mid - 1;
	}
	printf("%d\n", ans);
	return 0;
}