【知识小结】圆方树 && 广义圆方树
关于仙人掌的总结
immortalCO的博客
yyb的博客
模板
namespace T{
vector <int> e[maxn * 2];
int tag[maxn * 2];
void adde(int x,int y){ //cout<
e[x].PB(y) , e[y].PB(x); }
}
namespace G{
node e[maxn * 2];
int head[maxn],cnt,dfn[maxn],fa[maxn],dfstime,used[maxn];
void adde(int x,int y){
// if ( x < y ) cout<
e[++cnt].to = y;
e[cnt].next = head[x];
head[x] = cnt;
}
void dfs(int x){
dfn[x] = ++dfstime;
for(int i = head[x] ; i ; i = e[i].next){
if ( e[i].to == fa[x] ) continue;
if ( dfn[e[i].to] ){
if ( dfn[e[i].to] > dfn[x] ) continue;
//find_circle
++tot , T::tag[tot] = 1;
int z = x;
while ( z != e[i].to ){
T::adde(tot,z);
used[z] = 1;
z = fa[z];
}
T::adde(e[i].to,tot);
}
else{
fa[e[i].to] = x, dfs(e[i].to);
if ( !used[e[i].to] ) used[e[i].to] = 1 , T::adde(x,e[i].to);
}
}
}
}
例题
仙人掌同构
po姐题解
注意环上点是有序的，而树上子树无序。
环有翻转同构，环为根的话还有旋转同构
代码是NEERC 13 C
数据范围大一点
//仙人掌同构数
#include
e[x].PB(y) , e[y].PB(x); }
void find_g(int x,int fa){
sz[x] = 1;
for (auto y : e[x]){
if ( y == fa ) continue;
find_g(y,x);
sz[x] += sz[y];
mx[x] = max(mx[x],sz[y]);
}
}
ll get_hash_tree(vector <ll> &vec){
ll res = 1240198;
for (auto x : vec){
res = (res + power(C1,x)) % MOD;
}
return res;
}
ll get_hash_cyc(vector <ll> &vec){
ll res = 3982003; ll q = 1;
for (auto x : vec){
q = mul(q,C2);
res = (res + mul(q,power(C3,x))) % MOD;
}
return res;
}
vector <ll> all_cycles(vector <ll> &vec){
vector <ll> item(SZ(vec) * 2);
for (int i = 0 ; i < SZ(vec) ; i++){
item[i] = mul(mul_C2[i + 1],power(C3,vec[i]));
}
for (int i = 0 ; i < SZ(vec) ; i++){
item[i + SZ(vec)] = mul(mul_C2[i + 1 + SZ(vec)],power(C3,vec[i]));
}
for (int i = 1 ; i < SZ(item) ; i++) item[i] = (item[i] + item[i - 1]) % MOD;
vector <ll> res(SZ(vec));
for (int i = 0 ; i < SZ(vec) ; i++){
ll d = (item[i + SZ(vec) - 1] - (i ? item[i - 1] : 0) + MOD) % MOD;
d = mul(inv_C2[i],d);
res[i] = d;
}
return res;
}
ll hash_cactus(int x,int fa){
vector <ll> vec;
if ( tag[x] ){
if ( fa == 0 ){
for (auto y : e[x])
vec.PB(hash_cactus(y,x));
vector <ll> h = all_cycles(vec);
int num = 0;
for (auto y : h) if ( y == h[0] ) num++;
reverse(vec.begin(),vec.end());
vector <ll> h2 = all_cycles(vec);
for (auto y : h2) if ( y == h[0] ) num++;
if ( num > 1 ) add_to_ans(num);
return 0;
}
else{
int up = -1;
for (int i = 0 ; i < SZ(e[x]) ; i++){
if ( e[x][i] == fa ){ up = i; break; }
}
assert(up != -1);
for (int i = up + 1 ; i < SZ(e[x]) ; i++){
vec.PB(hash_cactus(e[x][i],x));
}
for (int i = 0 ; i < up ; i++){
vec.PB(hash_cactus(e[x][i],x));
}
ll h1 = get_hash_cyc(vec);
reverse(vec.begin(),vec.end());
ll h2 = get_hash_cyc(vec);
if ( h1 == h2 ) ans[2]++;
return min(h1,h2);
}
}
else{
for (auto y : e[x]){
if ( y == fa ) continue;
vec.PB(hash_cactus(y,x));
}
sort(vec.begin(),vec.end());
int cur = 1;
for (int i = 1 ; i < SZ(vec) ; i++){
if ( vec[i] != vec[i - 1] ) cur = 0;
++cur;
if ( cur > 1 ) add_to_ans(cur);
}
return get_hash_tree(vec);
}
}
void solve(){
find_g(1,0);
for (int i = 1 ; i <= tot ; i++){
int cur = max(mx[i],tot - sz[i]);
if ( id[0].fi > cur ) id[0] = {cur,i} , id[1] = {inf,0};
else if ( id[1].fi > cur ) id[1] = {cur,i};
}
// cout<
if ( id[0].fi == id[1].fi ){
ll h1 = hash_cactus(id[0].se,id[1].se);
ll h2 = hash_cactus(id[1].se,id[0].se);
if ( h1 == h2 ) ans[2]++;
}
else hash_cactus(id[0].se,0);
}
}
namespace G{
node e[maxn * 2];
int head[maxn],cnt,dfn[maxn],fa[maxn],dfstime,used[maxn];
void adde(int x,int y){
// if ( x < y ) cout<
e[++cnt].to = y;
e[cnt].next = head[x];
head[x] = cnt;
}
void dfs(int x){
dfn[x] = ++dfstime;
for(int i = head[x] ; i ; i = e[i].next){
if ( e[i].to == fa[x] ) continue;
if ( dfn[e[i].to] ){
if ( dfn[e[i].to] > dfn[x] ) continue;
//find_circle
++tot , T::tag[tot] = 1;
int z = x;
while ( z != e[i].to ){
T::adde(tot,z);
used[z] = 1;
z = fa[z];
}
T::adde(e[i].to,tot);
}
else{
fa[e[i].to] = x, dfs(e[i].to);
if ( !used[e[i].to] ) used[e[i].to] = 1 , T::adde(x,e[i].to);
}
}
}
}
int main(){
freopen("cactus.in","r",stdin);
freopen("cactus.out","w",stdout);
// freopen("input.txt","r",stdin);
scanf("%d %d",&n,&m);
for (int i = 1 ; i <= m ; i++){
/* int x,y;
scanf("%d %d",&x,&y);
G::adde(x,y) , G::adde(y,x);*/
int k,x,y = 0;
scanf("%d",&k);
for (int i = 1 ; i <= k ; i++){
scanf("%d",&x);
if ( y ) G::adde(x,y) , G::adde(y,x);
y = x;
}
}
init();
tot = n;
G::dfs(1);
//cout<<"solve\n";
T::solve();
if ( ans.size() == 0 ) puts("0");
else{
printf("%d\n",ans.size());
for (auto y : ans){
printf("%d %d\n",y.fi,y.se);
}
/* ll as = 1 , mod = 1e9 + 3;
for (auto y : ans){
for (int k = 0 ; k < y.se ; k++) as = as * y.fi % mod;
}
cout<
}
}
杂事：datamaker
感觉仙人掌好难造，我是在树上随机加边
#include1. 求仙人掌最大权独立集
一道40min就应该ok的圆方树DP。
犯了两个SB错误！
1. 特判方点忘了往下继续dfs
2. 往下dfs但是环上DP用的数组标号，数组被下面的点修改
都是非常简单。但我经常犯的错误！
从训练到下来总共调了1小时20分钟，非常浪费时间！
#include
// cout<<"end circle\n";
}
}
}
else{
low[x] = min(dfn[to],low[x]);
}
}
}
#define e e2
void adde2(int x,int y){
// cout<
e[x].pb(y);
e[y].pb(x);
}
void build(){
tot = n;
for (int i = 1 ; i <= n ; i++) if ( !dfn[i] ) tarjian(i,0);
//for (int i = 1 ; i <= n ; i++) cout<
}
void dfs(int x,int fa){
if ( !tag[x] ){
f[x][0] = 0 , f[x][1] = val[x];
for (int i = 0 ; i < e[x].size() ; i++){
int to = e[x][i];
if ( to == fa ) continue;
dfs(to,x);
if ( !tag[to] ){
int w = max(f[to][1],f[to][0]);
//g[x][0] = f[x][0] , g[x][1] = f[x][1] , f[x][0] = f[x][1] = 0;
f[x][0] += w;
f[x][1] += f[to][0];
}
else{
f[x][0] += f[to][0];
f[x][1] += f[to][1] - val[x];
}
}
return;
}
h[x][0][1] = h[x][1][0] = -inf , h[x][0][0] = 0 , h[x][1][1] = val[e[x][0]];
for (int i = 1 ; i < e[x].size() ; i++){
int to = e[x][i];
dfs(to,x); //忘了往下递归！
memcpy(g,h[x],sizeof(h[x])); //一定要注意递归下去的数组会改变。不能用标号h[i - 1]！！！
h[x][1][1] = g[1][0] + f[to][1];
h[x][1][0] = max(g[1][0],g[1][1]) + f[to][0];
h[x][0][1] = g[0][0] + f[to][1];
h[x][0][0] = max(g[0][0],g[0][1]) + f[to][0];
}
f[x][0] = max(h[x][0][1],h[x][0][0]);
f[x][1] = h[x][1][0];
}
#undef e
int main(){
// freopen("input.txt","r",stdin);
scanf("%d %d",&n,&m);
for (int i = 1 ; i <= n ; i++) scanf("%d",&val[i]);
for (int i = 1 ; i <= m ; i++){
int x,y;
scanf("%d %d",&x,&y);
adde(x,y);
}
build();
dfs(1,0);
// for (int i = 1; i <= tot ; i++) cout<
cout<<max(f[1][0],f[1][1])<<endl;
}
bzoj 2125: 最短路
仙人掌上最短路
直接圆方树
注意建树的时候处理环长的 问题。
我利用了仙人掌的性质：每条边只在一个环上，如果有返祖边，最多只有一条，并且一定成环，所以把这条边提前放入栈中。这样算环长很方面。
在圆方树上求lca要讨论是圆点还是方点，如果是方点，则表示在环上合并，要取距离较小的一段
倍增被卡常。悄悄开了O3
#include
len[cur[i].fi] = curd;
if ( i ) adde2(tot,cur[i].fi,min(tot_L - curd,curd));
curd += cur[i].se;
}
//cout<<"end circle\n";
}
}
}
else{
//low[x] = min(dfn[to],low[x]);
if ( dfn[to] < dfn[x] ) low[x] = dfn[to];
else tops--;
}
}
}
#define e e2
#define w w2
void adde2(int x,int y,ll d){
//cout<
e[x].pb(y);
w[x].pb(d);
e[y].pb(x);
w[y].pb(d);
}
void build(){
tot = n;
for (int i = 1 ; i <= n ; i++) if ( !dfn[i] ) tarjian(i,0);
//for (int i = 1 ; i <= n ; i++) cout<
}
void dfs(int x){
//cout<<"dfs : "<
for (int i = 0 ; i < e[x].size() ; i++){
int to = e[x][i];
if ( to == fa[x] || dth[to] ) continue;
fa[to] = jump[0][to] = x;
dth[to] = dth[x] + 1 , dis[0][to] = w[x][i];
//cout<
dfs(to);
}
}
void init(){
// rep(i,1,tot) cout<
// cout<
//for (int i = 1 ; i <= n ; i++){ rvc(j,e[i]) cout<
dth[1] = 1 , dfs(1);
rep(i,1,13)
rep(j,1,tot)
jump[i][j] = jump[i - 1][jump[i - 1][j]] , dis[i][j] = dis[i - 1][j] + dis[i - 1][jump[i - 1][j]];
// rep(i,1,tot){ cout<<"check : "<
}
inline int lca(int &x,int &y,ll &res){
if ( dth[x] < dth[y] ) swap(x,y);
int d = dth[x] - dth[y];
rep(i,0,13) if ( (1 << i) & d ) res += dis[i][x], x = jump[i][x];
if ( x == y ) return x;
repd(i,13,0) if ( jump[i][x] != jump[i][y] ) res += dis[i][x] + dis[i][y] , x = jump[i][x] , y = jump[i][y];
//res += dis[0][x] + dis[0][y];
return fa[x];
}
ll query(int x,int y){
ll res = 0;
int d = lca(x,y,res);
//cout<
if ( tag[d] ){
res += min(abs(len[x] - len[y]) , len[d] - abs(len[x] - len[y]));
}
else if ( x != y ) res += dis[0][x] + dis[0][y];
return res;
}
#undef e
#undef w
int main(){
//
// freopen("input.txt","r",stdin);
scanf("%d %d %d",&n,&m,&Q);
rep(i,1,m){
int x,y,w;
scanf("%d %d %d",&x,&y,&w);
adde(x,y,w);
//adde(y,x,w);
}
build();
init();
while ( Q-- ){
int x,y;
scanf("%d %d",&x,&y);
printf("%lld\n",query(x,y));
}
}