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：
- 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。
