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\n",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\n",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\n",a[i].x,a[i].y);}for(i=1;i<=n;i++)printf("%.8f %.8f\n",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\n",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\n",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\n" , 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$s\times p$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\n" , 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\n" , 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\n",a,b,c);//printf("res=%lld tot=%lld\n",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\n",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\n" , 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\n",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\n",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\n",ans[i]);
}

总结：