题目链接：Garden of Eden

题意：给定一颗n个节点的树，每个节点有一种颜色，颜色有k种，求树上有多少条路径包含这k种颜色，n<=50000，k<=10

思路：树上路径问题，用点分治求解，又由于k<=10，所以可以用二进制状态表示一条路径上包含的颜色集合，比如状态8转换成二进制为1000，那么就表示状态8表示的路径上含有第3种颜色(颜色标号从0开始)

那么考虑过重心向下的某一条路径，假设这条路径的二进制状态为d，设s表示含有所有颜色的集合，即s=(1<<k)-1，那么我们只需要找到二进制状态为s^d的集合的超集与d配对即可，所以需要求超集的和

1
2
3
4
5
6
7
8
9
// 超集和
for (int j = 0; j < k; j++)
        for (int i = s; i >= 0; i--)
            if (((1 << j) & i) == 0) tot[i] += tot[i | (1 << j)];

// 子集和
for (int j = 0; j < k; j++)
    for (int i = 0; i <= s; i++)
        if ((1 << j) & i) tot[i] ++ tot[i ^ (1 << j)];

求出超集后之后，按照点分治的一般步骤求解即可

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>

using namespace std;

typedef long long ll;

const int N = 50010;
const int M = 1050;

struct node {
    int to, nex;
};

node edge[2 * N];
int n, k, cnt, rt, sum, s, c, val[N], d[N];
int head[N], sz[N], son[N], vis[N], tot[M];
ll res;

inline void add_edge(int u, int v)
{
    edge[++cnt].to = v;
    edge[cnt].nex = head[u];
    head[u] = cnt;
}

void dfs(int u, int fa)
{
    sz[u] = 1;
    son[u] = 0;
    for (int i = head[u]; 0 != i; i = edge[i].nex) {
        int v = edge[i].to;
        if (v == fa || vis[v]) continue;
        dfs(v, u);
        sz[u] += sz[v];
        son[u] = max(son[u], sz[v]);
    }
    son[u] = max(son[u], sum - sz[u]);
    if (son[u] < son[rt]) rt = u;
}

void init()
{
    memset(vis, 0, sizeof(vis));
    memset(head, 0, sizeof(head));
    cnt = 0;
    res = 0;
    s = (1 << k) - 1;
}

void deep(int u, int fa, int now)
{
    d[++c] = now;
    tot[now]++;
    for (int i = head[u]; 0 != i; i = edge[i].nex) {
        int v = edge[i].to;
        if (vis[v] || v == fa) continue;
        deep(v, u, now | val[v]);
    }
}

ll calc(int u, int now)
{
    c = 0;
    memset(tot, 0, sizeof(tot));
    deep(u, 0, now);
    for (int j = 0; j < k; j++)
        for (int i = s; i >= 0; i--)
            if (((1 << j) & i) == 0) tot[i] += tot[i | (1 << j)];
    ll r = 0;
    for (int i = 1; i <= c; i++) r += tot[d[i] ^ s];
    return r;
}

void solve(int u)
{
    res += calc(u, val[u]);
    vis[u] = 1;
    for (int i = head[u]; 0 != i; i = edge[i].nex) {
        int v = edge[i].to;
        if (vis[v]) continue;
        res -= calc(v, val[u] | val[v]);
        sum = sz[v];
        rt = 0;
        dfs(v, -1);
        solve(rt);
    }
}

int main()
{
    while (scanf("%d%d", &n, &k) != EOF) {
        init();
        for (int i = 1; i <= n; i++) {
            int a;
            scanf("%d", &a);
            val[i] = 1 << (a - 1);
        }
        for (int i = 1; i <= n - 1; i++) {
            int u, v;
            scanf("%d%d", &u, &v);
            add_edge(u, v);
            add_edge(v, u);
        }
        rt = 0;
        sum = son[0] = n;
        dfs(1, -1);
        solve(rt);
        printf("%lld\n", res);
    }
    return 0;
}