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

A. Passage

枚举两个点,看看删掉之后剩下的图是否是二分图。

#include <bits/stdc++.h>using namespace std ;const int MAXN = 205 ;vector < int > G[MAXN] ;int vis[MAXN] , col[MAXN] ;int n ;int dfs ( int u ) {	for ( int i = 0 ; i < G[u].size () ; ++ i ) {		int v = G[u][i] ;		if ( vis[v] == 0 ) continue ;		if ( !col[v] ) {					col[v] = 3 - col[u] ;			if ( !dfs ( v ) ) return 0 ;		}		if ( col[u] + col[v] != 3 ) return 0 ;	}	return 1 ;}int check () {	for ( int i = 1 ; i <= n ; ++ i ) {		col[i] = 0 ;	}	for ( int i = 1 ; i <= n ; ++ i ) {		if ( col[i] || !vis[i] ) continue ;		col[i] = 1 ;		if ( !dfs ( i ) ) return 0 ;	}	return 1 ;}void solve () {	for ( int i = 1 ; i <= n ; ++ i ) {		int x , y ;		scanf ( "%d" , &x ) ;		for ( int j = 0 ; j < x ; ++ j ) {			scanf ( "%d" , &y ) ;			G[i].push_back ( y ) ;			G[y].push_back ( i ) ;		}		vis[i] = 1 ;	}	if ( n <= 3 ) {		printf ( "Hurrah!\n" ) ;		return ;	}	for ( int i = 1 ; i <= n ; ++ i ) {		for ( int j = i + 1 ; j <= n ; ++ j ) {			vis[i] = vis[j] = 0 ;			if ( check () ) {				printf ( "Hurrah!\n" ) ;				return ;			}			vis[i] = vis[j] = 1 ;		}	}	printf ( "Fired.\n" ) ;}int main () {	freopen ( "input.txt" , "r" , stdin ) ;	freopen ( "output.txt" , "w" , stdout ) ;	while ( ~scanf ( "%d" , &n ) ) solve () ;	return 0 ;}

  

B. Files list

按题意模拟。

#include <bits/stdc++.h>using namespace std ;const int MAXN = 10005 ;map < string , int > mp ;map < int , string > mp2 ;char s[MAXN] , p[MAXN] ;int n ;void solve () {	mp.clear () ;	mp2.clear () ;	int cnt = 0 ;	for ( int i = 0 ; i < n ; ++ i ) {		scanf ( "%s" , s ) ;		int t = 0 , f = 0 ;		for ( int j = 0 ; s[j] ; ++ j ) {			if ( f ) p[t ++] = s[j] ;			else if ( s[j] == ‘.‘ ) f = 1 ;		}		p[t] = 0 ;		if ( mp.count ( p ) ) mp[p] ++ ;		else {			mp[p] ++ ;			mp2[++ cnt] = p ;		}	}	for ( int i = 1 ; i <= cnt ; ++ i ) {		cout << mp2[i] << ": " << mp[mp2[i]] << endl ;	}}int main () {	freopen ( "input.txt" , "r" , stdin ) ;	freopen ( "output.txt" , "w" , stdout ) ;	while ( ~scanf ( "%d" , &n ) ) solve () ;	return 0 ;}

  

C. Graph optimization

将所有1类限制的边加入,分块bitset判定限制2是否都满足即可。

时间复杂度$O(\frac{nm}{64})$。

#include<algorithm>#include<cstdio>#include<bitset>#include<set>#include<ctime>using namespace std;typedef bitset<4096>B;typedef unsigned long long ll;const int N=300010;int n,m,i,j,k,x,y,z,ans[N][2],quailty;int g[N],G[N],v[N],nxt[N],ed;int vis[N],q[N],h,t,cnt,f[N],d[N];B dp[100010];//dp[x] : this can reach xset<int>SET[100010];struct E{int x,y,z;}e[N];const int BUF=5000000;char Buf[BUF],*buf=Buf;inline void read(int&a){for(a=0;*buf<48;buf++);while(*buf>47)a=a*10+*buf++-48;}inline bool cmp(const E&a,const E&b){return a.z<b.z;}inline void add(int x,int y){  v[++ed]=y;nxt[ed]=g[x];g[x]=ed;}inline void add2(int x,int y){  v[++ed]=y;nxt[ed]=G[x];G[x]=ed;}void dfs1(int x){  vis[x]=1;  for(int i=g[x];i;i=nxt[i])if(!vis[v[i]])dfs1(v[i]);  q[++t]=x;}void dfs2(int x,int y){  vis[x]=0;f[x]=y;  for(int i=G[x];i;i=nxt[i])if(vis[v[i]])dfs2(v[i],y);}inline bool solve(int L,int R){  if ( clock () > 2.95 * CLOCKS_PER_SEC ) return 1 ;  int z=e[L].z;  int i;  for(i=1;i<=cnt;i++){    dp[i].reset();    //if ( clock () > 2.95 * CLOCKS_PER_SEC ) return 1 ;  }  for(i=max(z<<12,1);(i>>12)==z;i++){    dp[f[i]][i&4095]=1;  }  for(i=1;i<=cnt;i++){    int x=q[i];    for(int j=g[x];j;j=nxt[j]){      dp[v[j]]|=dp[x];      //if ( clock () > 2.95 * CLOCKS_PER_SEC ) return 1 ;    }  }  for(i=L;i<=R;i++){    int x=e[i].x,y=e[i].y;    //printf("-> %d %d %d %llu\n",x,y,z,dp[f[y]]);    if(dp[f[y]][x&4095])return 0;  }  return 1;}int main (){  freopen ( "input.txt" , "r" , stdin ) ;  freopen ( "output.txt" , "w" , stdout ) ;  fread(Buf,1,BUF,stdin);  read(n),read(m);  for(i=1;i<=m;i++){    //x=i,y=i+1;    read(x),read(y);    ans[i][0]=x,ans[i][1]=y;    add(x,y);    add2(y,x);  }    for(i=1;i<=n;i++)if(!vis[i])dfs1(i);  for(i=n;i;i--)if(vis[q[i]])dfs2(q[i],++cnt);    for(ed=0,i=1;i<=cnt;i++)g[i]=0;  for(i=1;i<=m;i++){    x=ans[i][0],y=ans[i][1];    if(f[x]==f[y])continue;    if(SET[f[x]].find(f[y])!=SET[f[x]].end())continue;    SET[f[x]].insert(f[y]);    add(f[x],f[y]);    d[f[y]]++;  }  for(h=1,t=0,i=1;i<=cnt;i++)if(!d[i])q[++t]=i;  while(h<=t)for(i=g[q[h++]];i;i=nxt[i])if(!(--d[v[i]]))q[++t]=v[i];  //now q is topo sequence of SCC    read(quailty);  for(i=1;i<=quailty;i++){    //e[i].x=i+1;    //e[i].y=i;    read(e[i].x),read(e[i].y);    e[i].z=e[i].x>>12;  }  sort(e+1,e+quailty+1,cmp);  for(i=1;i<=quailty;i=j){    for(j=i;j<=quailty&&e[i].z==e[j].z;j++);    if(!solve(i,j-1))return puts("NO"),0;  }  puts("YES");  printf("%d\n",m);  for(i=1;i<=m;i++)printf("%d %d\n",ans[i][0],ans[i][1]);  return 0 ;}

  

D. Housing payments

设$f[i]$表示前$i$个月付清,且第$i$个月进行了交易的最小代价,那么因为债务指数级增长,所以可用决策只有$O(\log n)$项。

#include <bits/stdc++.h>using namespace std ;typedef pair < int , int > pii ;const int MAXN = 100005 ;const int INF = 0x3f3f3f3f ;int s[MAXN] , x[MAXN] , p[MAXN] ;double dp[MAXN] ;int n ;void solve () {	for ( int i = 1 ; i <= n ; ++ i ) {		scanf ( "%d%d%d" , &s[i] , &x[i] , &p[i] ) ;		dp[i] = 1e18 ;	}	dp[0] = 0 ;	for ( int i = 0 ; i <= n ; ++ i ) {		double sum1 = 0 ;		for ( int j = i + 1 ; j <= i + 100 && j <= n ; ++ j ) {			dp[j] = min ( dp[j] , dp[i] + sum1 + s[j] + x[j] ) ;			sum1 = ( sum1 + s[j] ) * ( 1 + 0.01 * p[j] ) ;		}	}	printf ( "%.10f\n" , dp[n] ) ;}int main () {	freopen ( "input.txt" , "r" , stdin ) ;	freopen ( "output.txt" , "w" , stdout ) ;	while ( ~scanf ( "%d" , &n ) ) solve () ;	return 0 ;}

  

E. Arithmetic expressions

$f[i][j]$表示长度为$i$的值为$j$的表达式个数,然后枚举后面接上什么转移即可。

#include <bits/stdc++.h>using namespace std ;const int mod=1e9+7;int n,m,p;int dp[55][222];int len[333];void up(int &x,int y){x+=y;if(x>=mod)x-=mod;}int main () {	freopen ( "input.txt" , "r" , stdin ) ;	freopen ( "output.txt" , "w" , stdout ) ;	while(scanf("%d%d%d",&n,&m,&p)!=EOF){		memset(dp,0,sizeof dp);		for(int i=0;i<m;i++){			int x=i;			if(!x){len[i]=1;continue;}			int l=0;			while(x)l++,x/=10;			len[i]=l;		}		for(int i=1;i<=n;i++){			for(int j=0;j<m;j++){				if(len[j]==i)up(dp[i][j],1);				if(i>=3)up(dp[i][j],dp[i-2][j]);			}			//if(i==2)up(dp[i][0],1);			for(int j=0;j<m;j++){				for(int k=1;k+1<i;k++){					int nj,ni=i-k-1;					for(int nm=0;nm<m;nm++){						for(int ty=0;ty<2;ty++){							if(!ty)nj=(j+nm)%m;							else nj=(j-nm+m)%m;							if(ni>2)up(dp[i][nj],1LL*dp[k][j]*dp[ni-2][nm]%mod);							if(len[nm]==ni)up(dp[i][nj],dp[k][j]);						}					}				}			}		}		printf("%d\n",dp[n][p]);	}        return 0;}

  

F. Sputnik

留坑。

 

G. Voting

设$f[i][j]$表示$i$的子树里选举情况为$j$的最小代价,转移则用另一个$dp[i][a][b][c]$表示考虑了前$i$个儿子,三个人选票各自为$a,b,c$的最小代价来转移。

#include <bits/stdc++.h>using namespace std ;const int mod=1e9+7;const int Maxn=10050,M=10,Inf=1e9;typedef pair<int,int>pi;int n,m,K;vector<int>G[Maxn];int f[Maxn][4];int g[23][23][23][23];int ori[Maxn];vector<pi>ans;vector<pi>res[Maxn][4];pi pre[23][23][23][23];void init(int cs){	for(int i=0;i<21;i++){		for(int j=0;j<21;j++){			for(int k=0;k<21;k++)				g[cs][i][j][k]=Inf;		}	}}struct State{	int x,y,z;	State(){}	State(int x,int y,int z):x(x),y(y),z(z){}}st[Maxn][4];int get(int a,int b,int c){	vector<pi>v;	v.push_back(pi(a,1));	v.push_back(pi(b,2));	v.push_back(pi(c,3));	sort(v.begin(),v.end());	reverse(v.begin(),v.end());	if(v[0].first==v[1].first)return 0;	return v[0].second;}void dfs(int u){	if(u>m){		for(int i=0;i<4;i++){			if(i>K)f[u][i]=Inf;			else {				res[u][i].push_back(pi(u,i));				f[u][i]=i==ori[u]?0:1;			}		}		return ;	}	for(int i=0;i<G[u].size();i++){		int v=G[u][i];		dfs(v);	}	int n=G[u].size();	init(0);	g[0][0][0][0]=0;	for(int i=0;i<n;i++){		int v=G[u][i];		init(i+1);		int it[4];		for(it[0]=0;it[0]<=20;it[0]++)		for(it[1]=0;it[1]<=20;it[1]++)		for(it[2]=0;it[2]<=20;it[2]++){			if(g[i][it[0]][it[1]][it[2]]==Inf)continue;			int w=g[i][it[0]][it[1]][it[2]];			for(int ty=0;ty<4;ty++){				if(f[v][ty]==Inf)continue;				if(ty)it[ty-1]++;				int nw=w+f[v][ty];				int &t=g[i+1][it[0]][it[1]][it[2]];				if(nw<t){					t=nw;					pre[i+1][it[0]][it[1]][it[2]]=pi(v,ty);				}				if(ty)it[ty-1]--;			}		}	}	for(int i=0;i<4;i++)f[u][i]=Inf;	for(int i=0;i<=20;i++){		for(int j=0;j<=20;j++){			for(int k=0;k<=20;k++){				if(g[n][i][j][k]==Inf)continue;				int ty=get(i,j,k);				if(f[u][ty]>g[n][i][j][k]){					f[u][ty]=g[n][i][j][k];					st[u][ty]=State(i,j,k);				}			}		}	}	for(int i=0;i<4;i++){		if(f[u][i]==Inf)continue;		res[u][i].clear();		State tmp=st[u][i];		for(int cur=n;cur>=1;cur--){			res[u][i].push_back(pre[cur][tmp.x][tmp.y][tmp.z]);			int ty=pre[cur][tmp.x][tmp.y][tmp.z].second;			if(ty==1)tmp.x--;			if(ty==2)tmp.y--;			if(ty==3)tmp.z--;		}		reverse(res[u][i].begin(),res[u][i].end());	}	//printf("u=%d\n",u);	//for(int i=0;i<4;i++)printf("%d ",f[u][i]);puts("");}void dfs2(int u,int ty){	//printf("u=%d ty=%d\n",u,ty);	if(u>m){		if(ori[u]!=ty)ans.push_back(pi(u,ty));		return;	}	//printf("sz=%d\n",(int)res[u][1].size());	for(int i=0;i<G[u].size();i++){		//printf("v=%d\n",G[u][i]);		dfs2(G[u][i],res[u][ty][i].second);	}}int main () {	freopen ( "input.txt" , "r" , stdin ) ;	freopen ( "output.txt" , "w" , stdout ) ;	scanf("%d%d%d",&n,&m,&K);	for(int i=1;i<=n;i++)scanf("%d",&ori[i+m]);	for(int i=1;i<=m;i++){		int k;scanf("%d",&k);		for(int j=0;j<k;j++){			int x;scanf("%d",&x);			if(x<0)x=-x;			else x=m+x;			G[i].push_back(x);		}	}	dfs(1);	//puts("ok1");	dfs2(1,1);	printf("%d\n",(int)ans.size());	for(int i=0;i<ans.size();i++)printf("%d %d\n",ans[i].first-m,ans[i].second);        return 0;}

  

H. Novice urbanist

枚举每个点和每个区间,那么它能贡献的距离是一段区间,差分前缀和即可。

#include <bits/stdc++.h>using namespace std ;int n,m,l,r,cnt,i,j;int a[11111],f[4444444];struct P{int l,r;P(){}P(int _l,int _r){l=_l,r=_r;}}b[1111],c[1111];inline bool cmp(const P&a,const P&b){return a.l<b.l;}int main () {  freopen ( "input.txt" , "r" , stdin ) ;  freopen ( "output.txt" , "w" , stdout ) ;  scanf("%d%d",&n,&m);  for(i=1;i<=n;i++)scanf("%d",&a[i]);  for(i=1;i<=m;i++)scanf("%d%d",&b[i].l,&b[i].r);  sort(b+1,b+m+1,cmp);  l=1e9;  r=-l;  for(i=1;i<=m;i++){    if(b[i].l>r+1){      if(l<=r)c[++cnt]=P(l,r);      l=b[i].l;    }    r=max(r,b[i].r);  }  if(l<=r)c[++cnt]=P(l,r);  for(i=1;i<=n;i++){    for(j=1;j<=cnt;j++){      f[c[j].l-a[i]+2000000]++;      f[c[j].r-a[i]+2000000+1]--;    }  }  for(i=1;i<4000000;i++)f[i]+=f[i-1];  int ans=-1,ans2;  for(i=0;i<2000000;i++){    int t=max(f[2000000+i],f[2000000-i]);    if(t>ans)ans=t,ans2=i;  }  printf("%d %d",ans2,ans);  return 0 ;}

  

I. Rangefinder

留坑。

 

J. Hive

求出方向向量$(x,y)$,顺时针旋转$60$°后是$(x+y,-x)$。

#include <bits/stdc++.h>using namespace std ;const int Maxn=100020;int n,x,y,a,b;int main () {	freopen ( "input.txt" , "r" , stdin ) ;	freopen ( "output.txt" , "w" , stdout ) ;	scanf("%d%d%d",&n,&x,&y);	while(n--){	  scanf("%d%d",&a,&b);	  a-=x,b-=y;	  printf("%d %d\n",x+a+b,y-a);        }        return 0;}

  

K. Side effects

每次加入一条边之后暴力染色,然后将经过的边删掉,均摊复杂度$O(n+m)$。

#include <bits/stdc++.h>using namespace std ;const int Maxn=100020;int n,m,q,ans;set<int>G[Maxn];int col[Maxn];void dfs(int u){	//printf("u=%d\n",u);	col[u]=1;	for(set<int>::iterator it=G[u].begin();it!=G[u].end();){		//printf("iv=%d\n",*it);		if(!col[*it]){			dfs(*it);			ans++;		}		G[u].erase(it++);	}}void solve(){	ans=0;	for(int i=1;i<=m;i++){		int x;scanf("%d",&x);		col[x]=1;		ans++;	}	while(q--){		//puts("ok");		int u,v;scanf("%d%d",&u,&v);		if(u!=v)G[v].insert(u);		if(col[v]==1)dfs(v);		printf("%d\n",ans);	}}int main () {	freopen ( "input.txt" , "r" , stdin ) ;	freopen ( "output.txt" , "w" , stdout ) ;	while ( ~scanf ( "%d%d%d" , &n,&m,&q ) ) solve () ;	return 0 ;}

  

L. Cypher

留坑。

 


总结:

  • 在分块bitset时,应尽量设大bitset的大小,减少bitset的操作次数,这样常数反而小。

 

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