RNQOJ [stupid]愚蠢的矿工(树形依赖背包)

题意

题目链接

Sol

树形依赖背包板子题

树形依赖背包大概就是说：对于一个点，只有选了它的父亲才能选自身

把dfs序建出来，倒过来考虑

设\(f[i][j]\)表示从第\(i\)个节点往后背包体积为\(j\)的最大价值

转移的时候，只有选了该点才能从子树中转移而来

\(f[i][j] = max(f[i + 1][j - w[i]] + val[i], f[i + siz[rev[i]]][j]);\)

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 3001, INF = 1e9 + 10;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-')f =- 1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
	return x * f;
}
int N, M, w[MAXN], val[MAXN], siz[MAXN], rev[MAXN], f[MAXN][MAXN], tot = 0;
vector<int> v[MAXN];
void dfs(int x, int _fa) {
	rev[++tot] = x; siz[x] = 1;
	for(int i = 0, to; i < v[x].size(); i++) {
		if((to = v[x][i]) == _fa) continue;
		dfs(to, x);
		siz[x] += siz[to];
	}
}
main() {
	N = read(); M = read();
	for(int i = 1; i <= N; i++) val[i] = read(), w[i] = 1;
	for(int i = 1; i <= N; i++) {
		int x = read(), y = read();
		if(x == 0) continue;
		v[x].push_back(y); v[y].push_back(x);
	}
	dfs(1, 0);
	for(int i = N; i >= 1; i--) {
		for(int j = 0; j <= M; j++) {
			f[i][j] = f[i + siz[rev[i]]][j];
			if(j >= w[i]) f[i][j] = max(f[i][j], f[i + 1][j - w[rev[i]]] + val[rev[i]]);
		//	printf("%d %d %d\n", i, j, f[i][j]);
		}
	}
	cout << f[1][M];
}
/*
*/