【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
1 #include <bits/stdc++.h> 2 using namespace std; 3 inline int read() { 4 int ret = 0, op = 1; 5 char c = getchar(); 6 while (c < '0' || c > '9') { 7 if (c == '-') op = -1; 8 c = getchar(); 9 } 10 while (c >= '0' && c <= '9') { 11 ret = ret * 10 + c - '0'; 12 c = getchar(); 13 } 14 return ret * op; 15 } 16 int n, p; 17 struct node { 18 int next, to; 19 } a[110 << 1]; 20 int head[110 << 1], num, c[110][15]; 21 int f[110][1 << 15], val[1 << 15]; 22 inline void add(int from, int to) { 23 a[++num].next = head[from]; 24 a[num].to = to; 25 head[from] = num; 26 } 27 void dfs(int now, int fa) { 28 for (register int i = head[now]; i != -1; i = a[i].next) 29 if (a[i].to != fa) { 30 dfs(a[i].to, now); 31 for (register int j = (1 << p) - 1; j; j--) { 32 for (register int k = j; k; k = j & (k - 1)) 33 f[now][j] = max(f[now][j], f[now][j ^ k] + f[a[i].to][k]); 34 } 35 } 36 for (register int i = (1 << p) - 1; i; i--) 37 f[now][i] = f[now][i] + val[i]; 38 } 39 int main() { 40 num = -1; 41 memset(head, -1, sizeof(head)); 42 n = read(); p = read(); 43 for (register int i = 1; i < n; ++i) { 44 int x = read(), y = read(); 45 add(x, y); 46 add(y, x); 47 } 48 for (register int i = 1; i <= n; ++i) { 49 for (register int j = 0; j < p; ++j) { 50 c[i][j] = read(); 51 } 52 f[i][0] = 0; 53 for (register int j = 1; j < (1 << p); ++j) { 54 int xx = j & (-j); 55 int xxx = log2(xx); 56 f[i][j] = f[i][j ^ xx] - c[i][xxx]; 57 } 58 } 59 int t = read(); 60 for (register int i = 1; i <= t; ++i) { 61 int v = read(), tot = read(), now = 0; 62 for (register int j = 1; j <= tot; ++j) { 63 int nn = read(); 64 now |= (1 << (nn - 1)); 65 } 66 val[now] += v; 67 int k = ((1 << p) - 1) ^ now; 68 for (register int l = k; l; l = (l - 1) & k) { 69 val[l | now] += v; 70 } 71 } 72 dfs(1, 0); 73 printf("%d\n", f[1][(1 << p) - 1]); 74 return 0; 75 }
