RNQOJ [stupid]愚蠢的矿工(树形依赖背包)

题意

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];
}
/*
*/