/ Vijos / 题库 / 线图 /

题解

2 条题解

  • 1
    @ 2023-07-12 17:46:30 
    #include<cstdio>
#include<map>
#include<vector>
#include<cassert>
#include<bitset>
#include<ctime>
#include<iostream>
#include<algorithm>
using namespace std;
vector<int> E[5005];
vector<int> kind[1250];
vector<int> T[15];
int rnd[2005];
int dp[5005][1250];
int crt[20] , ntd[1250];
int C[5005][20];
int inv[21];
const int mod = 998244353;
const int inv6 = 166374059;
int n , kk , ans = 0 , limit , cnt = 0;
int p[20] , tot = 0 , e_tot = 0;
map<int,int> mp;
map<long long,int> hashh;
map<pair<int,int> , int> edge;
int seed = 91478513 , a = 16554871 , b = 35659598;
int power(int a,int b)
{
int temp = a , ans = 1;
while(b){
if(b&1) ans = (1LL * ans * temp) % mod;
temp = (1LL * temp * temp) % mod;
b >>= 1;
}
return ans;
}
inline int rand()
{
return seed = (1LL * (seed ^ a) * b) % mod;
}
int get_hash(int fa,int u)
{
long long h = 1;
vector<int> hh;
for(int i = 0;i < T[u].size();i++){
if(T[u][i] != fa){
int g = get_hash(u , T[u][i]);
h += rnd[g];
hh.push_back(g);
}
}
map<long long , int>::iterator it = hashh.find(h);
if(it != hashh.end()) return it->second;
hashh.insert(pair<long long,int>{h , ++tot});
kind[tot] = hh;ntd[tot] = 1;
sort(kind[tot].begin() , kind[tot].end());
int qq = 1 , start = 0;ntd[tot] = 1;
for(;start < kind[tot].size() && kind[tot][start] == 1;start++);
for(int i = start + 1;i < kind[tot].size();i++){
if(kind[tot][i] == 1) continue;
if(kind[tot][i] == kind[tot][i - 1]) qq++;
else{
ntd[tot] = (1LL * ntd[tot] * inv[qq]) % mod;
qq = 1;
}
}
ntd[tot] = (1LL * ntd[tot] * inv[qq]) % mod;
return tot;
}
int get_p(int fa,int u)
{
long long h = 1;
for(int i = 0;i < T[u].size();i++){
if(T[u][i] != fa){
int g = get_p(u , T[u][i]);
if(g == -1) return -1;
h += rnd[g];
}
}
map<long long , int>::iterator it = hashh.find(h);
if(it != hashh.end()) return it->second;
return -1;
}
inline int get_num(int u,int v)
{
if(u > v) swap(u , v);
map<pair<int,int> , int>::iterator it = edge.find(pair<int,int>{u , v});
if(it != edge.end()) return it->second;
e_tot++;
edge.insert(pair<pair<int,int>,int>{pair<int,int>{u , v} , e_tot});
return e_tot;
}
int get_node(int cnode , int k , vector<int> G[])
{
edge.clear();e_tot = 0;
int q = 0;
for(int i = 1;i <= cnode;i++) q += G[i].size();
q /= 2;
if(k == 2){
int ans = 0;
for(int i = 1;i <= cnode;i++) ans = (ans + 1LL * (G[i].size() - 1) * G[i].size()) % mod;
return (1LL * ans * inv[2]);
}
if(k == 3){
int ans = 0;
for(int i = 1;i <= cnode;i++){
for(int j = 0;j < G[i].size();j++){
if(i > G[i][j]) continue;
if(G[i].size() + G[G[i][j]].size() < 4) continue;
int d = G[i].size() + G[G[i][j]].size() - 2;
ans = (ans + 1LL * d * (d - 1) % mod * inv[2]) % mod;
}
}
return ans;
}
if(k == 4){
vector<int> ed[cnode + 1];
int ans = 0;
for(int i = 1;i <= cnode;i++){
for(int j = 0;j < G[i].size();j++){
if(i > G[i][j]) continue;
int d0 = G[i].size()+G[G[i][j]].size()-2;
ed[i].push_back (d0 - 1), ed[G[i][j]].push_back (d0 - 1);
if (d0 > 1) ans = (ans + 1LL * d0 * (d0 - 1) % mod * (d0 - 2) % mod) % mod;
}
}
for(int i = 1;i <= cnode;i++){
long long x = 0, y = 0;
for (int j = 0; j < ed[i].size(); ++ j) if (ed[i][j]>0)
x = (x + ed[i][j]) % mod, y = (y+1LL * (ed[i][j] * ed[i][j]) % mod) % mod;
ans = (ans + (x * x % mod - y + mod) % mod) % mod;
}
return 1LL * ans * inv[2] % mod;
}
vector<int> G2[q + 1];
for(int i = 1;i <= cnode;i++){
if(G[i].size() < 2) continue;
int pst[G[i].size()];
for(int j = 0;j < G[i].size();j++) pst[j] = get_num(i , G[i][j]);
for(int j = 0;j < G[i].size();j++){
for(int k = j + 1;k < G[i].size();k++){
G2[pst[j]].push_back(pst[k]);G2[pst[k]].push_back(pst[j]);
}
}
}
return get_node(e_tot , k - 1 , G2);
}
bool connect[(1<<10) + 1];
int neigh[20];
int brout(vector<int> G[])
{
for(int i = 0;i < (1<<limit);i++) connect[i] = 0;
for(int i = 1;i <= limit;i++){
neigh[i] = 0;
for(int j = 0;j < G[i].size();j++){
neigh[i] |= (1<<G[i][j]-1);
}
connect[1<<i-1] = 1;
}
int ans = 0;
for(int i = 1;i < (1<<limit) - 1;i++){
if(!connect[i]) continue;
int mask = 0 , pp = 0;
for(int j = 1;j <= limit;j++){
if((i >> j-1) & 1) mask |= neigh[j];
}
for(int j = 1;j <= limit;j++){
if((mask >> j - 1) & 1) {connect[i | (1<<j-1)] = 1;}
}
if(i == (i&-i)) continue;
for(int j = 1;j <= limit;j++){
T[j].clear();
if(((i >> j-1) & 1) == 0) continue;
if(!pp) pp = j;
for(int k = 0;k < G[j].size();k++){
if((i>>G[j][k]-1) & 1) {T[j].push_back(G[j][k]);}
}
}
ans = (ans + mp[get_hash(0 , pp)]) % mod;
}
return ans;
}
void count()
{
for(int i = 1;i <= limit;i++) T[i].clear();
for(int i = 2;i <= limit;i++){
T[i].push_back(p[i]);T[p[i]].push_back(i);
}
int h = get_hash(0 , 1);
map<int,int>::iterator it = mp.find(h);
if(it != mp.end()) return;
for(int i = 2;i <= limit;i++){
int g = get_p(0 , i);
if(g == -1) continue;
map<int,int>::iterator it = mp.find(g);
if(it != mp.end()) {mp.insert(pair<int,int>{h , it->second});return;}
}
vector<int> W[limit + 1];
for(int i = 1;i <= limit;i++){
for(int j = 0;j < T[i].size();j++) W[i].push_back(T[i][j]);
}
int t = get_node(limit , kk , T);
int g = brout(W);
t = (t + mod - g) % mod;
mp.insert(pair<int,int>{h , t});
return;
}
void dfs(int x)
{
if(x == limit){
count();return;
}
for(int i = 1;i <= x;i++){
p[x + 1] = i;dfs(x + 1);
}
return;
}
void find(int fa,int u)
{
dp[u][1] = 1;
for(int i = 0;i < E[u].size();i++){
if(E[u][i] != fa) find(u , E[u][i]);
}
if(E[u].size() == 1 && fa != 0) return;
int siz = (fa == 0) ? E[u].size() : E[u].size() - 1;
int cop[siz + 1][(1<<kk)+1];
for(int i = 2;i <= cnt;i++){
int len = 0 , g = 1 , pcnt = 0;
for(int j = 0;j < kind[i].size();j++){
if(kind[i][j] != 1){
crt[++pcnt] = j;
++len;
}
}
for(int j = 0;j <= siz;j++){
for(int k = 0;k < (1<<len);k++) cop[j][k] = 0;
}
cop[0][0] = 1;
for(int j = 0;j < E[u].size();j++){
if(E[u][j] == fa) continue;
cop[g][0] = 1;
for(int k = 1;k < (1<<len);k++){
cop[g][k] = cop[g - 1][k];
for(int p = 0;p < len;p++){
if((k>>p) & 1){
cop[g][k] = (cop[g][k] + 1LL * cop[g - 1][k ^ (1<<p)] * dp[E[u][j]][kind[i][crt[p+1]]]) % mod;
}
}
}
g++;
}
if(siz >= kind[i].size()) dp[u][i] = (1LL * cop[siz][(1<<len) - 1] * C[siz - len][kind[i].size() - len]) % mod;
dp[u][i] = (1LL * dp[u][i] * ntd[i]) % mod;
}
return;
}
int main()
{
scanf("%d%d",&n,&kk);
C[0][0] = 1;
for(int i = 1;i <= n;i++){
C[i][0] = 1;
for(int j = 1;j <= i && j <= 15;j++){
C[i][j] = (C[i-1][j] + C[i-1][j - 1]) % mod;
}
}
int g = 1;inv[1] = inv[0] = 1;
for(int i = 2;i <= 20;i++){
g = (1LL * g * i) % mod;
inv[i] = power(g , mod - 2);
}
for(int i = 0;i <= 2000;i++) rnd[i] = rand();
for(int i = 1;i < n;i++){
int u , v;scanf("%d%d",&u,&v);
E[u].push_back(v);
E[v].push_back(u);
}
mp.insert(pair<int,int>{1 , 0});
for(int i = 2;i <= kk + 1;i++){
limit = i;
dfs(1);
}cnt = tot;
find(0 , 1);
int ans = 0;
for(int i = 1;i <= n;i++){
for(int j = 1;j <= cnt;j++){
ans = (ans + 1LL * dp[i][j] * mp[j]) % mod;
}
}
printf("%d\n",ans);
return 0;
}
  • -1
    @ 2020-04-30 18:37:54

    想要知道正解的请去看另外两位大佬的题解。

    我只是想说一下如何用人类智慧，步步套娃，来骗到一些部分分。

    先说一个事情，这个图一定没有重边与自环。我的个人能力只想到了30-50分的办法，不过作为ZJOIZJOI，6个题每个题骗50，就进省队了。

    算法一：k = 1，n = 5000嘛，你可以暴力把图建出来，然后就可以获得0分的好成绩了。

    算法二：k = 2，不难发现一个图的线图的点数就是这个图的边数，利用算法一里的图，输出边数，期望得分10分。

    算法三：k = 3，我们想要L^2(G)L
    2
    (G)的点数，就是需要知道L(G)L(G)的边数，不难发现L(G)L(G)的边数与G每个点的度数有关，点ii的度数为d[i]d[i]，对答案的贡献是C_{d[i]}^2C
    d[i]
    2
    ​ ，期望得分20分。聪明的你一定要预处理逆元的，要不然常数不优秀的话就会收到TLETLE好礼。

    算法四：k = 4，我想要L(G)L(G)的点的度数，不难发现，这与GG的共点边数有关。对于每个边，它变成的点的度数就是与它相邻的边（与它有公共点的边）的数量，期望得分30分。共点边是我自己口胡定义的，不过我相信聪明的你一定可以理解的。

    算法五: k = 5，这个是重头戏，值20分呢。我想要L(G)L(G)的共点边数，不难发现这可以枚举G的两条邻边进行统计，不过复杂度是糟糕的O(n ^ 4n
    4
    )，就算是你的复杂度是O(松)的，都过不去。

    然后这个时候你需要一些信仰，还需要一些卡常能力，你要相信图上的边和点会很少，O(n ^ 2lognn
    2
    logn)是能过的，因为没有写过，我也不确定能不能过，如果有人写完，请告诉我一声，非常感谢。

    不过我们不难发现，对于同一个点，共点边数量的相同的边，是等价的。而一对相邻的公共点为x的边的贡献是g[i] + g[j] - 2g[i]+g[j]−2, g[x]g[x]含义是x的共点边数，不难发现这个答案是可以NTTNTT的, g[x] <= ng[x]<=n，所以复杂度O(n ^ 2lognn
    2
    logn)的。相信信仰的力量，奥利给一下就完事了。

    最后没有代码，因为懒得敲了，要是在考场上就敲了。
    cpp
    #include<cstdio>
    #include<map>
    #include<vector>
    #include<cassert>
    #include<bitset>
    #include<ctime>
    #include<iostream>
    #include<algorithm>
    using namespace std;
    vector<int> E[5005];
    vector<int> kind[1250];
    vector<int> T[15];
    int rnd[2005];
    int dp[5005][1250];
    int crt[20] , ntd[1250];
    int C[5005][20];
    int inv[21];
    const int mod = 998244353;
    const int inv6 = 166374059;
    int n , kk , ans = 0 , limit , cnt = 0;
    int p[20] , tot = 0 , e_tot = 0;
    map<int,int> mp;
    map<long long,int> hashh;
    map<pair<int,int> , int> edge;
    int seed = 91478513 , a = 16554871 , b = 35659598;
    int power(int a,int b)
    {
    int temp = a , ans = 1;
    while(b){
    if(b&1) ans = (1LL * ans * temp) % mod;
    temp = (1LL * temp * temp) % mod;
    b >>= 1;
    }
    return ans;
    }
    inline int rand()
    {
    return seed = (1LL * (seed ^ a) * b) % mod;
    }
    int get_hash(int fa,int u)
    {
    long long h = 1;
    vector<int> hh;
    for(int i = 0;i < T[u].size();i++){
    if(T[u][i] != fa){
    int g = get_hash(u , T[u][i]);
    h += rnd[g];
    hh.push_back(g);
    }
    }
    map<long long , int>::iterator it = hashh.find(h);
    if(it != hashh.end()) return it->second;
    hashh.insert(pair<long long,int>{h , ++tot});
    kind[tot] = hh;ntd[tot] = 1;
    sort(kind[tot].begin() , kind[tot].end());
    int qq = 1 , start = 0;ntd[tot] = 1;
    for(;start < kind[tot].size() && kind[tot][start] == 1;start++);
    for(int i = start + 1;i < kind[tot].size();i++){
    if(kind[tot][i] == 1) continue;
    if(kind[tot][i] == kind[tot][i - 1]) qq++;
    else{
    ntd[tot] = (1LL * ntd[tot] * inv[qq]) % mod;
    qq = 1;
    }
    }
    ntd[tot] = (1LL * ntd[tot] * inv[qq]) % mod;
    return tot;
    }
    int get_p(int fa,int u)
    {
    long long h = 1;
    for(int i = 0;i < T[u].size();i++){
    if(T[u][i] != fa){
    int g = get_p(u , T[u][i]);
    if(g == -1) return -1;
    h += rnd[g];
    }
    }
    map<long long , int>::iterator it = hashh.find(h);
    if(it != hashh.end()) return it->second;
    return -1;
    }
    inline int get_num(int u,int v)
    {
    if(u > v) swap(u , v);
    map<pair<int,int> , int>::iterator it = edge.find(pair<int,int>{u , v});
    if(it != edge.end()) return it->second;
    e_tot++;
    edge.insert(pair<pair<int,int>,int>{pair<int,int>{u , v} , e_tot});
    return e_tot;
    }
    int get_node(int cnode , int k , vector<int> G[])
    {
    edge.clear();e_tot = 0;
    int q = 0;
    for(int i = 1;i <= cnode;i++) q += G[i].size();
    q /= 2;
    if(k == 2){
    int ans = 0;
    for(int i = 1;i <= cnode;i++) ans = (ans + 1LL * (G[i].size() - 1) * G[i].size()) % mod;
    return (1LL * ans * inv[2]);
    }
    if(k == 3){
    int ans = 0;
    for(int i = 1;i <= cnode;i++){
    for(int j = 0;j < G[i].size();j++){
    if(i > G[i][j]) continue;
    if(G[i].size() + G[G[i][j]].size() < 4) continue;
    int d = G[i].size() + G[G[i][j]].size() - 2;
    ans = (ans + 1LL * d * (d - 1) % mod * inv[2]) % mod;
    }
    }
    return ans;
    }
    if(k == 4){
    vector<int> ed[cnode + 1];
    int ans = 0;
    for(int i = 1;i <= cnode;i++){
    for(int j = 0;j < G[i].size();j++){
    if(i > G[i][j]) continue;
    int d0 = G[i].size()+G[G[i][j]].size()-2;
    ed[i].push_back (d0 - 1), ed[G[i][j]].push_back (d0 - 1);
    if (d0 > 1) ans = (ans + 1LL * d0 * (d0 - 1) % mod * (d0 - 2) % mod) % mod;
    }
    }
    for(int i = 1;i <= cnode;i++){
    long long x = 0, y = 0;
    for (int j = 0; j < ed[i].size(); ++ j) if (ed[i][j]>0)
    x = (x + ed[i][j]) % mod, y = (y+1LL * (ed[i][j] * ed[i][j]) % mod) % mod;
    ans = (ans + (x * x % mod - y + mod) % mod) % mod;
    }
    return 1LL * ans * inv[2] % mod;
    }
    vector<int> G2[q + 1];
    for(int i = 1;i <= cnode;i++){
    if(G[i].size() < 2) continue;
    int pst[G[i].size()];
    for(int j = 0;j < G[i].size();j++) pst[j] = get_num(i , G[i][j]);
    for(int j = 0;j < G[i].size();j++){
    for(int k = j + 1;k < G[i].size();k++){
    G2[pst[j]].push_back(pst[k]);G2[pst[k]].push_back(pst[j]);
    }
    }
    }
    return get_node(e_tot , k - 1 , G2);
    }
    bool connect[(1<<10) + 1];
    int neigh[20];
    int brout(vector<int> G[])
    {
    for(int i = 0;i < (1<<limit);i++) connect[i] = 0;
    for(int i = 1;i <= limit;i++){
    neigh[i] = 0;
    for(int j = 0;j < G[i].size();j++){
    neigh[i] |= (1<<G[i][j]-1);
    }
    connect[1<<i-1] = 1;
    }
    int ans = 0;
    for(int i = 1;i < (1<<limit) - 1;i++){
    if(!connect[i]) continue;
    int mask = 0 , pp = 0;
    for(int j = 1;j <= limit;j++){
    if((i >> j-1) & 1) mask |= neigh[j];
    }
    for(int j = 1;j <= limit;j++){
    if((mask >> j - 1) & 1) {connect[i | (1<<j-1)] = 1;}
    }
    if(i == (i&-i)) continue;
    for(int j = 1;j <= limit;j++){
    T[j].clear();
    if(((i >> j-1) & 1) == 0) continue;
    if(!pp) pp = j;
    for(int k = 0;k < G[j].size();k++){
    if((i>>G[j][k]-1) & 1) {T[j].push_back(G[j][k]);}
    }
    }
    ans = (ans + mp[get_hash(0 , pp)]) % mod;
    }
    return ans;
    }
    void count()
    {
    for(int i = 1;i <= limit;i++) T[i].clear();
    for(int i = 2;i <= limit;i++){
    T[i].push_back(p[i]);T[p[i]].push_back(i);
    }
    int h = get_hash(0 , 1);
    map<int,int>::iterator it = mp.find(h);
    if(it != mp.end()) return;
    for(int i = 2;i <= limit;i++){
    int g = get_p(0 , i);
    if(g == -1) continue;
    map<int,int>::iterator it = mp.find(g);
    if(it != mp.end()) {mp.insert(pair<int,int>{h , it->second});return;}
    }
    vector<int> W[limit + 1];
    for(int i = 1;i <= limit;i++){
    for(int j = 0;j < T[i].size();j++) W[i].push_back(T[i][j]);
    }
    int t = get_node(limit , kk , T);
    int g = brout(W);
    t = (t + mod - g) % mod;
    mp.insert(pair<int,int>{h , t});
    return;
    }
    void dfs(int x)
    {
    if(x == limit){
    count();return;
    }
    for(int i = 1;i <= x;i++){
    p[x + 1] = i;dfs(x + 1);
    }
    return;
    }
    void find(int fa,int u)
    {
    dp[u][1] = 1;
    for(int i = 0;i < E[u].size();i++){
    if(E[u][i] != fa) find(u , E[u][i]);
    }
    if(E[u].size() == 1 && fa != 0) return;
    int siz = (fa == 0) ? E[u].size() : E[u].size() - 1;
    int cop[siz + 1][(1<<kk)+1];
    for(int i = 2;i <= cnt;i++){
    int len = 0 , g = 1 , pcnt = 0;
    for(int j = 0;j < kind[i].size();j++){
    if(kind[i][j] != 1){
    crt[++pcnt] = j;
    ++len;
    }
    }
    for(int j = 0;j <= siz;j++){
    for(int k = 0;k < (1<<len);k++) cop[j][k] = 0;
    }
    cop[0][0] = 1;
    for(int j = 0;j < E[u].size();j++){
    if(E[u][j] == fa) continue;
    cop[g][0] = 1;
    for(int k = 1;k < (1<<len);k++){
    cop[g][k] = cop[g - 1][k];
    for(int p = 0;p < len;p++){
    if((k>>p) & 1){
    cop[g][k] = (cop[g][k] + 1LL * cop[g - 1][k ^ (1<<p)] * dp[E[u][j]][kind[i][crt[p+1]]]) % mod;
    }
    }
    }
    g++;
    }
    if(siz >= kind[i].size()) dp[u][i] = (1LL * cop[siz][(1<<len) - 1] * C[siz - len][kind[i].size() - len]) % mod;
    dp[u][i] = (1LL * dp[u][i] * ntd[i]) % mod;
    }
    return;
    }
    int main()
    {
    scanf("%d%d",&n,&kk);
    C[0][0] = 1;
    for(int i = 1;i <= n;i++){
    C[i][0] = 1;
    for(int j = 1;j <= i && j <= 15;j++){
    C[i][j] = (C[i-1][j] + C[i-1][j - 1]) % mod;
    }
    }
    int g = 1;inv[1] = inv[0] = 1;
    for(int i = 2;i <= 20;i++){
    g = (1LL * g * i) % mod;
    inv[i] = power(g , mod - 2);
    }
    for(int i = 0;i <= 2000;i++) rnd[i] = rand();
    for(int i = 1;i < n;i++){
    int u , v;scanf("%d%d",&u,&v);
    E[u].push_back(v);
    E[v].push_back(u);
    }
    mp.insert(pair<int,int>{1 , 0});
    for(int i = 2;i <= kk + 1;i++){
    limit = i;
    dfs(1);
    }cnt = tot;
    find(0 , 1);
    int ans = 0;
    for(int i = 1;i <= n;i++){
    for(int j = 1;j <= cnt;j++){
    ans = (ans + 1LL * dp[i][j] * mp[j]) % mod;
    }
    }
    printf("%d\n",ans);
    return 0;
    }

  • 1

线图

信息

ID
2041
难度
4
分类
(无)
标签
递交数
20
已通过
16
通过率
80%
被复制
2
上传者
  1. 关于
  2. 联系我们
  3. 隐私
  4. 服务条款
  5. 版权申诉
  6. Language
  1. © 2005 - 2023 Vijos.org
  2. uibuild-2f1065f
  3. 沪ICP备14040537号-1