「CF815C」Karen and Supermarket

传送门

Luogu

解题思路

树形背包。

设 \(f[i][j][0/1]\) 表示在以 \(i\) 为根的子树中选 \(j\) 件商品的最少花费。

边界条件：

\(f[i][j][0] = \min\limits_{0\le k\le siz[son]}\left\{f[i][j - k][0]+f[son][k][0]\right\}\)

\(f[i][j][1] = \min\limits_{0\le k\le siz[son]}\left\{f[i][j - k][1]+f[son][k][0]\right\}\)

\(f[i][j][1] = \min\limits_{0\le k\le siz[son]}\left\{f[i][j - k][1]+f[son][k][1]\right\}\)

最后输出最大的满足 \(\min\left\{f[1][i][0],f[1][i][1]\right\}\le b\) 的 \(i\) 即可。

细节注意事项

• 咕咕咕

参考代码

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <cctype>
#include <cmath>
#include <ctime>
#include <vector>
#define rg register
using namespace std;
template < typename T > inline void read(T& s) {
 	s = 0; int f = 0; char c = getchar();
 	while (!isdigit(c)) f |= (c == '-'), c = getchar();
 	while (isdigit(c)) s = s * 10 + (c ^ 48), c = getchar();
 	s = f ? -s : s;
}

const int _ = 1000010;
const int __ = 2000010;

int tot, head[_], nxt[__], ver[__];
inline void Add_edge(int u, int v)
{ nxt[++tot] = head[u], head[u] = tot, ver[tot] = v; }

int n, k, dgr[_], res;

inline int dfs(int u, int f) {
	if (dgr[u] == 1) return 0;
	vector < int > t;
	for (rg int i = head[u]; i; i = nxt[i])
		if (ver[i] != f) t.push_back(dfs(ver[i], u) + 1);
	sort(t.begin(), t.end());
	int len = t.size() - 1;
	for (; len > 0; --len)
		if (t[len] + t[len - 1] <= k) break; else ++res;
	return t[len];
}

int main() {
#ifndef ONLINE_JUDGE
	freopen("in.in", "r", stdin);
#endif
	read(n), read(k);
	for (rg int u, v, i = 1; i < n; ++i)
		read(u), read(v), Add_edge(u, v), Add_edge(v, u), ++dgr[u], ++dgr[v];
	int rt = 1;
	for (rg int i = 1; i <= n; ++i)
		if (dgr[i] > 1) { rt = i; break; }
	dfs(rt, 0);
	printf("%d\n", res + 1);
	return 0;
}

完结撒花 \(qwq\)