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

A. Nanoassembly

首先用叉积判断是否在指定向量右侧，然后解出法线与给定直线的交点，再关于交点对称即可。

#include<bits/stdc++.h>
using namespace std;
const int Maxn=300020;
typedef long long LL;
typedef pair<int,int>pi;
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);}
	double len(){return hypot(x,y);}
	double len_sqr(){return x*x+y*y;}
	P rot90(){return P(-y,x);}
}a[111111],A,B;
const double eps=1e-8;
int sgn(double x){
	if(x<-eps)return -1;
	if(x>eps)return 1;
	return 0;
}
double cross(P a,P b){
	return a.x*b.y-a.y*b.x;
}
P line_intersection(P a,P b,P p,P q){
	double U=cross(p-a,q-p),D=cross(b-a,q-p);
	return a+(b-a)*(U/D);
}

int main(){
	freopen("input.txt","r",stdin);
	freopen("output.txt","w",stdout);
	int n,m,i;
	scanf("%d%d",&n,&m);
	for(i=1;i<=n;i++)scanf("%lf%lf",&a[i].x,&a[i].y);
	while(m--){
		scanf("%lf%lf%lf%lf",&A.x,&A.y,&B.x,&B.y);
		P t=B-A;
		t=t.rot90();
		//printf("t=%.4f %.4f
",t.x,t.y);
		for(i=1;i<=n;i++)if(cross(a[i]-A,B-A)>0){
			P o=line_intersection(A,B,a[i],a[i]+t);
			//printf("->%d %.4f %.4f
",i,(a[i]+t).x,(a[i]+t).y);
			a[i]=(o*2.0)-a[i];
		}
		//for(i=1;i<=n;i++)printf("%.4f %.4f
",a[i].x,a[i].y);
	}
	for(i=1;i<=n;i++)printf("%.8f %.8f
",a[i].x,a[i].y);
	return 0;
}

　　

B. Playoff

建树根据dfs括号序列判断是否成祖孙关系即可。

#include<bits/stdc++.h>
using namespace std;
const int Maxn=300020;
int n;
string name[Maxn];
map<string,int>id;
int a[Maxn<<2];

char s[Maxn];
bool check(int id1,int id2){
	id1+=n;id2+=n;
	for(int i=id2;i;i>>=1)if(a[i]==id1)return 1;
	return 0;
}
int main(){
	freopen("input.txt","r",stdin);
	freopen("output.txt","w",stdout);
	while(scanf("%d",&n)!=EOF){
		n=1<<n;
		id.clear();
		for(int i=0;i<n;i++)
		cin>>name[i],id[name[i]]=i,a[n+i]=n+i;
		int cur=n>>1,tot=0;
		scanf("%s",s);
		//printf("sss=%s
",s);
		for(;cur;cur>>=1){
			for(int i=cur;i<cur<<1;i++){
				char c=s[tot++];
				if(c=='W')a[i]=a[i<<1];
				else a[i]=a[i<<1|1];
			}
		}
		int q;scanf("%d",&q);
		//printf("q=%d
",q);
		while(q--){
			string s1,s2;
			cin>>s1>>s2;
			int id1=id[s1],id2=id[s2];
			if(check(id1,id2)){
				puts("Win");
			}
			else if(check(id2,id1)){puts("Lose");}
			else puts("Unknown");
		}
	}
}

　　

C. Inequalities

差分约束系统，下界直接作为初始值，然后判断是否出现正环或者超过上限，需要SLF优化。

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

typedef long long LL ;

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

const int MAXN = 1000005 ;
const int MAXE = 1000005 ;

struct Edge {
	int v , c , n ;
	Edge () {}
	Edge ( int v , int c , int n ) : v ( v ) , c ( c ) , n ( n ) {}
} ;

Edge E[MAXE] ;
int H[MAXN] , cntE ;
int d[MAXN] , vis[MAXN] , cnt[MAXN] , Q[MAXN] , head , tail ;
int maxv[MAXN] ;
int n , m ;

void init () {
	cntE = 0 ;
	clr ( H , -1 ) ;
}

void addedge ( int u , int v , int c ) {
	E[cntE] = Edge ( v , c , H[u] ) ;
	H[u] = cntE ++ ;
}

int spfa () {
	while ( head != tail ) {
		int u = Q[head ++] ;
		if ( head == MAXN ) head = 0 ;
		vis[u] = 0 ;
		for ( int i = H[u] ; ~i ; i = E[i].n ) {
			//if ( clock () > 1.99 * CLOCKS_PER_SEC ) return 0 ;
			int v = E[i].v ;
			if ( d[v] < d[u] + E[i].c ) {
				d[v] = d[u] + E[i].c ;
				if ( d[v] > maxv[v] ) return 0 ;
				if ( !vis[v] ) {
					vis[v] = 1 ;
					cnt[v] ++ ;
					if ( cnt[v] == n + 1 ) return 0 ;
					if ( d[Q[head]] < d[v] ) {
						-- head ;
						if ( head < 0 ) head = MAXN - 1 ;
						Q[head] = v ;
					} else {
						Q[tail ++] = v ;
						if ( tail == MAXN ) tail = 0 ;
					}
				}
			}
		}
	}
	return 1 ;
}

void solve () {
	init () ;
	int ok = 1 ;
	head = tail = 0 ;
	for ( int i = 1 ; i <= n ; ++ i ) {
		d[i] = -2e9 ;
		maxv[i] = 2e9 ;
		Q[tail ++] = i ;
		vis[i] = 1 ;
		cnt[i] = 0 ;
	}
	for ( int i = 0 ; i < m ; ++ i ) {
		int op , x , xv , y , yv ;
		scanf ( "%d%d%d%d%d" , &op , &x , &xv , &y , &yv ) ;
		if ( x == 0 ) {
			if ( y == 0 ) addedge ( xv , yv , op ) ;
			else maxv[xv] = min ( maxv[xv] , yv - op ) ;
		} else {
			if ( y == 0 ) d[yv] = max ( d[yv] , xv + op ) ;
			else if ( xv + op > yv ) ok = 0 ;
		}
	}
	if ( !ok || !spfa () ) {
		puts ( "NO" ) ;
		return ;
	}
	puts ( "YES" ) ;
	for ( int i = 1 ; i <= n ; ++ i ) {
		printf ( "%d
" , d[i] ) ;
	}
}

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

　　

D. How to measure the Ocean?

按$s imes p$从小到大排序，然后二分答案，尽量延伸每条线段的长度，看看是否达到$d$即可。

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

const int MAXN = 100005 ;

struct Node {
	int p , a ;
	bool operator < ( const Node& t ) const {
		return p < t.p ;
		//return min ( a - p , t.a - p - t.p ) > min ( a - p - t.p , t.a - t.p ) ;
	}
} ;

Node a[MAXN] ;
int d , n ;

void solve () {
	int S , P , A ;
	for ( int i = 1 ; i <= n ; ++ i ) {
		scanf ( "%d%d%d" , &S , &P , &A ) ;
		a[i].p = S * P ;
		a[i].a = S * A ;
	}
	sort ( a + 1 , a + n + 1 ) ;
	double l = 0 , r = 1e6 ;
	for ( int o = 0 ; o <= 100 ; ++ o ) {
		double x = ( l + r ) / 2 , mid = x ;
		double D = 0 ;
		int ok = 0 ;
		for ( int i = 1 ; i <= n ; ++ i ) {
			double l = max ( 0.0 , 1.0 * ( a[i].a - x ) / a[i].p ) ;
			D += l ;
			x += l * a[i].p ;
			if ( D >= d ) {
				ok = 1 ;
				break ;
			}
		}
		if ( ok ) l = mid ;
		else r = mid ;
	}
	printf ( "%.10f
" , l ) ;
}
	
int main () {
	freopen ( "input.txt" , "r" , stdin ) ;
	freopen ( "output.txt" , "w" , stdout ) ;
	while ( ~scanf ( "%d%d" , &d , &n ) ) solve () ;
	return 0 ;
}

　　

E. Navigation

建图跑最短路即可。

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

typedef long long LL ;

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

const int MAXN = 1605 ;
const double INF = 1e50 ;

int n , m , k , vr , vf ;
double d[MAXN] , G[MAXN][MAXN] ;
int vis[MAXN] , p[MAXN] ;
int x[MAXN] , y[MAXN] ;
vector < int > S ;

double get_dis ( int x , int y ) {
	return sqrt ( 1.0 * x * x + 1.0 * y * y ) ;
}

void dij ( int s ) {
	for ( int i = 1 ; i <= n ; ++ i ) {
		d[i] = INF ;
		vis[i] = 0 ;
		p[i] = 0 ;
	}
	d[s] = 0 ;
	for ( int i = 1 ; i < n ; ++ i ) {
		double minv = INF ;
		int u = s ;
		for ( int j = 1 ; j <= n ; ++ j ) {
			if ( !vis[j] && d[j] < minv ) {
				minv = d[j] ;
				u = j ;
			}
		}
		vis[u] = 1 ;
		for ( int j = 1 ; j <= n ; ++ j ) {
			if ( !vis[j] && d[u] + G[u][j] < d[j] ) {
				d[j] = d[u] + G[u][j] ;
				p[j] = u ;
			}
		}
	}
}

void insert ( int o ) {
	if ( p[o] ) {
		insert ( p[o] ) ;
		S.push_back ( p[o] ) ;
	}
}

void solve () {
	S.clear () ;
	for ( int i = 1 ; i <= n ; ++ i ) {
		scanf ( "%d%d" , &x[i] , &y[i] ) ;
		for ( int j = 1 ; j <= i ; ++ j ) {
			G[i][j] = G[j][i] = get_dis ( x[i] - x[j] , y[i] - y[j] ) / vf ;
		}
	}
	for ( int i = 0 ; i < m ; ++ i ) {
		int u , v ;
		scanf ( "%d%d" , &u , &v ) ;
		G[u][v] = G[v][u] = G[u][v] * vf / vr ;
	}
	int pre = 1 , now ;
	double ans = 0 ;
	for ( int i = 1 ; i <= k ; ++ i ) {
		scanf ( "%d" , &now ) ;
		dij ( pre ) ;
		insert ( now ) ;
		ans += d[now] ;
		pre = now ;
	}
	dij ( now ) ;
	insert ( n ) ;
	ans += d[n] ;
	S.push_back ( n ) ;
	printf ( "%.10f
" , ans ) ;
	for ( int i = 0 ; i < S.size () ; ++ i ) {
		i && putchar ( ' ' ) ;
		printf ( "%d" , S[i] ) ;
	}
	puts ( "" ) ;
}

int main () {
	freopen ( "input.txt" , "r" , stdin ) ;
	freopen ( "output.txt" , "w" , stdout ) ;
	while ( ~scanf ( "%d%d%d%d%d" , &n , &m , &k , &vr , &vf ) ) solve () ;
	return 0 ;
}

　　

F. Bets

根据题意模拟即可。

#include<bits/stdc++.h>
using namespace std;
const int Maxn=300020;
typedef long long LL;
typedef pair<int,int>pi;
void scan(LL &x){
	char s[10];scanf("%s",s);
	int ned=5;
	int has=0;
	x=0;
	for(int i=0;s[i];i++){
		if(s[i]=='.'){
			has=1;
			continue;
		}
		x=x*10+s[i]-'0';
		if(has)ned--;
	}
	for(int i=0;i<ned;i++)x=x*10;
}
int main(){
	
	freopen("input.txt","r",stdin);
	freopen("output.txt","w",stdout);
	int _;scanf("%d",&_);
	while(_--){
		LL a,b,c;
		scan(a);
		scan(b);
		scan(c);
		LL tot=a*b*c;
		LL res=tot/a+tot/b+tot/c;
		//printf("a=%lld b=%lld c=%lld
",a,b,c);
		//printf("res=%lld tot=%lld
",res,tot);
		if(res*100000<tot)puts("YES");
		else puts("NO");
	}
}

　　

G. Ant on the road

留坑。

H. Bouquet

将花按照颜色排序，设$f[i][j][k]$表示考虑了前$i$朵花，总价值为$j$，第$i$种花所在的颜色是否需要计入答案为$k$时颜色数的最大值，然后DP即可。

时间复杂度$O(nS)$。

#include<bits/stdc++.h>
using namespace std;
const int Maxn=300020;
typedef long long LL;
typedef pair<int,int>pi;
int n,S;
pi a[Maxn];
int dp[2][50020][2];
void up(int &x,int y){x=max(x,y);}
int main(){
	freopen("input.txt","r",stdin);
	freopen("output.txt","w",stdout);
	while(scanf("%d%d",&n,&S)!=EOF){
		for(int i=0;i<n;i++){
			scanf("%d%d",&a[i].first,&a[i].second);
		}
		sort(a,a+n);
		memset(dp,-1,sizeof dp);
		dp[0][0][0]=0;
		int cs=0;
		for(int i=0;i<n;i++){
			int islast=(i==n-1)||(a[i+1].first!=a[i].first);
			memset(dp[cs^1],-1,sizeof dp[cs^1]);
			int w=a[i].second;
			for(int j=0;j<=S;j++){
				int nw=w+j;
				//if(nw>S)continue;
				for(int k=0;k<2;k++){
					int val=dp[cs][j][k];
					if(val<0)continue;
					up(dp[cs^1][j][islast?0:k],val);
					if(nw>S)continue;
					up(dp[cs^1][nw][islast?0:1],val+(k!=1));
				}
			}
			cs^=1;
		}
		int ans=max(dp[cs][S][0],dp[cs][S][1]);
		if(ans<0)puts("Impossible");
		else printf("%d
",ans);
	}
}

　　

I. Hash function

倒着解出初始值即可。

#include<cstdio>

typedef unsigned int UI ;

UI Hash ( UI v ) {
	v = v + ( v << 10 ) ;
	v = v ^ ( v >> 6 ) ;
	v = v + ( v << 3 ) ;
	v = v ^ ( v >> 11 ) ;
	v = v + ( v << 16 ) ;
	return v;
}

typedef long long ll;

ll exgcd(ll a,ll b,ll&x,ll&y){
  if(!b)return x=1,y=0,a;
  ll d=exgcd(b,a%b,x,y),t=x;
  return x=y,y=t-a/b*y,d;
}
ll cal(ll a,ll b,ll n){
  ll x,y,d=exgcd(a,n,x,y);
  x=(x%n+n)%n;
  return x*(b/d)%(n/d);
}

UI F(UI v,int B){
  ll a=(1U<<B)+1;
  ll b=v;
  ll n=1LL<<32;
  return cal(a,b,n);
}

UI G11(UI v){
  UI G=v>>21,H=(v>>10)&((1U<<11)-1),I=v&((1U<<10)-1);
  UI A=G;
  UI B=H^A;
  UI C=I^(B>>1);
  return (A<<21)|(B<<10)|(C);
}

UI G6(UI v){
  UI a=v>>26,b=(v>>20)&((1U<<6)-1),c=(v>>14)&((1U<<6)-1),
     d=(v>>8)&((1U<<6)-1),e=(v>>2)&((1U<<6)-1),f=v&((1U<<2)-1);
  UI A=a;
  UI B=A^b;
  UI C=B^c;
  UI D=C^d;
  UI E=D^e;
  UI F=(E>>4)^f;
  return (A<<26)|(B<<20)|(C<<14)|(D<<8)|(E<<2)|F;
}

UI Hash2 ( UI v ) {
	v = F(v,16);
	v = G11(v);
	v = F(v,3);
	v = G6(v);
	v = F(v,10);
	return v;
}

int cal(UI v){
  UI t=v;
  for(int i=1;;i++){
    t=Hash(t);
    if(t==v)return i;
  }
}

UI n ;

int main () {
	freopen ( "input.txt" , "r" , stdin ) ;
	freopen ( "output.txt" , "w" , stdout ) ;
	int T ;
	scanf ( "%d" , &T ) ;
	while ( T -- ) {
		scanf ( "%u" , &n ) ;
		printf ( "%u
" , Hash2 ( n ) ) ;
	}
	return 0 ;
}

　　

J. Civilization

留坑。

K. Master Gambs chairs

每个集合取最小的即可。

#include<bits/stdc++.h>
using namespace std;
const int Maxn=300020;
typedef long long LL;
int n,S;
vector<int>V[Maxn];
int main(){
	freopen("input.txt","r",stdin);
	freopen("output.txt","w",stdout);
	while(scanf("%d%d",&n,&S)!=EOF){
		vector<int>tmp;
		for(int i=1;i<=n;i++)V[i].clear();
		for(int i=1;i<=n;i++){
			int x,y;scanf("%d%d",&x,&y);
			V[x].push_back(y);
		}
		for(int i=1;i<=n;i++){
			if(!V[i].size())continue;
			sort(V[i].begin(),V[i].end());
			tmp.push_back(V[i][0]);
		}
		sort(tmp.begin(),tmp.end());
		LL cur=0;
		int ans=0;
		for(int i=0;i<tmp.size();i++){
			if(cur+tmp[i]<=S){
				cur+=tmp[i];
				ans++;
			}
			else break;
		}
		printf("%d
",ans);
	}
}

　　

L. Scrabble

根据题意模拟即可。

#include<bits/stdc++.h>
using namespace std;
const int Maxn=300020;
typedef long long LL;
typedef pair<int,int>pi;
int n,m;
int di[2][2]={{1,0},{0,1}};
int Mp[22][22];
int col[22][22];
int ltc[]={1,1,2,3,1,1};
int wc[]={1,1,1,1,2,3};
int ans[11];
int base[]={0,1,3,2,3,2,1,5,5,1,2,2,2,2,1,1,2,2,2,2,3,10,5,10,5,10,10,10
	    ,5,5,10,10,3};
int main(){

	freopen("input.txt","r",stdin);
	freopen("output.txt","w",stdout);
	col[1][1]=5;
	for(int i=2;i<=5;i++){
		col[i][i]=4;
	}
	col[6][2]=col[2][6]=3;
	col[4][1]=col[1][4]=col[3][7]=col[7][3]=col[7][7]=2;
	for(int i=1;i<=7;i++){
		for(int j=1;j<=7;j++){
			col[16-i][j]=col[i][16-j]=col[i][j];
		}
	}
	col[8][1]=col[1][8]=col[8][15]=col[15][8]=5;
	col[8][4]=col[4][8]=col[8][12]=col[12][8]=2;
	col[8][8]=1;
	for(int i=1;i<=7;i++){
		for(int j=9;j<=15;j++){
			col[16-i][j]=col[i][j];
		}
	}
	scanf("%d%d",&n,&m);
	//puts("ok");
	for(int i=1,turn=0;i<=m;i++,(turn=(turn+1)%n)){
		int k;scanf("%d",&k);
		//printf("kk=%d
",k);
		int cnt=0;
		for(int it=0;it<k;it++){
			char d;
			int curx,cury;
			int num;scanf("%d %c%d%d",&num,&d,&curx,&cury);
			int ty=d=='h'?0:1;
			int tmp=0,mul=1;
			for(int it2=0;it2<num;it2++){
				int x;scanf("%d",&x);
				if(!Mp[curx][cury])cnt++;
				Mp[curx][cury]=x;
				tmp+=base[x]*ltc[col[curx][cury]];
				mul*=wc[col[curx][cury]];
				curx+=di[ty][0];
				cury+=di[ty][1];
			}
			ans[turn]+=tmp*mul;			
		}
		if(cnt>=7)ans[turn]+=15;
	}
	for(int i=0;i<n;i++)printf("%d
",ans[i]);
}

---

总结：

• D题想复杂了，在错误的道路上越走越远，碰到这种情况应该换人想。
• 读题速度需要提高，没有来得及阅读的J题其实也可做。