标签:namespace return init online eth geo mes 最小 费用流
套路:找出重心,如果有两个就新建一个点
然后把这棵树hash一下
设\(f[i][j]\)表示第一颗树到\(i\)第二棵树到\(j\),子树\(i,j\)同构的付出的最小代价
转移:每次把这一层hash值相同的点做一边二分图权匹配(KM/费用流)就好了
一遍AC
# include <bits/stdc++.h>
# define IL inline
# define RG register
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
IL int Input(){
RG int x = 0, z = 1; RG char c = getchar();
for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
return x * z;
}
const int maxn(1005);
const int inf(1e9);
const ll seed1(9377);
const ll seed2(10007);
namespace MCMF{
int dis[maxn], pre1[maxn], pre2[maxn], vis[maxn], first[maxn], cnt, ans, s, t;
queue <int> q;
struct Edge{
int to, next, f, w;
} edge[maxn * maxn * 2];
IL void Init(RG int a, RG int b){
s = a, t = b, ans = cnt = 0;
for(RG int i = a; i <= b; ++i) first[i] = -1;
}
IL void Add(RG int u, RG int v, RG int f, RG int w){
edge[cnt] = (Edge){v, first[u], f, w}, first[u] = cnt++;
edge[cnt] = (Edge){u, first[v], 0, -w}, first[v] = cnt++;
}
IL int Aug(){
for(RG int i = s; i <= t; ++i) dis[i] = inf;
q.push(s), dis[s] = 0, vis[s] = 1;
while(!q.empty()){
RG int u = q.front(); q.pop();
for(RG int e = first[u]; e != -1; e = edge[e].next){
RG int v = edge[e].to;
if(edge[e].f && dis[v] > dis[u] + edge[e].w){
dis[v] = dis[u] + edge[e].w;
pre1[v] = e, pre2[v] = u;
if(!vis[v]) q.push(v), vis[v] = 1;
}
}
vis[u] = 0;
}
if(dis[t] == inf) return 0;
RG int ret = inf;
for(RG int p = t; p; p = pre2[p]) ret = min(ret, edge[pre1[p]].f);
ans += ret * dis[t];
for(RG int p = t; p; p = pre2[p])
edge[pre1[p]].f -= ret, edge[pre1[p] ^ 1].f += ret;
return 1;
}
IL int Calc(){
for(ans = 0; Aug(); );
return ans;
}
}
int cnt, first[maxn];
struct Edge{
int to, next;
} edge[maxn << 1];
namespace Hash{
struct Val{
unsigned long long f1, f2;
IL int operator ==(RG Val b) const{
return f1 == b.f1 && f2 == b.f2;
}
} v[maxn];
IL int Cmp(RG int a, RG int b){
return v[a].f1 != v[b].f1 ? v[a].f1 < v[b].f1 : v[a].f2 < v[b].f2;
}
int size[maxn];
IL void GetHash(RG int u, RG int ff){
vector <int> son; son.clear(), size[u] = 1;
for(RG int e = first[u]; e != -1; e = edge[e].next){
RG int v = edge[e].to;
if(v != ff){
GetHash(v, u);
size[u] += size[v], son.push_back(v);
}
}
sort(son.begin(), son.end(), Cmp);
for(RG int i = 0, l = son.size(); i < l; ++i){
v[u].f1 = v[u].f1 * seed1 + v[son[i]].f1 * seed1 + v[son[i]].f2 * seed2;
v[u].f2 = v[u].f2 * seed2 + v[son[i]].f1 * seed2 + v[son[i]].f2 * seed1;
}
v[u].f1 = v[u].f1 * seed1 + size[u];
v[u].f2 = v[u].f2 * seed2 + size[u];
}
};
int n, size[maxn], c1[maxn], c2[maxn], rt, tmp;
int f[maxn][maxn];
IL void Add(RG int u, RG int v){
edge[cnt] = (Edge){v, first[u]}, first[u] = cnt++;
}
IL void GetRoot(RG int u){
size[u] = 1; RG int mx = 0;
for(RG int e = first[u]; e != -1; e = edge[e].next){
RG int v = edge[e].to;
if(!size[v]){
GetRoot(v);
size[u] += size[v];
mx = max(mx, size[v]);
}
}
mx = max(mx, n - size[u]);
if((mx << 1) <= n){
if(!rt) rt = u;
else tmp = u;
}
}
IL void Solve(RG int u1, RG int u2, RG int ff1, RG int ff2){
f[u1][u2] = c1[u1] != c2[u2];
vector <int> son1, son2; son1.clear(), son2.clear();
for(RG int e = first[u1]; e != -1; e = edge[e].next)
if(edge[e].to != ff1) son1.push_back(edge[e].to);
for(RG int e = first[u2]; e != -1; e = edge[e].next)
if(edge[e].to != ff2) son2.push_back(edge[e].to);
RG int l1 = son1.size(), l2 = son2.size(), s = 0, t = l1 + l2 + 1;
for(RG int i = 0; i < l1; ++i)
for(RG int j = 0; j < l2; ++j)
if(Hash::v[son1[i]] == Hash::v[son2[j]]) Solve(son1[i], son2[j], u1, u2);
MCMF::Init(0, t);
for(RG int i = 0; i < l1; ++i)
for(RG int j = 0; j < l2; ++j)
if(Hash::v[son1[i]] == Hash::v[son2[j]]) MCMF::Add(i + 1, j + 1 + l1, 1, f[son1[i]][son2[j]]);
for(RG int i = 0; i < l1; ++i) MCMF::Add(s, i + 1, 1, 0);
for(RG int i = 0; i < l2; ++i) MCMF::Add(i + 1 + l1, t, 1, 0);
RG int ret = MCMF::Calc();
f[u1][u2] += ret;
}
int main(){
n = Input();
for(RG int i = 1; i <= n + 1; ++i) first[i] = -1;
for(RG int i = 1; i < n; ++i){
RG int u = Input(), v = Input();
Add(u, v), Add(v, u);
}
for(RG int i = 1; i <= n; ++i) c1[i] = Input();
for(RG int i = 1; i <= n; ++i) c2[i] = Input();
GetRoot(1);
if(tmp){
for(RG int e = first[rt]; e != -1; e = edge[e].next)
if(edge[e].to == tmp) edge[e].to = n + 1;
for(RG int e = first[tmp]; e != -1; e = edge[e].next)
if(edge[e].to == rt) edge[e].to = n + 1;
Add(n + 1, rt), Add(n + 1, tmp), rt = n + 1;
}
Hash::GetHash(rt, 0);
Solve(rt, rt, 0, 0);
printf("%d\n", f[rt][rt]);
return 0;
}
标签:namespace return init online eth geo mes 最小 费用流
原文地址:https://www.cnblogs.com/cjoieryl/p/9113950.html