【SDOI2008】山贼集团

这是一道树形dp和状压dp结合的题目，思考量和代码细节都不少。

我们首先定义f[i][j]表示在以i为根的子树当中，建立分部的节点状压之后为j的最大收益

那么转移是显然的，定义k为i的儿子，那么f[i][j] = max(f[k][l] + f[i][j ^ l] + val). 其中val表示其他的收益

我们可以预处理val数组，剩下的就是细节问题了

#include <bits/stdc++.h>
using namespace std;
inline int read() {
    int ret = 0, op = 1;
    char c = getchar();
    while (c < '0' || c > '9') {
        if (c == '-') op = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9') {
        ret = ret * 10 + c - '0';
        c = getchar();
    }
    return ret * op;
}
int n, p;
struct node {
    int next, to; 
} a[110 << 1];
int head[110 << 1], num, c[110][15];
int f[110][1 << 15], val[1 << 15];
inline void add(int from, int to) {
    a[++num].next = head[from];
    a[num].to = to;
    head[from] = num;
}
void dfs(int now, int fa) {
    for (register int i = head[now]; i != -1; i = a[i].next)
        if (a[i].to != fa) {
            dfs(a[i].to, now);
            for (register int j = (1 << p) - 1; j; j--) {
                for (register int k = j; k; k = j & (k - 1))
                    f[now][j] = max(f[now][j], f[now][j ^ k] + f[a[i].to][k]);
            }
        }
    for (register int i = (1 << p) - 1; i; i--)
        f[now][i] = f[now][i] + val[i];
}
int main() {
    num = -1;
    memset(head, -1, sizeof(head));
    n = read(); p = read();
    for (register int i = 1; i < n; ++i) {
        int x = read(), y = read();
        add(x, y);
        add(y, x);
    }
    for (register int i = 1; i <= n; ++i) {
        for (register int j = 0; j < p; ++j) {
            c[i][j] = read();
        }
        f[i][0] = 0;
        for (register int j = 1; j < (1 << p); ++j) {
            int xx = j & (-j);
            int xxx = log2(xx);
            f[i][j] = f[i][j ^ xx] - c[i][xxx];
        }
    }
    int t = read();
    for (register int i = 1; i <= t; ++i) {
        int v = read(), tot = read(), now = 0;
        for (register int j = 1; j <= tot; ++j) {
            int nn = read();
            now |= (1 << (nn - 1));
        }
        val[now] += v;
        int k = ((1 << p) - 1) ^ now;
        for (register int l = k; l; l = (l - 1) & k) {
            val[l | now] += v;
        }
    }
    dfs(1, 0);
    printf("%d
", f[1][(1 << p) - 1]);
    return 0;
}