JZOJ 5344. 【NOIP2017模拟9.3A组】摘果子

题目

略

分析

又是一个显然的树形依赖背包
然而可以 (O(nm)) 依靠 (dfs) 序来 (dp)

(Code)

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;

const int N = 2005;
int v[N] , p[N] , f[N][10005] , n , m , tot , h[N] , dfn[N] , dfc , siz[N];
struct edge{
	int to , nxt;
}e[N * 2 + 5];

void add(int x , int y){e[++tot] = edge{y , h[x]} , h[x] = tot;}
void dfs(int x , int fa)
{
	dfn[++dfc] = x , siz[x] = 1;
	for(register int i = h[x]; i; i = e[i].nxt)
	{
		int y = e[i].to;
		if (y == fa) continue;
		dfs(y , x);
		siz[x] += siz[y];
	}
}

int main()
{
	scanf("%d%d" , &n , &m);
	int x , y;
	for(register int i = 1; i <= n; i++) scanf("%d%d" , &v[i] , &p[i]);
	for(register int i = 1; i < n; i++) scanf("%d%d" , &x , &y) , add(x , y) , add(y , x);
	dfs(1 , 0);
	for(register int i = n; i >= 1; i--)
		for(register int j = m; j >= 0; j--)
		{
			if (j >= p[dfn[i]]) f[i][j] = max(f[i][j] , f[i + 1][j - p[dfn[i]]] + v[dfn[i]]);
			f[i][j] = max(f[i][j] , f[i + siz[dfn[i]]][j]);
		}
	printf("%d" , f[1][m]);
}