Codeforces Round #354 (Div. 2)

A Nicholas and Permutation

题意：给你一个大小为n的数组,保证数组里的数是1-n.可以任意交换一次位置,求1的位置和n的位置的最大差.

/*************************************************************************     File Name: cf_A.cpp     ID: obsoles1     PROG:      LANG: C++      Mail: [e-mail]      Created Time: 三  5/25 23:08:09 2016 ************************************************************************/#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<map>#include<queue>#include<stack>#include<cmath>#include<cctype>#include<ctime>#include<cstdlib>#include<string>#include<vector>#include<set>#include<bitset>#define Max(x,y) ((x)>(y)?(x):(y))#define Min(x,y) ((x)<(y)?(x):(y))#define each(it,v) for(__typeof((v).begin()) it=(v).begin();it!=(v).end();++it)#define Abs(x,y) ((x)>(y)?((x)-(y)):((y)-(x)))#define ll long long#define Mem0(x) memset(x,0,sizeof(x))#define Mem1(x) memset(x,-1,sizeof(x))#define MemX(x) memset(x,0x3f,sizeof(x))#define pb push_backusing namespace std;const int N=110;int pos[N];int main(){  int n,x,i;  scanf("%d",&n);  for(i=1;i<=n;++i){    scanf("%d",&x);    pos[x]=i;  }  int ans;  if(pos[1]==1 || pos[1]==n || pos[n]==1 || pos[n]==n)ans=n-1;  else ans=Max(Max(Abs(pos[1],n),Abs(pos[1],1)),Max(Abs(pos[n],n),Abs(pos[n],1)));  //if(ans<0)ans=-ans;  //ans++;  //cout<<"ans="<<ans<<endl;  //if((pos[1]!=1 && pos[1]!=n) || (pos[n]!=1 && pos[n]!=n))ans++;  printf("%d\n",ans);}

B Pyramid of Glasses

题意：宴会上金字塔酒杯,给出n层数,t时间.第一秒能灌满顶层的杯子,之后每过1s,溢出的部分往下流.问t时间后灌满的杯子有几个.

/*************************************************************************     File Name: cf_B.cpp     ID: obsoles1     PROG:      LANG: C++      Mail: [e-mail]      Created Time: 三  5/25 23:52:58 2016 ************************************************************************/#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<map>#include<queue>#include<stack>#include<cmath>#include<cctype>#include<ctime>#include<cstdlib>#include<string>#include<vector>#include<set>#include<bitset>#define Max(x,y) ((x)>(y)?(x):(y))#define Min(x,y) ((x)<(y)?(x):(y))#define each(it,v) for(__typeof((v).begin()) it=(v).begin();it!=(v).end();++it)#define Abs(x,y) ((x)>(y)?((x)-(y)):((y)-(x)))#define ll long long#define Mem0(x) memset(x,0,sizeof(x))#define Mem1(x) memset(x,-1,sizeof(x))#define MemX(x) memset(x,0x3f,sizeof(x))#define pb push_backusing namespace std;const int N=15;int bot[N][N];bool f[N][N];int main(){  int n,t,i,j,ans;  scanf("%d%d",&n,&t);  ans=0;  bot[1][1]=2048*t;  for(i=1;i<=n;++i){    for(j=1;j<=i;++j){      //cout<<"t="<<k<<endl;      if(bot[i][j]>=2048){        bot[i+1][j]+=(bot[i][j]-2048)/2;        bot[i+1][j+1]+=(bot[i][j]-2048)/2;        ans++;        //cout<<"bot["<<i+1<<"]["<<j<<"]="<<bot[i+1][j]<<endl;        //cout<<"bot["<<i+1<<"]["<<j+1<<"]="<<bot[i+1][j+1]<<endl;      }    }  }  printf("%d\n",ans);}

C Vasya and String

题意：给一个只含有a,b的字符串,给你为n长度的字符串,可以翻转m个位置(即a->b,b->a),问最长相同的字母的连续子串长度.

/*************************************************************************     File Name: cf_C.cpp     ID: obsoles1     PROG:      LANG: C++      Mail: [e-mail]      Created Time: 三  5/25 23:32:36 2016 ************************************************************************/#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<map>#include<queue>#include<stack>#include<cmath>#include<cctype>#include<ctime>#include<cstdlib>#include<string>#include<vector>#include<set>#include<bitset>#define Max(x,y) ((x)>(y)?(x):(y))#define Min(x,y) ((x)<(y)?(x):(y))#define each(it,v) for(__typeof((v).begin()) it=(v).begin();it!=(v).end();++it)#define Abs(x,y) ((x)>(y)?((x)-(y)):((y)-(x)))#define ll long long#define Mem0(x) memset(x,0,sizeof(x))#define Mem1(x) memset(x,-1,sizeof(x))#define MemX(x) memset(x,0x3f,sizeof(x))#define pb push_backusing namespace std;const int N=1e5+10;char s[N];int pre_a[N],pre_b[N];int main(){  int n,i,k;  scanf("%d%d%s",&n,&k,s);  pre_a[0]=pre_b[0]=0;  for(i=1;i<=n;++i){    if(!(s[i-1]-'a')){      pre_a[i]=pre_a[i-1]+1;      pre_b[i]=pre_b[i-1];    }else{      pre_a[i]=pre_a[i-1];      pre_b[i]=pre_b[i-1]+1;    }  }  int low,high,mid,ans=0;  for(i=0;i<=n;++i){    low=i,high=n;    while(low<=high){      mid=(low+high)>>1;      if(pre_a[mid]-pre_a[i]>k && pre_b[mid]-pre_b[i]>k)high=mid-1;      else low=mid+1;    }    ans=Max(ans,high-i);  }  printf("%d\n",ans);}

D Theseus and labyrinth

题意：给出一个n*m的地图,每个格子代表一个房间,规定房间上哪几个方向有门,之后要想走到相邻房间,该房间必须也要有通向当前房间的门,也可以将地图上各个房间顺时针旋转90度。

/*************************************************************************     File Name: cf_D.cpp     ID: obsoles1     PROG:      LANG: C++      Mail: [e-mail]      Created Time: 二  5/31 21:56:49 2016 ************************************************************************/#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<map>#include<queue>#include<stack>#include<cmath>#include<cctype>#include<ctime>#include<cstdlib>#include<string>#include<vector>#include<set>#include<bitset>#define Max(x,y) ((x)>(y)?(x):(y))#define Min(x,y) ((x)<(y)?(x):(y))#define each(it,v) for(__typeof((v).begin()) it=(v).begin();it!=(v).end();++it)#define Abs(x,y) ((x)>(y)?((x)-(y)):((y)-(x)))#define ll long long#define Mem0(x) memset(x,0,sizeof(x))#define Mem1(x) memset(x,-1,sizeof(x))#define MemX(x) memset(x,0x3f,sizeof(x))#define pb push_backusing namespace std;const int N=1010;char mp[4][N][N];struct node{  int step,x,y,k;}s,t;queue<node> q;bool vis[4][N][N];int n,m,sx,sy,ex,ey,mx[]={0,0,1,-1},my[]={1,-1,0,0};char change(char x){  if(x=='-')return '|';  if(x=='|')return '-';  if(x=='^')return '>';  if(x=='>')return 'v';  if(x=='v')return '<';  if(x=='<')return '^';  if(x=='L')return 'U';  if(x=='U')return 'R';  if(x=='R')return 'D';  if(x=='D')return 'L';  return x;}bool left(char x){  if(x=='-' || x=='>' || x=='L' || x=='U' || x=='D' || x=='+')return 1;  return 0;}bool right(char x){  if(x=='-' || x=='<' || x=='R' || x=='U' || x=='D' || x=='+')return 1;  return 0;}bool up(char x){  if(x=='|' || x=='v' || x=='L' || x=='R' || x=='U' || x=='+')return 1;  return 0;}bool down(char x){  if(x=='|' || x=='^' || x=='L' || x=='R' || x=='D' || x=='+')return 1;  return 0;}bool deal(node s,node t){  char sw=mp[s.k][s.x][s.y],tw=mp[t.k][t.x][t.y];  //cout<<"sw="<<sw<<" tw="<<tw<<endl;  if(sw=='+'){    if(s.x-1==t.x)return up(tw);    if(s.x+1==t.x)return down(tw);    if(s.y-1==t.y)return left(tw);    if(s.y+1==t.y)return right(tw);  }else if(sw=='-'){    if(s.y-1==t.y)return left(tw);    if(s.y+1==t.y)return right(tw);  }else if(sw=='|'){    if(s.x-1==t.x)return up(tw);    if(s.x+1==t.x)return down(tw);  }else if(sw=='^' && s.x-1==t.x)    return up(tw);  else if(sw=='v' && s.x+1==t.x)    return down(tw);  else if(sw=='>' && s.y+1==t.y)    return right(tw);  else if(sw=='<' && s.y-1==t.y)    return left(tw);  else if(sw=='L'){    if(s.x-1==t.x)return up(tw);    if(s.x+1==t.x)return down(tw);    if(s.y+1==t.y)return right(tw);  }else if(sw=='R'){    if(s.x-1==t.x)return up(tw);    if(s.x+1==t.x)return down(tw);    if(s.y-1==t.y)return left(tw);  }else if(sw=='U'){    if(s.x+1==t.x)return down(tw);    if(s.y-1==t.y)return left(tw);    if(s.y+1==t.y)return right(tw);  }else if(sw=='D'){    if(s.x-1==t.x)return up(tw);    if(s.y-1==t.y)return left(tw);    if(s.y+1==t.y)return right(tw);  }  return 0;}void bfs(){  Mem0(vis);  while(!q.empty())q.pop();  s.x=sx,s.y=sy,s.step=0,s.k=0;  vis[0][sx][sy]=1;  q.push(s);  while(!q.empty()){    s=q.front();    //cout<<"s("<<s.x<<','<<s.y<<")   step="<<s.step<<"   k="<<s.k<<endl;    q.pop();    if(s.x==ex && s.y==ey){      printf("%d\n",s.step);      return;    }    for(int i=0;i<4;++i){      t=s;      t.x+=mx[i],t.y+=my[i];      t.step++;      if(t.x<0 || t.x>=n || t.y<0 || t.y>=m || vis[t.k][t.x][t.y] || mp[t.k][t.x][t.y]=='*')continue;      if(deal(s,t)){        q.push(t);        vis[t.k][t.x][t.y]=1;      }    }    t=s;    t.k=(t.k+1)%4;    t.step++;    if(vis[t.k][t.x][t.y])continue;    vis[t.k][t.x][t.y]=1;    q.push(t);  }  puts("-1");}int main(){  int i,j,k;  scanf("%d%d",&n,&m);  for(i=0;i<n;++i){    scanf("%s",mp[0][i]);    for(k=1;k<4;++k){      for(j=0;j<m;++j)        mp[k][i][j]=change(mp[k-1][i][j]);    }  }  scanf("%d%d%d%d",&sx,&sy,&ex,&ey);  sx--,sy--,ex--,ey--;  bfs();}

E The Last Fight Between Human and AI