2015 ACM Syrian Collegiate Programming Contest

A. My Friend of Misery

计算出答案的上下界即可。

时间复杂度$O(n)$。

#include<bits/stdc++.h>
using namespace std;
const int N=100010;
typedef long long LL;
int main(){int _;scanf("%d",&_);while(_--){int n;scanf("%d",&n);LL cur=0,r=1LL<<60;while(n--){int x;char c;scanf("%d %c",&x,&c);if(c=='-'){r=min(r,cur+x+25);}else{cur+=x+25;}//printf("%lld\n",r-cur);}printf("%lld\n",r-cur);}
}

B. Brother Louie

纵坐标根据深度即可计算，比较麻烦的是横坐标。

设最左边的叶子的横坐标为$t$，那么每个点的横坐标都能表示为$t+A_ir+B_ih$，根据$x_{root}=0$，可以解出$t$，然后就可以计算出对应点的横坐标。

时间复杂度$O(n)$。

#include<cstdio>
#include<algorithm>
using namespace std;
const int N=100010;
int T,n,m,i,j,a[N],f[N],son[N][2],root;
int d[N],cnt;
double A[N],B[N];//t+A*r+B*h
void dfs(int x,int y){d[x]=y;if(!a[x]){A[x]=cnt*2;B[x]=cnt;cnt++;return;}for(int i=0;i<2;i++)dfs(son[x][i],y+1);A[x]=(A[son[x][0]]+A[son[x][1]])/2.0;B[x]=(B[son[x][0]]+B[son[x][1]])/2.0;
}
inline double askx(double r,double h,int x){return (A[x]-A[root])*r+(B[x]-B[root])*h;
}
inline double asky(double r,double v,int x){return -(r*2+v)*d[x];
}
int main(){scanf("%d",&T);while(T--){scanf("%d%d",&n,&m);for(i=1;i<=n;i++){scanf("%d",&a[i]);if(a[i]){for(j=0;j<2;j++){scanf("%d",&son[i][j]);f[son[i][j]]=i;}}}for(i=1;i<=n;i++)if(!f[i])root=i;dfs(root,0);while(m--){double r,v,h;int x;scanf("%lf%lf%lf%d",&r,&v,&h,&x);printf("%.4f %.4f\n",askx(r,h,x),asky(r,v,x));}root=cnt=0;for(i=0;i<=n;i++){a[i]=f[i]=son[i][0]=son[i][1]=d[i]=0;A[i]=B[i]=0;}}return 0;
}

C. Everything

建立Trie树，dfs一遍得到每个串的字典序，然后枚举前缀，根据排名更新答案即可。

时间复杂度$O(n\log n)$。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=100010,M=500010;
int T,n,i,j,pos[N],rk[N],cnt,ans[N];
int tot,son[M][26],id[M],d[M],f[M];
int m,q[N],g[M],v[M],nxt[M],ed;
char s[N];
inline bool cmp(int x,int y){return rk[x]<rk[y];}
inline void ins(int p){scanf("%s",s);int l=strlen(s),x,i;for(x=0,i=0;i<l;i++){int w=s[i]-'a';if(!son[x][w]){son[x][w]=++tot;f[tot]=x;}x=son[x][w];}pos[p]=x;id[x]=p;
}
void dfs(int x,int y){d[x]=y;if(id[x])rk[id[x]]=++cnt;for(int i=0;i<26;i++)if(son[x][i])dfs(son[x][i],y+1);
}
inline void deal(int x){for(int i=pos[x];i;i=f[i]){v[++ed]=x;nxt[ed]=g[i];g[i]=ed;}
}
inline int cal(int n,int x){return min(x,n-x+2);
}
inline void solve(int x){m=0;for(int i=g[x];i;i=nxt[i])q[++m]=v[i];if(!m)return;sort(q+1,q+m+1,cmp);for(int i=1;i<=m;i++)ans[q[i]]=min(ans[q[i]],d[x]+cal(m,i));
}
int main(){scanf("%d",&T);while(T--){scanf("%d",&n);for(i=1;i<=n;i++)ins(i);dfs(0,0);for(i=1;i<=n;i++)ans[i]=cal(n,rk[i]);for(i=1;i<=n;i++)deal(i);for(i=1;i<=tot;i++)solve(i);for(i=1;i<=n;i++)printf("%d%c",ans[i],i<n?' ':'\n');for(i=0;i<=tot;i++){g[i]=id[i]=d[i]=f[i]=0;for(j=0;j<26;j++)son[i][j]=0;}cnt=ed=tot=0;}return 0;
}

D. Secure but True

首先计算出答案串的长度，然后看成11进制下的高精度加法即可。

#include <bits/stdc++.h>
using namespace std ;const int MAXN = 100005 ;
typedef long long LL;
char s[]={'A', 'H', 'I', 'M', 'O', 'T', 'U', 'V', 'W', 'X', 'Y'};
int id[333];
char ss[1020];
int a[1020],b[1020];
int c[1020];
LL sum[1020],pww[1020];
const LL Inf=1LL<<50;
void upd(LL &pw){pw*=11;if(pw>=Inf)pw=Inf;
}
void solve () {int x;scanf("%d",&x);scanf("%s",ss);int cnt2=strlen(ss);for(int i=0;ss[i];i++)b[i]=id[ss[i]];LL pw=1,cursum=0,sel=-1;sum[cnt2]=0;for(int i=cnt2-1;i>=0;i--,upd(pw)){LL cost=(10-b[i])*pw;cursum+=cost;sum[i]=cursum;pww[i]=pw;if(cursum>=x){sel=i;break;}}//printf("sel=%lld\n",sel);if(sel>=0){x-=sum[sel+1]+1;b[sel]+=x/pww[sel]+1;x%=pww[sel];for(int i=sel+1;i<cnt2;i++)b[i]=0;int curloc=cnt2-1;while(x){b[curloc--]=x%11;x/=11;}for(int i=0;i<cnt2;i++)printf("%c",s[b[i]]);puts("");return ;}x-=cursum;//printf("x=%d\n",x);pw=pww[0];upd(pw);upd(pw);//printf("pw=%lld\n",pw);int len=cnt2;if(x>1){while(x>pw)x-=pw,pw*=11,len++;}len++;//printf("x=%d\n",x);for(int i=0;i<len;i++){c[i]=0;}x--;int cnt3=0;while(x){c[cnt3++]=x%11;x/=11;}reverse(c,c+len);for(int i=0;i<len;i++)printf("%c",s[c[i]]);puts("");
}int main () {for(int i=0;i<11;i++)id[s[i]]=i;int T ;scanf ( "%d" , &T ) ;for ( int i = 1 ; i <= T ; ++ i ) {solve () ;}return 0 ;
}

E. Going in Circles

切割点只可能是2个，3个或者4个。

首先特判掉整个环全部都是小写字母的情况，然后将每段小写字符压缩成一块，并计算出在每块内部切割的方案数。

枚举每种大写字符，可以得到要切割的两部分应该分布在哪些范围内，设$f[i][j]$为第$i$个字符下某个切割块所属的连通块编号，对$f$进行Hash，那么一定是在Hash值相同的里面选取两个块，每块切1~2刀。

时间复杂度$O(n(26+\log n))$。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=50010;
const int BASE=10007;
typedef long long ll;
typedef unsigned long long ull;
char a[N];
int block;
int T,n,m,i,j,k,f[N],q[N];
int id[N],size[N],vis[N];
ll C2[N],C3[N],C4[N],ans;
ull h[N];
inline bool cmp(int x,int y){return h[x]<h[y];}
inline bool small(char x){return x>='a'&&x<='z';}
int F(int x){return f[x]==x?x:f[x]=F(f[x]);}
inline void merge(int x,int y){if(F(x)!=F(y))f[f[x]]=f[y];
}
inline bool special(){for(int i=0;i<n;i++)if(!small(a[i]))return 0;return 1;
}
inline ll solvespecial(){return C2[n]+C2[3]*C3[n]+2LL*C4[n];
}
inline void update(int l,int r,int p){for(int i=l;i<=r;i++){int x=id[f[i]];if(vis[x])continue;vis[x]=1;h[x]+=p;}
}
inline ll solveoneblock(int n){return C2[n]+2LL*C3[n]+C4[n];
}
int main(){for(i=1;i<N;i++){C2[i]=1LL*i*(i-1)/2;C3[i]=1LL*i*(i-1)*(i-2)/6;C4[i]=1LL*i*(i-1)*(i-2)*(i-3)/24;}scanf("%d",&T);while(T--){scanf("%s",a);n=strlen(a);if(special()){printf("%lld\n",2LL*solvespecial());continue;}for(i=0;i<n;i++)f[i]=i;for(i=0;i<n;i++){//the divide between i and (i+1)%nif(small(a[i]))merge((i-1+n)%n,i);}for(i=0;i<n;i++)id[i]=0;for(i=1;i<=n;i++)size[i]=0;block=0;for(i=0;i<n;i++){if(!id[F(i)]){id[f[i]]=++block;}size[id[f[i]]]++;}ans=0;for(i=1;i<=block;i++)h[i]=0;for(char O='A';O<='Z';O++){for(i=1;i<=block;i++)h[i]*=BASE,vis[i]=0;m=0;for(i=0;i<n;i++)if(a[i]==O)q[m++]=i;for(i=1;i<m;i++){update(q[i-1],q[i]-1,i);}}//printf("block=%d\n",block);//for(i=0;i<n;i++)printf("%d ",f[i]);puts("");for(i=1;i<=block;i++){ans+=solveoneblock(size[i]);// printf("%d %d %lld\n",i,size[i],solveoneblock(size[i]));}//ans=0;for(i=1;i<=block;i++)q[i]=i;sort(q+1,q+block+1,cmp);for(i=1;i<=block;i=j){ll sum=0;for(j=i;j<=block&&h[q[i]]==h[q[j]];j++);for(k=i;k<j;k++){ll ret=C2[size[q[k]]]+size[q[k]];//printf("%d %d %lld\n",q[k],size[q[k]],ret);//printf("%llu %d %d %lld %lld\n",h[q[k]],q[k],size[q[k]],ret,sum);ans+=ret*sum;sum+=ret;}//[i,j)}printf("%lld\n",2LL*ans);}return 0;
}

F. Hey JUDgE

暴力枚举所有合并情况然后检验即可。

时间复杂度$O(n^5)$。

#include<cstdio>
#include<iostream>
#include<string>
using namespace std;
string s;
int T;
inline bool check(string s){static bool v[11];for(int i=0;i<11;i++)v[i]=0;for(int i=0;i<7;i++)v[s[i]]=1;for(int i=1;i<=5;i++)if(!v[i])return 0;return 1;
}
bool solve(){int i,j,a,b;cin>>s;for(i=0;i<7;i++)s[i]=s[i]-'A'+1;if(check(s))return 1;for(i=0;i<7;i++)for(j=0;j<i;j++){string t=s;t[i]+=t[j];t[j]=0;if(check(t))return 1;}for(i=0;i<7;i++)for(j=0;j<i;j++)for(a=0;a<7;a++)for(b=0;b<a;b++){if(i==a)continue;if(i==b)continue;if(j==a)continue;if(j==b)continue;string t=s;t[i]+=t[j];t[j]=0;t[a]+=t[b];t[b]=0;if(check(t))return 1;}return 0;
}
int main(){cin>>T;while(T--){puts(solve()?"YES":"NO");}
}

G. Paradise City

按题意模拟即可。

#include <bits/stdc++.h>
using namespace std ;const int MAXN = 303 ;char s[MAXN][MAXN] ;void solve () {int n ;scanf ( "%d" , &n ) ;for ( int i = 1 ; i <= 3 ; ++ i ) {scanf ( "%s" , s[i] + 1 ) ;}int ans = 0 ;for ( int i = 1 ; i <= 3 ; ++ i ) {for ( int j = 1 ; j <= 3 * n ; ++ j ) if ( s[i][j] == 'X' ) {int t = 0 ;for ( int x = -1 ; x <= 1 ; ++ x ) {for ( int y = -1 ; y <= 1 ; ++ y ) if ( x || y ) {int nx = i + x ;int ny = j + y ;if ( nx < 1 || nx > 3 || ny < 1 || ny > 3 * n ) continue ;if ( s[nx][ny] == '*' ) t += 4 ;}}ans = max ( ans , t ) ;}}printf ( "%d\n" , ans ) ;
}int main () {int T ;scanf ( "%d" , &T ) ;for ( int i = 1 ; i <= T ; ++ i ) {solve () ;}return 0 ;
}

H. Another Square in the Floor

$ans=\max^2(X,Y)$。

#include<bits/stdc++.h>
using namespace std;
int main(){int _;scanf("%d",&_);while(_--){int n,m;scanf("%d%d",&n,&m);printf("%d\n",max(n,m)*max(n,m));}
}

I. Home Sweet Home

枚举前导零的个数，然后从两边同时开始数位DP即可，一边从高位到低位转移，另一边从低位到高位转移。

#include <bits/stdc++.h>
using namespace std ;const int MAXN = 100005 ;typedef long long LL;
int a[44],b[44];
LL dp1[66][2][2];
LL dp2[50][50][2][2][3][3];
int get(int x,int y){if(x==y)return 0;if(x<y)return 1;return 2;
}
int check(int can1,int can2,int ncan1,int ncan2){if(!can1&&ncan1==2)return 0; if(!can2&&ncan2==2)return 0;  return 1;
}
LL dfs2(int cur,int totlen,int can1,int can2,int lcan1,int lcan2){if(!totlen)return check(can1,can2,lcan1,lcan2);if(totlen%2==0&&cur>=totlen/2)return check(can1,can2,lcan1,lcan2);LL &ret=dp2[cur][totlen][can1][can2][lcan1][lcan2];if(ret>=0)return ret;ret=0;int x1=a[totlen-cur],x2=b[totlen-cur];int lx1=a[cur+1],lx2=b[cur+1];if((totlen%2==1)&&(cur==totlen/2)){for(int t1=0;t1<2;t1++)for(int t2=0;t2<2;t2++){if(t1>x1&&!can1)continue;if(t2>x2&&!can2)continue;int ncan1=can1||(t1<x1);int ncan2=can2||(t2<x2);ret+=check(ncan1,ncan2,lcan1,lcan2);}return ret;}for(int t1=0;t1<2;t1++)for(int t2=0;t2<2;t2++){if(t1>x1&&!can1)continue;if(t2>x2&&!can2)continue;int ncan1=can1||(t1<x1);int ncan2=can2||(t2<x2);for(int lt1=0;lt1<2;lt1++)for(int lt2=0;lt2<2;lt2++){if((lt1^lt2)!=(t1^t2))continue;int nlcan1=get(lt1,lx1);if(!nlcan1)nlcan1=lcan1;int nlcan2=get(lt2,lx2);if(!nlcan2)nlcan2=lcan2;ret+=dfs2(cur+1,totlen,ncan1,ncan2,nlcan1,nlcan2);}}return ret;
}
LL dfs1(int cur,int can1,int can2){if(cur<0)return 1;LL &ret=dp1[cur][can1][can2];if(ret>=0)return ret;ret=0;for(int t1=0;t1<2;t1++)for(int t2=0;t2<2;t2++){if(t1>a[cur]&&!can1)continue;if(t2>b[cur]&&!can2)continue;int ncan1=can1||(t1<a[cur]);int ncan2=can2||(t2<b[cur]);if(!(t1^t2)){ret+=dfs1(cur-1,ncan1,ncan2);}else{if(!cur){ret++;continue;}for(int lt1=0;lt1<2;lt1++)for(int lt2=0;lt2<2;lt2++){if((lt1^lt2)!=1)continue;int nlt1=get(lt1,a[0]);int nlt2=get(lt2,b[0]);ret+=dfs2(0,cur-1,ncan1,ncan2,nlt1,nlt2);}}}return ret;
}
LL solve(LL x,LL y){for(int i=0;i<43;i++){a[i]=x&1;x>>=1;}for(int i=0;i<43;i++){b[i]=y&1;y>>=1;}memset(dp1,-1,sizeof dp1);memset(dp2,-1,sizeof dp2);LL ret=dfs1(42,0,0);return ret;
}
void go () {LL L,R;scanf("%lld%lld",&L,&R);//printf("ans1=%lld\n",solve(0,4));//printf("ans2=%lld\n",solve(4,0));//return ;LL ans=solve(R,R);//printf("ans=%lld\n",ans);if(L)ans-=solve(L-1,R);//printf("ans=%lld\n",ans);if(L)ans-=solve(R,L-1);//printf("ans=%lld\n",ans);if(L)ans+=solve(L-1,L-1);printf("%lld\n",ans);
}           int main () {int T ;scanf ( "%d" , &T ) ;for ( int i = 1 ; i <= T ; ++ i ) {go () ;}return 0 ;
}

J. Smooth Developer

按题意模拟即可。

#include <bits/stdc++.h>
using namespace std ;const int MAXN = 100005 ;map < string , int > mp1 ;
map < int , string > mp2 ;
int n , m , cnt ;
vector < int > G[MAXN] ;
int vis[MAXN] ;
char s[20] ;int get () {scanf ( "%s" , s ) ;if ( mp1.count ( s ) ) return mp1[s] ;++ cnt ;mp1[s] = cnt ;mp2[cnt] = s ;return cnt ;
}void dfs ( int u ) {if ( vis[u] ) return ;vis[u] = 1 ;for ( int i = 0 ; i < G[u].size () ; ++ i ) {int v = G[u][i] ;dfs ( v ) ;}
}void solve () {cnt = 0 ;mp1.clear () ;mp2.clear () ;scanf ( "%d%d" , &n , &m ) ;for ( int i = 1 ; i <= n ; ++ i ) {G[i].clear () ;vis[i] = 0 ;}for ( int i = 1 ; i <= n ; ++ i ) {int u = get () , c ;scanf ( "%d" , &c ) ;for ( int j = 0 ; j < c ; ++ j ) {int v = get () ;G[u].push_back ( v ) ;}}for ( int i = 1 ; i <= m ; ++ i ) {int u = get () ;dfs ( u ) ;}for ( int i = 1 ; i <= n ; ++ i ) {if ( vis[i] ) cout << mp2[i] << endl ;}
}int main () {int T ;scanf ( "%d" , &T ) ;for ( int i = 1 ; i <= T ; ++ i ) {solve () ;}return 0 ;
}

K. Betrayed

首先对于每棵树两遍DP求出以每个点为根的最大深度，就可以得到每棵树爆栈的概率，递推求出期望步数即可。

时间复杂度$O(Cn)$。

#include <bits/stdc++.h>
using namespace std ;const int MAXN = 100005 ;typedef long long LL;vector < int > G[MAXN] ;
int dp[MAXN] , dp2[MAXN] ;
double p[22] ;
double f[22] ;
//double a[22][22] ;
int C , K , n ;void dfs2 ( int u , int f ) {int maxv = 0 ;for ( int i = 0 ; i < G[u].size () ; ++ i ) {int v = G[u][i] ;if ( v == f ) continue ;dp2[v] = max ( dp2[u] + 1 , maxv ) ;maxv = max ( maxv , dp[v] + 2 ) ;}maxv = 0 ;for ( int i = G[u].size () - 1 ; ~i ; -- i ) {int v = G[u][i] ;if ( v == f ) continue ;dp2[v] = max ( dp2[v] , maxv ) ;maxv = max ( maxv , dp[v] + 2 ) ;}for ( int i = 0 ; i < G[u].size () ; ++ i ) {int v = G[u][i] ;if ( v == f ) continue ;dfs2 ( v , u ) ;}
}void dfs ( int u , int f ) {for ( int i = 0 ; i < G[u].size () ; ++ i ) {int v = G[u][i] ;if ( v == f ) continue ;dfs ( v , u ) ;dp[u] = max ( dp[u] , dp[v] + 1 ) ;}
}
/*
void gauss ( int n , int m ) {int r = 0 , c = 0 ;for ( ; r < n && c < m ; ++ c ) {int tr = r ;for ( int i = r ; i < n ; ++ i ) {if ( a[i][c] > a[tr][c] ) tr = i ;}if ( fabs ( a[tr][c] ) < 1e-8 ) continue ;for ( int i = c ; i <= C ; ++ i ) {swap ( a[r][i] , a[tr][i] ) ;}for ( int i = C ; i >= c ; -- i ) a[r][i] /= a[r][c] ;for ( int i = 0 ; i < n ; ++ i ) if ( i != r && fabs ( a[i][c] ) > 1e-8 ) {for ( int j = C ; j >= c ; -- j ) {a[i][j] -= a[r][j] * a[i][c] ;}}++ r ;}for ( int i = r - 1 ; i ; -- i ) {for ( int j = }
}
*/
void solve () {scanf ( "%d%d" , &C , &K ) ;for ( int i = 1 ; i <= C ; ++ i ) {scanf ( "%d" , &n ) ;for ( int j = 1 ; j <= n ; ++ j ) {G[j].clear () ;dp[j] = dp2[j] = 0  ;}for ( int j = 2 , x ; j <= n ; ++ j ) {scanf ( "%d" , &x ) ;G[j].push_back ( x ) ;G[x].push_back ( j ) ;}dfs ( 1 , 1 ) ;dfs2 ( 1 , 1 ) ;int cnt = 0 ;for ( int j = 1 ; j <= n ; ++ j ) {if ( max ( dp2[j] , dp[j] ) <= K ) ++ cnt ;}p[i] = 1.0 * cnt / n ; //a[i][i] = 1 ;//a[i][i + 1] = -p[i] ;//a[i][0] = ( 1 - p[i] ) ;//a[i][C] = 4 - 3 * p[i] ;}double pp = 1 ;f[1] = -3 ;for ( int i = C ; i >= 1 ; -- i ) {pp *= p[i] ;if ( i > 1 ) f[1] += 1.0 / pp ;else f[1] += 4.0 / pp ;}printf ( "%.4f\n" , f[1] ) ;//a[C][C] = 1 ;//gauss ( C + 1 , C + 1 ) ;
}           int main () {int T ;scanf ( "%d" , &T ) ;for ( int i = 1 ; i <= T ; ++ i ) {solve () ;}return 0 ;
}

L. Chance

预处理出$f[i]$表示$[1,i]$中二进制下1的个数为素数的个数，然后$O(1)$回答即可。

#include<bits/stdc++.h>
using namespace std;
const int N=100010;
int f[N],vis[N];
int main(){vis[1]=1;for(int i=2;i<N;i++)if(!vis[i])for(int j=i+i;j<N;j+=i)vis[j]=1;int _;scanf("%d",&_);for(int i=1;i<=100000;i++){f[i]=f[i-1];if(!vis[__builtin_popcount(i)])f[i]++;}while(_--){int l,r;scanf("%d%d",&l,&r);if(!l)l=1;printf("%d\n",f[r]-f[l-1]);}
}

M. ACPC Headquarters : AASTMT (Stairway to Heaven)

设$f[i][j]$表示志愿者$i$在第$j$天需要服务几个赛事，然后检查是否存在$j$满足$f[i][j]>1$即可。

#include<iostream>
#include<string>
#include<map>
using namespace std;
int Case,n;
string s;
map<string,int>T;
int m,f[10010][370];
inline int getid(){cin>>s;int&t=T[s];if(t)return t;return t=++m;
}
inline void make(int x,int l,int r){for(int i=l;i<=r;i++)f[x][i]++;
}
inline bool check(int x){for(int i=1;i<=365;i++)if(f[x][i]>1)return 1;return 0;
}
int main(){cin>>Case;while(Case--){cin>>n;T.clear();int ans=0;while(n--){cin>>s;int l,r,k;cin>>l>>r>>k;while(k--){int x=getid();make(x,l,r);}}for(map<string,int>::iterator it=T.begin();it!=T.end();it++){if(check(it->second))ans++;}cout<<ans<<endl;for(map<string,int>::iterator it=T.begin();it!=T.end();it++){if(check(it->second))cout<<it->first<<endl;}for(int i=1;i<=m;i++)for(int j=1;j<=365;j++)f[i][j]=0;m=0;}return 0;
}

总结：

D题_ilovelife对Claris提供的做法理解上出现了偏差，导致写了一个小时，还交了3发才通过。间接导致I题没有写完，以致没有AK。发生这种情况应考虑换人写。
E题poursoul在Claris写代码时进行误导，致使原来正确的公式被改成错误的公式，无故增加一发罚时，下次要注意。

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