Solution -「AGC 010C」「AT 2304」Cleaning

$\mathcal{Description}$

  Link.

  给定一棵 $n$ 个点的无根树，点有点权，每次选择两个不同的叶子，使它们间的简单路径的所有点权 $-1$，问能否将所有点权变成 $0$。

  $n\le10^5$。

$\mathcal{Solution}$

  这不就是我那道题的削弱版么 www。

  考虑叶子 $u$，显然有 $val_u$ 条路径经过 $u$。DFS 回溯时考虑合并儿子们的路径。假设儿子们向上的路径共 $s$ 条，单个儿子向上路径最多有 $m$ 条；我们拿 $a$ 对路径在当前结点 $u$ 处匹配成完整路径，剩下 $b$ 条继续向上，那么有：

$$\begin{cases} a+b=val_u\\ 2a+b=s\\ a\le\min\{\lfloor\frac{s}2\rfloor,s-m\} \end{cases}$$

  解出 $a,b$，判断是否合法即可。

$\mathcal{Code}$

#include <cstdio>
#include <cstdlib>
#include <iostream>

typedef long long LL;

const int MAXN = 1e5;
int n, ecnt, val[MAXN + 5], head[MAXN + 5], d[MAXN + 5];

struct Edge { int to, nxt; } graph[MAXN * 2 + 5];

inline void link ( const int s, const int t ) {
	graph[++ ecnt] = { t, head[s] };
	head[s] = ecnt;
}

inline int DFS ( const int u, const int f ) {
	if ( d[u] == 1 ) return val[u];
	LL sum = 0; int mx = 0;
	for ( int i = head[u], v; i; i = graph[i].nxt ) {
		if ( ( v = graph[i].to ) ^ f ) {
			int up = DFS ( v, u );
			sum += up, mx = std::max ( mx, up );
		}
	}
	if ( sum - val[u] > std::min ( sum >> 1, sum - mx )
		|| 2 * val[u] < sum || sum < val[u] ) puts ( "NO" ), exit ( 0 );
	return 2 * val[u] - sum;
	// a+b=val, 2a+b=sum, a<=min{sum/2,sum-mx}
}

int main () {
	scanf ( "%d", &n );
	for ( int i = 1; i <= n; ++ i ) scanf ( "%d", &val[i] );
	for ( int i = 1, u, v; i < n; ++ i ) {
		scanf ( "%d %d", &u, &v ), ++ d[u], ++ d[v];
		link ( u, v ), link ( v, u );
	}
	if ( n == 2 ) return puts ( val[1] == val[2] ? "YES" : "NO" ), 0;
	for ( int i = 1; i <= n; ++ i ) if ( d[i] > 1 ) return puts ( DFS ( i, 0 ) ? "NO" : "YES" ), 0;
	return 0;
}