题意:给一棵节点数为n,节点种类为k的无根树,问其中有多少种不同的简单路径,可以满足路径上经过所有k种类型的点?
析:对于路径,就是两类,第一种情况,就是跨过根结点,第二种是不跨过根结点,分别讨论就好,由于结点比较大,所以采用分治来进行处理,优先选取重点作为划分的依据。
代码如下:
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define pu push_up
#define pd push_down
#define FOR(x,n) for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std; typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 50000 + 10;
const LL mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
return r > 0 && r <= n && c > 0 && c <= m;
}
int all; struct Edge{
int to, next;
}; Edge edge[maxn<<1];
int head[maxn], cnt;
int val[maxn]; void addEdge(int u, int v){
edge[cnt].to = v;
edge[cnt].next = head[u];
head[u] = cnt++;
} int root, num, f[maxn];
LL dp[1<<10];
int sum[maxn];
bool vis[maxn];
LL ans; void dfs_for_root(int u, int fa){
sum[u] = 1; f[u] = 0;
for(int i = head[u]; ~i; i = edge[i].next){
int v = edge[i].to;
if(v == fa || vis[v]) continue;
dfs_for_root(v, u);
sum[u] += sum[v];
f[u] = max(f[u], sum[v]);
}
f[u] = max(f[u], num - sum[u]);
if(f[root] > f[u]) root = u;
} void dfs_for_color(int u, int fa, int s){
for(int i = head[u]; ~i; i = edge[i].next){
int v = edge[i].to;
if(v == fa || vis[v]) continue;
++dp[s|1<<val[v]];
dfs_for_color(v, u, s|1<<val[v]);
}
} LL solve(int u, int s){
ms(dp, 0);
++dp[s];
dfs_for_color(u, -1, s);
LL ans = 0;
for(int i = 0; i <= all; ++i){
if(!dp[i]) continue;
int tmp = 0;
tmp += dp[all];
for(int j = i; j; j = (j-1)&i)
tmp += dp[all^j];
ans += (LL)tmp * dp[i];
}
return ans;
} void dfs_for_ans(int u){
ans += solve(u, 1<<val[u]);
vis[u] = true;
for(int i = head[u]; ~i; i = edge[i].next){
int v = edge[i].to;
if(vis[v]) continue;
ans -= solve(v, 1<<val[u]|1<<val[v]);
root = 0;
f[0] = num = sum[v];
dfs_for_root(v, u);
dfs_for_ans(root);
}
} int main(){
while(scanf("%d %d", &n, &m) == 2){
for(int i = 1; i <= n; ++i){
scanf("%d", val+i);
--val[i];
}
all = (1<<m) - 1;
ms(head, -1); cnt = 0;
for(int i = 1; i < n; ++i){
int u, v;
scanf("%d %d", &u, &v);
addEdge(u, v);
addEdge(v, u);
}
ms(vis, 0);
root = 0; ans = 0LL;
f[0] = num = n;
dfs_for_root(1, -1);
dfs_for_ans(root);
printf("%I64d\n", ans);
}
return 0;
}