XVI Open Cup named after E.V. Pankratiev. GP of SPB

A. Bubbles

枚举两个点,求出垂直平分线与$x$轴的交点,答案=交点数+1。

时间复杂度$O(n^2\log n)$。

#include<cstdio>
#include<algorithm>
#include<cmath>
using namespace std;
const double eps=1e-9;
int sgn(double x){
  if(x<-eps)return -1;
  if(x>eps)return 1;
  return 0;
}

struct P{
  double x,y;
  P(){x=y=0;}
  P(double _x,double _y){x=_x,y=_y;}
  P operator+(P v){return P(x+v.x,y+v.y);}
  P operator-(P v){return P(x-v.x,y-v.y);}
  P operator*(double v){return P(x*v,y*v);}
  P operator/(double v){return P(x/v,y/v);}
  P rot90(){return P(-y,x);}
}a[2000];

double q[1111111];

double cross(P a,P b){return a.x*b.y-a.y*b.x;}

int line_intersection(P a,P b,P p,P q,P&o){
  if(!sgn(a.x-b.x)&&!sgn(a.y-b.y))return 0;
  if(!sgn(p.x-q.x)&&!sgn(p.y-q.y))return 0;  
  double U=cross(p-a,q-p),
         D=cross(b-a,q-p);
  if(sgn(D)==0)return 0;
  o=a+(b-a)*(U/D);
  return 1;
}

int main () {
  freopen ( "bubbles.in" , "r" , stdin ) ;
  freopen ( "bubbles.out" , "w" , stdout ) ;
  int n;
  scanf("%d",&n);
  for(int i=1;i<=n;i++){
    scanf("%lf%lf",&a[i].x,&a[i].y);
  }
  P A(-100,0),B(100,0);
  int cnt=0;
  for(int i=1;i<=n;i++)for(int j=1;j<i;j++){
    P t=(a[i]+a[j])/2.0;
    P v=a[i]-a[j];
    v=v.rot90();
    P tmp;
    if(line_intersection(t,t+v,A,B,tmp)){
      q[++cnt]=tmp.x;
    }
  }
  sort(q+1,q+cnt+1);
  int ans=1;
  for(int i=1;i<=cnt;i++)if(i==1||sgn(q[i]-q[i-1]))ans++;
  printf("%d",ans);
  return 0;
}

  

B. Drop7

留坑。

 

C. Eulerian Graphs

留坑。

 

D. At Least Half

枚举所有质数$p$,找出所有$p$的倍数,设$s[i]$表示前$i$个数里$p$的倍数$\times2$,枚举$i$,要找到最小的$j$满足$s[i]-i\geq s[j]-j$,维护$s[i]-i$的前缀最小值,然后二分查找即可。找到$j$之后贪心计算往左往右能扩展多少。

时间复杂度$O(n\log n\log\log n)$。

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<int,int>pi;
const int Maxn=1000020;
int a[Maxn];
vector<int>V[Maxn/10];
vector<int>pri;
int n;
bool isp[Maxn];
void getp(){
	for(int i=2;i<Maxn;i++){
		if(!isp[i]){
			pri.push_back(i);
		}
		for(int j=0;j<pri.size();j++){
			if(i*pri[j]>=Maxn)break;
			isp[i*pri[j]]=1;
			if(i%pri[j]==0)break;
		}
	}
}
int ans,ansr;
int g[Maxn],gmx[Maxn];
void check(vector<int>&V){
	//for(int i=0;i<V.size();i++)printf("%d ",V[i]);puts("");
	for(int i=0;i<V.size();i++)g[i]=V[i]-2*(i+1);
	for(int i=0;i<V.size();i++)gmx[i]=V[i]-1-2*(i);
	for(int i=1;i<V.size();i++)gmx[i]=max(gmx[i],gmx[i-1]);
	for(int i=0;i<V.size();i++){
		int j=lower_bound(gmx,gmx+V.size(),g[i])-gmx;
		int has=i-j+1,haslen=V[i]-V[j]+1;
		//printf("val=1%d val2=%d\n",V[i-1],j);
		int rr=i==V.size()-1?n:(V[i+1]-1);
		int ll=j==0?1:(V[j-1]+1);
		int tmpans=min(rr-ll+1,has*2);
		int csr=V[i]+min(tmpans-haslen,rr-V[i]);
		//printf("%d %d %d %d\n",ll,rr,has,tmpans);
		if(tmpans>ans){
			ans=tmpans;
			ansr=csr;
		}
	}
}
void solve(vector<int>&V){
	check(V);
}
int main()
{
freopen("halfgcd.in","r",stdin);
freopen("halfgcd.out","w",stdout);
	getp();
	scanf("%d",&n);
	for(int i=1;i<=n;i++)scanf("%d",a+i);
	for(int i=1;i<=n;i++){
		int x=a[i];
		for(int j=0;j<pri.size();j++){
			if(1LL*pri[j]*pri[j]>x)break;
			if(x%pri[j])continue;
			V[j].push_back(i);
			while(x%pri[j]==0)x/=pri[j];
		}
		if(x>1)V[lower_bound(pri.begin(),pri.end(),x)-pri.begin()].push_back(i);
	}
	ans=0,ansr=-1;
	for(int it=0;it<pri.size();it++){
		if(!V[it].size())continue;
		solve(V[it]);
	}
	if(!ans){
		puts("0 0");
	}
	else printf("%d %d\n",ansr-ans+1,ansr);
	return 0;
}

  

E. Next Partition in RLE Notation

只有最后几个数是有用的,分类讨论构造即可。

#include <bits/stdc++.h>
using namespace std ;

typedef long long LL ;
typedef pair < int , int > pii ;
typedef pair < LL , LL > pll ;

#define clr( a , x ) memset ( a , x , sizeof a )

const int MAXN = 100005 ;

vector < pll > G ;
int n ;

void get ( LL N , LL K , LL val ) {
	LL tmp1 = ( N - K ) / ( val - 1 ) ;
	LL tmp2 = ( N - K ) % ( val - 1 ) ;
	LL tmp3 = K - tmp1 - ( tmp2 > 0 ) ;
	if ( tmp1 ) G.push_back ( pll ( tmp1 , val ) ) ;
	if ( tmp2 ) G.push_back ( pll ( 1 , tmp2 + 1 ) ) ;
	if ( tmp3 ) G.push_back ( pll ( tmp3 , 1 ) ) ;
}

void solve () {
	G.clear () ;
	for ( int i = 1 ; i <= n ; ++ i ) {
		LL x , y ;
		scanf ( "%lld%lld" , &x , &y ) ;
		G.push_back ( pll ( x , y ) ) ;
	}
	if ( n == 1 || n == 2 && G[0].second - G[1].second == 1 ) {
		printf ( "-1\n" ) ;
		return ;
	}
	pll a = G[n - 1] ;
	-- n ;
	G.pop_back () ;
	pll b = G[n - 1] ;
	if ( G[n - 1].second - a.second > 1 ) {
		G[n - 1].first -- ;
		if ( !G[n - 1].first ) {
			G.pop_back () ;
			-- n ;
		}
		//printf ( "%lld %lld %lld %lld\n" , a.first , a.second , b.first , b.second ) ;
		get ( a.first * a.second + b.second , a.first + 1 , b.second - 1 ) ;
	} else {
		G.pop_back () ;
		-- n ;
		pll c = G[n - 1] ;
		G[n - 1].first -- ;
		if ( !G[n - 1].first ) {
			G.pop_back () ;
			-- n ;
		}
		get ( a.first * a.second + b.first * b.second + c.second , a.first + b.first + 1 , c.second - 1 ) ;
	}
	printf ( "%d\n" , ( int ) G.size () ) ;
	for ( int i = 0 ; i < G.size () ; ++ i ) {
		printf ( "%lld %lld\n" , G[i].first , G[i].second ) ;
	}
}

int main () {
	freopen ( "next-partition-rle.in" , "r" , stdin ) ;
	freopen ( "next-partition-rle.out" , "w" , stdout ) ;
	while ( ~scanf ( "%d" , &n ) ) solve () ;
	return 0 ;
}

  

F. Equation

留坑。

 

G. “Swap-Plus” Puzzle

显然存在一种最优操作序列,先做行交换,在做列交换,最后再交换格子。

枚举行交换、列交换的顺序,然后贪心交换即可。

#include<cstdio>
#include<algorithm>
#include<cmath>
#include<string>
#include<iostream>
using namespace std;


int i,j,a[9][9],b[9],c[9],d[9][9],m,e[100];
int ans=~0U>>1,cnt;


string q[100];

int now;


int cal(int*a,int n){
  int t=0;
  int i,j;
  static int b[100];
  for(i=1;i<=n;i++)b[i]=a[i];
  for(i=1;i<=n;i++){
    for(j=1;j<=n;j++)if(b[j]==i){
      if(i!=j)t++;
      swap(b[j],b[i]);
      break;
    }
  }
  return t;
}

int cal2(int*a,int n){
  int t=0;
  int i,j;
  static int b[100];
  for(i=1;i<=n;i++)b[i]=i;
  for(i=1;i<=n;i++){
    for(j=1;j<=n;j++)if(b[j]==a[i]){
      if(i!=j)t++;
      swap(b[j],b[i]);
      break;
    }
  }
  return t;
}

void conb(int*a,char st){//[1,2,3,4]->a
  static int b[10];
  int i,j;
  for(i=1;i<=4;i++)b[i]=i;
  for(i=1;i<=4;i++){
    for(j=1;j<=4;j++)if(b[j]==a[i]){
      if(i!=j){
        string t="";
        t.push_back(b[i]+st-1);
        t.push_back('-');
        t.push_back(b[j]+st-1);
        q[++cnt]=t;
      }
      swap(b[j],b[i]);
      break;
    }
  }
}
void cond(){//a->[1,2,3,4,...,16]
  static int a[9][9];
  int i,j,x,y;
  for(i=1;i<=4;i++)for(j=1;j<=4;j++)a[i][j]=d[i][j];
  for(i=1;i<=4;i++)for(j=1;j<=4;j++){
    int now=(i-1)*4+j;
    if(a[i][j]==now)continue;
    bool flag=0;
    for(x=1;x<=4;x++){
      for(y=1;y<=4;y++)if(a[x][y]==now){
        flag=1;
        swap(a[x][y],a[i][j]);
        string t="";
        t.push_back(i+'a'-1);
        t.push_back(j+'0');
        t.push_back('-');
        t.push_back(x+'a'-1);
        t.push_back(y+'0');
        q[++cnt]=t;
        break;
      }
      if(flag)break;
    }
  }
}


int main (){
  freopen ( "puzzle-swap-plus.in" , "r" , stdin ) ;
  freopen ( "puzzle-swap-plus.out" , "w" , stdout ) ;
  for(i=1;i<=4;i++)
    for(j=1;j<=4;j++)
      scanf("%d",&a[i][j]);
  for(i=1;i<=4;i++)b[i]=i;
  
  do{
    for(i=1;i<=4;i++)c[i]=i;
    do{
      now=cal2(b,4)+cal2(c,4);
      for(i=1;i<=4;i++)for(j=1;j<=4;j++)d[b[i]][c[j]]=a[i][j];
      m=0;
      for(i=1;i<=4;i++)for(j=1;j<=4;j++)e[++m]=d[i][j];
      now+=cal(e,16);
      if(now<ans){
        ans=now;
        cnt=0;
        conb(b,'a');
        conb(c,'1');
        cond();
      }
    }while(next_permutation(c+1,c+4+1));
  }while(next_permutation(b+1,b+4+1));
  printf("%d\n",ans);
  for(i=1;i<=cnt;i++)cout<<q[i]<<endl;
  return 0;
}

  

H. Wrong Sieve

考虑每一轮与上一轮位置的递推关系即可。

#include<bits/stdc++.h>
using namespace std;
const int Maxn=105;
typedef long long LL;
typedef pair<int,int>pi;
LL solve(LL x){
	for(LL i=1;x>=(i+1);i++){
		if(x%(i+1)==0)return -1;
		x=x-x/(i+1);
	}
	return x;
}
int main()
{
freopen("sieve.in","r",stdin);
freopen("sieve.out","w",stdout);
	int _;scanf("%d",&_);
	while(_--){
		LL x;scanf("%lld",&x);
		LL ans=solve(x);
		printf("%lld\n",ans);
	}
	return 0;
}

  

I. Space Cat

DP,$f[i][j]$表示在第$i$条线段,重心方向为$j$的最小代价,注意要特判图不联通的情况。

时间复杂度$O(n)$。

#include <bits/stdc++.h>
using namespace std ;

typedef pair < int , int > pii ;

#define clr( a , x ) memset ( a , x , sizeof a )

typedef long long ll;
const ll inf=1LL<<60;
const int N=100010;

int n,i,a[N],b[N],c[N];
ll f[N][2];

inline void up(ll&x,ll y){if(x>y)x=y;}


int main () {
	int T ;
	freopen ( "space-cat.in" , "r" , stdin ) ;
	freopen ( "space-cat.out" , "w" , stdout ) ;
	scanf("%d",&n);
	for(i=1;i<=n;i++)scanf("%d",&a[i]);
	for(i=1;i<=n;i++)scanf("%d",&b[i]),c[i]=a[i]-b[i];
	for(i=2;i<=n;i++)if(b[i]>=a[i-1]||b[i-1]>=a[i])return puts("-1"),0;
	for(i=1;i<=n;i++)f[i][0]=f[i][1]=inf;
	f[1][0]=0;
	for(i=1;i<=n;i++){
	  if(i>1){
	    if(a[i-1]<=a[i])up(f[i][1],f[i-1][1]);
	    if(b[i-1]>=b[i])up(f[i][0],f[i-1][0]);
          }
          for(int j=0;j<4;j++){
            up(f[i][0],f[i][1]+c[i]);
            up(f[i][1],f[i][0]+c[i]);
          }
	}
	if(f[n][0]>=inf)f[n][0]=-1;
	printf("%lld\n",f[n][0]);
	return 0 ;
}

  

J. Tic-Tac-Toe Variation

第一步填中间,然后只有两种情况,分类讨论构造策略即可。

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<int,int>pi;
const int Maxn=1000020;
char Mp[4][4];
bool getMp(){
	bool ret=0;
	for(int i=0;i<3;i++){
		scanf("%s",Mp[i]);
		for(int j=0;j<3;j++){
			if(Mp[i][j]!='x')ret=1;
		}
	}
	return ret;
}
int idx[3][3]={
	{0,1,2},
	{7,-1,3},
	{6,5,4}
};
int dx[8],dy[8];
int curMp[3][3];
int cg(char c){
	if(c=='.')return 0;
	if(c=='x')return 1;
	return 2;
}
void tran(){
	for(int i=0;i<3;i++){
		for(int j=0;j<3;j++){
			curMp[i][j]=cg(Mp[i][j]);
		}
	}
}
void pt(){
	for(int i=0;i<3;i++){
		for(int j=0;j<3;j++){
			int c=curMp[i][j];
			if(!c)putchar('.');
			if(c==1)putchar('x');
			if(c==2)putchar('o');
		}
		puts("");
	}
	fflush(stdout);
}
int getst(){
	for(int i=0;i<3;i++){
		for(int j=0;j<3;j++){
			if(curMp[i][j]==2){
				return idx[i][j];
			}
		}
	}
}
int main()
{
//freopen("halfgcd.in","r",stdin);
//freopen("halfgcd.out","w",stdout);
	for(int i=0;i<3;i++){
		for(int j=0;j<3;j++){
			if(i==1&&j==1)continue;
			dx[idx[i][j]]=i;
			dy[idx[i][j]]=j;
		}
	}
	while(getMp()){
		int st;
		memset(curMp,0,sizeof curMp);
		for(int it=0;;it++){
			if(it==0){
				curMp[1][1]=1;
				pt();
				continue;
			}
			if(it==1){
				getMp();
				tran();
				st=getst();
				int nst=(st+2)%8;
				curMp[dx[nst]][dy[nst]]=1;
				pt();
				continue;
			}
			if(it==2){
				getMp();
				tran();
				if(curMp[dx[(st+6)%8]][dy[(st+6)%8]]!=2){
					curMp[dx[(st+6)%8]][dy[(st+6)%8]]=1;
					pt();
					break;
				}
				else{
					curMp[dx[(st+3)%8]][dy[(st+3)%8]]=1;
					pt();
				}
			}
			if(it==3){
				getMp();
				tran();
				int nst=st&1?((st+1)%8):((st+4)%8);
				if(!curMp[dx[nst]][dy[nst]]){
					curMp[dx[nst]][dy[nst]]=1;
					pt();
					break;
				}
				else{
					nst=(st+7)%8;
					curMp[dx[nst]][dy[nst]]=1;
					pt();
					break;
				}
			}
		}
	}
	return 0;
}

  

K. Captain Tarjan

对于每个询问,将路径上的边权都加1。

那么问题就变成了,给定一棵树,找到一个点,然后随意剖分,使得所有轻边的权值和最小,两遍DP即可。

时间复杂度$O(m\log n)$。

#include <bits/stdc++.h>
using namespace std ;

typedef long long LL ;
typedef pair < int , int > pii ;

#define clr( a , x ) memset ( a , x , sizeof a )

const int MAXN = 100005 ;

vector < pii > G[MAXN] ;
LL dp[MAXN] , dp2[MAXN] , maxc[MAXN] ;
LL Ldp[MAXN] , Rdp[MAXN] , Lmaxc[MAXN] , Rmaxc[MAXN] ;
int c[MAXN] ;
int siz[MAXN] ;
int son[MAXN] ;
int pre[MAXN] ;
int top[MAXN] ;
int dep[MAXN] ;
int n , m ;
LL ans ;

void predfs ( int u , int f ) {
	siz[u] = 1 ;
	son[u] = 0 ;
	for ( int i = 0 ; i < G[u].size () ; ++ i ) {
		int v = G[u][i].first ;
		if ( v == f ) continue ;
		pre[v] = u ;
		dep[v] = dep[u] + 1 ;
		predfs ( v , u ) ;
		siz[u] += siz[v] ;
		if ( siz[v] > siz[son[u]] ) son[u] = v ;
	}
}

void rebuild ( int u , int top_element ) {
	top[u] = top_element ;
	if ( son[u] ) rebuild ( son[u] , top_element ) ;
	for ( int i = 0 ; i < G[u].size () ; ++ i ) {
		int v = G[u][i].first ;
		if ( v != pre[u] && v != son[u] ) rebuild ( v , v ) ;
	}
}

int get_lca ( int x , int y ) {
	while ( top[x] != top[y] ) {
		if ( dep[top[x]] > dep[top[y]] ) x = pre[top[x]] ;
		else y = pre[top[y]] ;
	}
	return dep[x] < dep[y] ? x : y ;
}

void dfs1 ( int u ) {
	for ( int i = 0 ; i < G[u].size () ; ++ i ) {
		int v = G[u][i].first ;
		if ( v == pre[u] ) continue ;
		dfs1 ( v ) ;
		G[u][i].second = c[v] ;
		c[u] += c[v] ;
	}
}

void dfs2 ( int u ) {
	dp[u] = maxc[u] = 0 ;
	for ( int i = 0 ; i < G[u].size () ; ++ i ) {
		int v = G[u][i].first ;
		if ( v == pre[u] ) continue ;
		dfs2 ( v ) ;
		dp[u] += dp[v] + G[u][i].second ;
		maxc[u] = max ( maxc[u] , G[u][i].second + 0LL ) ;
	}
	dp[u] -= maxc[u] ;
}

void dfs3 ( int u , int cost ) {
	int n = G[u].size () ;
	Ldp[0] = 0 ;
	Rdp[n - 1] = 0 ;
	Lmaxc[0] = 0 ;
	Rmaxc[n - 1] = 0 ;
	for ( int i = 1 ; i < n ; ++ i ) {
		int v = G[u][i - 1].first ;
		if ( v == pre[u] ) {
			Ldp[i] = Ldp[i - 1] ;
			Lmaxc[i] = Lmaxc[i - 1] ;
		} else {
			Ldp[i] = Ldp[i - 1] + dp[v] + G[u][i - 1].second ;
			Lmaxc[i] = max ( Lmaxc[i - 1] , G[u][i - 1].second + 0LL ) ;
		}
	}
	for ( int i = n - 2 ; i >= 0 ; -- i ) {
		int v = G[u][i + 1].first ;
		if ( v == pre[u] ) {
			Rdp[i] = Rdp[i + 1] ;
			Rmaxc[i] = Rmaxc[i + 1] ;
		} else {
			Rdp[i] = Rdp[i + 1] + dp[v] + G[u][i + 1].second ;
			Rmaxc[i] = max ( Rmaxc[i + 1] , G[u][i + 1].second + 0LL ) ;
		}
	}
	for ( int i = 0 ; i < n ; ++ i ) {
		int v = G[u][i].first ;
		if ( v == pre[u] ) continue ;
		dp2[v] = dp2[u] + Ldp[i] + Rdp[i] + cost - max ( max ( Lmaxc[i] , Rmaxc[i] ) , cost + 0LL ) ;
		ans = min ( ans , dp2[v] + dp[v] + maxc[v] + G[u][i].second - max ( maxc[v] , 0LL + G[u][i].second ) ) ;
	}
	for ( int i = 0 ; i < n ; ++ i ) {
		int v = G[u][i].first ;
		if ( v == pre[u] ) continue ;
		dfs3 ( v , G[u][i].second ) ;
	}
}

void solve () {
	for ( int i = 1 ; i <= n ; ++ i ) {
		G[i].clear () ;
		c[i] = 0 ;
	}
	for ( int i = 1 ; i < n ; ++ i ) {
		int u , v ;
		scanf ( "%d%d" , &u , &v ) ;
		G[u].push_back ( pii ( v , 0 ) ) ;
		G[v].push_back ( pii ( u , 0 ) ) ;
	}
	predfs ( 1 , 1 ) ;
	rebuild ( 1 , 1 ) ;
	for ( int i = 0 ; i < m ; ++ i ) {
		int u , v ;
		scanf ( "%d%d" , &u , &v ) ;
		c[u] ++ ;
		c[v] ++ ;
		int lca = get_lca ( u , v ) ;
		c[lca] -= 2 ;
	}
	dp2[1] = 0 ;
	dfs1 ( 1 ) ;
	dfs2 ( 1 ) ;
	ans = dp[1] ;
	dfs3 ( 1 , 0 ) ;
	printf ( "%lld\n" , ans ) ;
}

int main () {
	freopen ( "treepaths.in" , "r" , stdin ) ;
	freopen ( "treepaths.out" , "w" , stdout ) ;
	while ( ~scanf ( "%d%d" , &n , &m ) ) solve () ;
	return 0 ;
}

  

L. Gardening Lesson

求出两棵树的重心,枚举重心作为根,然后求出每棵子树的Hash值,在两棵树上一起dfs即可。

时间复杂度$O(n\log n)$。

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<int,int>pi;
const int Maxn = 100005 , x = 123 , M = 998244353 , Left = 23 , Right = 371 ;
const int Hleaf=Left*Right,Inf=1e9;
vector < int > G1[Maxn] , G2[Maxn] ;
int H1[Maxn] , H2[Maxn] ;
int sz[Maxn],sonsz[Maxn],ans ;
void dfs(int u , vector<int>*G,int p){
	sz[u]= 1;
	sonsz[u]=0;
	for(int i=0;i<G[u].size ();++i){
		int v =G[u][i];
		if(v==p)continue ;
		dfs(v,G,u);
		sz[u]+=sz[v];
		if(sz[v]>sonsz[u])sonsz[u]=sz[v];
	}
}
vector<int>getroot(vector<int>*G,int n){
	dfs(1,G,0);
	vector<int>ret;
	for(int i=1;i<=n;++i){
		if(max(n-sz[i],sonsz[i])<=n/2){
			ret.push_back(i);
		}
	}
	return ret;
}
bool cmp1(int a, int b){return H1[a]<H1[b];}
bool cmp2(int a, int b){return H2[a]<H2[b];}
void dfs2(int u,int p,vector<int>*G,int*H,int ty){
	H[u]=Left;
	for(int i =0;i<G[u].size();++i){
		int v=G[u][i];if(v==p)continue;
		dfs2(v,u,G,H,ty);
	}
	sort(G[u].begin(),G[u].end(),ty==1?cmp1:cmp2);
	for(int i=0;i<G[u].size();++i){
		int v=G[u][i];if(v==p)continue;
		H[u]=((1LL*H[u]*x)^H[v])%M;
	}
	H[u]=1LL*H[u]*Right%M;
}
void check(int ua,int pa,int ub,int pb){
	//printf("%d %d %d %d\n",ua,pa,ub,pb);
	set<pi>S;
	for(int i=0;i<G2[ub].size();i++){
		int v=G2[ub][i];
		if(v==pb)continue;
		S.insert(pi(H2[v],v));
	}
	vector<int>rest;
	for(int i=0;i<G1[ua].size();i++){
		int v=G1[ua][i];
		if(v==pa)continue;
		set<pi>::iterator it=S.lower_bound(pi(H1[v],-1));
		if(it==S.end()||(it->first!=H1[v]))rest.push_back(v);
		else S.erase(it);
	}
	if(rest.size()>1||S.size()>1)return;
	if(S.size()==1&&!rest.size()){
		if(S.begin()->first!=Hleaf)return;
		for(int i=0;i<G2[ub].size();i++){
			int v=G2[ub][i];if(v==pb)continue;
			if(H2[v]==Hleaf){ans=min(ans,v);}
		}
		return;
	}
	if(S.size()==1&&rest.size()==1){
		for(int i=0;i<G2[ub].size();i++){
			int v=G2[ub][i];if(v==pb)continue;
			if(H2[v]==S.begin()->first)check(rest[0],ua,v,ub);
		}
	}
}
void solve( int rt1 , int rt2){
	dfs2(rt1,0,G1,H1,1);
	dfs2(rt2,0,G2,H2,2);
	check(rt1,0,rt2,0);
}
void pt(vector<int>&v){
	for(int i=0;i<v.size();i++)printf("%d ",v[i]);
	puts("");
}
int main()
{
freopen("unexpected-leaf.in","r",stdin);
freopen("unexpected-leaf.out","w",stdout);
	int n;
	while(scanf("%d",&n)!=EOF){
		for(int i=1;i<=n+1;i++){
			G1[i].clear();
			G2[i].clear();
		}
		for(int i=1;i<n;i++){
			int u,v;scanf("%d%d",&u,&v);
			G1[u].push_back(v);
			G1[v].push_back(u);
		}
		for(int i=1;i<=n;i++){
			int u,v;scanf("%d%d",&u,&v);
			G2[u].push_back(v);
			G2[v].push_back(u);
		}
		if(n==1){
			printf("1\n");
			continue;
		}
		if(n==2){
			if(G2[1].size()==2)printf("2\n");
			else printf("1\n");
			continue;
		}
		vector<int>tmp1=getroot(G1,n);
		vector<int>tmp2=getroot(G2,n+1);
		//pt(tmp1);
		//pt(tmp2);
		ans=Inf;
		//solve(1,3);
		for(int i=0;i<tmp1.size();i++){
			for(int j=0;j<tmp2.size();j++){
				solve(tmp1[i],tmp2[j]);
			}
		}
		printf("%d\n",ans);
	}
	return 0;
}

  


总结:

  • L题发现树同构模板有误。

 

posted @ 2016-11-04 23:27  Claris  阅读(...)  评论(... 编辑 收藏