题面
题解
（除了代码均摘自喻队的博客，可是他退役了）
首先固定一棵树,枚举另一棵树,显然另一棵树只有与这棵树同构才有可能产生贡献 如果固定的树以重心为根,那么另一棵树最多就只有重心为根才有可能同构了(可能有两个) 然后就是求改动次数最小值,设$f[x][y]$表示以第一棵树$x$为根的子树内和第二棵树内$y$为根的子树内,达到目标最少需要改动的次数 我们发现只有同构的子树需要决策,我们把同构的子树分别拿出来,我们要做的就是做一个匹配,跑一遍$KM$或者费用流就好了。因为要最小化$f[x][y]$，所以是跑最小完美匹配。 $f[x][y]$要记忆化一下,判断同构用树哈希即可
#include <cstdio>
#include <vector>
#include <cstring>
#include <algorithm>
using std::sort; using std::vector;
typedef long long ll;
template<typename T>
void read(T &x) {
int flag = 1; x = 0; char ch = getchar();
while(ch < '0' || ch > '9') { if(ch == '-') flag = -flag; ch = getchar(); }
while(ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar(); x *= flag;
}
const int N = 1.4e3 + 10, Inf = 1e9 + 7;
void upt0(int &x, int y) { if(x < y) x = y; }
void upt1(int &x, int y) { if(x > y) x = y; }
namespace KM {
int n, w[N][N], match[N], ret, lx[N], ly[N];
bool visx[N], visy[N];
bool Hungary(int x) {
visx[x] = 1;
for(int y = 1; y <= n; ++y)
if(!visy[y] && lx[x] + ly[y] == w[x][y]) {
visy[y] = true;
if(!match[y] || Hungary(match[y])) { match[y] = x; return 1; }
}
return 0;
}
int main(int opt) {
for(int i = 1; i <= n; ++i)
for(int j = 1; j <= n; ++j)
w[i][j] = opt * w[i][j];
for(int i = 1; i <= n; ++i) {
lx[i] = -Inf, ly[i] = 0;
for(int j = 1; j <= n; ++j) upt0(lx[i], w[i][j]);
}
memset(match, 0, sizeof match);
for(int x = 1; x <= n; ++x)
while(1) {
memset(visx, 0, sizeof visx), memset(visy, 0, sizeof visy);
if(Hungary(x)) break;
int inc = Inf;
for(int i = 1; i <= n; ++i)
if(visx[i])
for(int j = 1; j <= n; ++j)
if(!visy[j]) upt1(inc, lx[i] + ly[j] - w[i][j]);
for(int i = 1; i <= n; ++i) {
if(visx[i]) lx[i] -= inc;
if(visy[i]) ly[i] += inc;
}
}
for(int i = 1; i <= n; ++i)
if(match[i]) ret += w[match[i]][i];
return opt * ret;
}
}//KM模板
int n, rt, fir[N], sec[N], f[N][N], c[N][N]; ll hash[N];
int from[N], cnt, to[N << 1], nxt[N << 1];
inline void addEdge(int u, int v) {
to[++cnt] = v, nxt[cnt] = from[u], from[u] = cnt;
}
int tmp, siz[N];
vector<int> v1[N], v2[N];
inline bool cmp(const int &i, const int &j) { return hash[i] < hash[j]; }
void getrt(int u, int fa) {
int max_part = 0; siz[u] = 1;
for(int i = from[u]; i; i = nxt[i]) {
int v = to[i]; if(v == fa) continue;
getrt(v, u), siz[u] += siz[v], upt0(max_part, siz[v]);
} upt0(max_part, n - siz[u]);
if(max_part < tmp) tmp = max_part, rt = u;
}//寻找树的重心
void dfs(int u, int fa, vector<int> *V) {
siz[u] = 1, hash[u] = 0, V[u].clear();
for(int i = from[u]; i; i = nxt[i]) {
int v = to[i]; if(v == fa) continue;
dfs(v, u, V), siz[u] += siz[v], V[u].push_back(v);
} sort(V[u].begin(), V[u].end(), cmp);
for(int i = V[u].size() - 1; ~i; --i)
hash[u] = hash[u] * N + hash[V[u][i]];
hash[u] = hash[u] * N + siz[u];
}//处理各子树hash值以及儿子（将儿子放进一个vector里面）
int dp(int x, int y) {
if(f[x][y] != -1) return f[x][y];
f[x][y] = fir[x] ^ sec[y]; int lim = v1[x].size() - 1;
for(int i = 0; i <= lim; ++i) {
int j = i;
while(j < lim && hash[v1[x][j + 1]] == hash[v1[x][i]]) ++j;
for(int k = i; k <= j; ++k)
for(int l = i; l <= j; ++l)
dp(v1[x][k], v2[y][l]);
for(int k = i; k <= j; ++k)
for(int l = i; l <= j; ++l)
KM::w[k - i + 1][l - i + 1] = dp(v1[x][k], v2[y][l]);
//初始化边权
KM::ret = 0, KM::n = j - i + 1, f[x][y] += KM::main(-1), i = j;
//最小化f[x][y]
} return f[x][y];
}
int main () {
read(n);
for(int i = 1, u, v; i < n; ++i)
read(u), read(v), addEdge(u, v), addEdge(v, u);
for(int i = 1; i <= n; ++i) read(fir[i]);
for(int i = 1; i <= n; ++i) read(sec[i]);
tmp = Inf, getrt(1, 0), dfs(rt, 0, v2); ll tmp = hash[rt];
int ans = Inf;
for(int i = 1; i <= n; ++i) {//暴力枚举重心
dfs(i, 0, v1);
if(hash[i] == tmp) memset(f, -1, sizeof f), upt1(ans, dp(i, rt));
} printf("%d
", ans);
return 0;
}