2016-2017 CT S03E06: Codeforces Trainings Season 3 Episode 6

比赛连接：

http://codeforces.com/gym/101124/

代码地址

http://git.oschina.net/qscqesze/Acm/tree/master/contest/2016-2017%20CT%20S03E06%20Codeforces%20Trainings%20Season%203%20Episode%206?dir=1&filepath=contest%2F2016-2017+CT+S03E06+Codeforces+Trainings+Season+3+Episode+6&oid=25d818184806af64cbeb5a0e6235eeb6f1252004&sha=bdab61c2245418e3be90bd32b6dccc4df0995d8f

Problem A. The Baguette Master

题意

给你一个四边形，让你给这个四边形加上一个长度为2的框，问你这个框的最大的周长是多少。

题解

框最多就往上移动或者往下移动，那么2^4枚举一下，然后取一个最大值就好了。

不过我队友好像不是这样写的。。。

代码

#include <bits/stdc++.h>using namespace std;const double pi=acos(-1);
const double eps=1e-9;int main()
{double tha,thb,thc,thd,w,ab,bc,cd,da,ac,ans,AB,BC,CD,DA,AC,BD;scanf("%lf%lf%lf%lf%lf%lf",&w,&ab,&bc,&cd,&da,&ac);thb=acos((ab*ab+bc*bc-ac*ac)/(2.0*ab*bc));thd=acos((cd*cd+da*da-ac*ac)/(2.0*cd*da));tha=acos((ab*ab+ac*ac-bc*bc)/(2.0*ab*ac))+acos((da*da+ac*ac-cd*cd)/(2.0*da*ac));thc=acos((bc*bc+ac*ac-ab*ab)/(2.0*bc*ac))+acos((cd*cd+ac*ac-da*da)/(2.0*cd*ac));if(fabs(tha-pi)<eps){BC=bc+(w*cos(thb)/sin(thb)+w*cos(thc)/sin(thc));CD=cd+(w*cos(thc)/sin(thc)+w*cos(thd)/sin(thd));BD=ab+da+(w*cos(thd)/sin(thd)+w*cos(thb)/sin(thb));ans=CD+BD+BC+2.0*(w/sin(thb)+w/sin(thc)+w/sin(thd));}else if(fabs(thb-pi)<eps){CD=cd+(w*cos(thc)/sin(thc)+w*cos(thd)/sin(thd));DA=da+(w*cos(thd)/sin(thd)+w*cos(tha)/sin(tha));AC=ab+bc+(w*cos(tha)/sin(tha)+w*cos(thc)/sin(thc));ans=CD+DA+AC+2.0*(w/sin(tha)+w/sin(thc)+w/sin(thd));}else if(fabs(thc-pi)<eps){AB=ab+(w*cos(tha)/sin(tha)+w*cos(thb)/sin(thb));DA=da+(w*cos(thd)/sin(thd)+w*cos(tha)/sin(tha));BD=bc+cd+(w*cos(thb)/sin(thb)+w*cos(thd)/sin(thd));ans=AB+BD+DA+2.0*(w/sin(tha)+w/sin(thb)+w/sin(thd));}else if(fabs(thd-pi)<eps){AB=ab+(w*cos(tha)/sin(tha)+w*cos(thb)/sin(thb));BC=bc+(w*cos(thb)/sin(thb)+w*cos(thc)/sin(thc));AC=cd+da+(w*cos(thc)/sin(thc)+w*cos(tha)/sin(tha));ans=AB+BC+AC+2.0*(w/sin(tha)+w/sin(thb)+w/sin(thc));}else{AB=ab+(w*cos(tha)/sin(tha)+w*cos(thb)/sin(thb));BC=bc+(w*cos(thb)/sin(thb)+w*cos(thc)/sin(thc));CD=cd+(w*cos(thc)/sin(thc)+w*cos(thd)/sin(thd));DA=da+(w*cos(thd)/sin(thd)+w*cos(tha)/sin(tha));ans=AB+BC+CD+DA+2.0*(w/sin(tha)+w/sin(thb)+w/sin(thc)+w/sin(thd));}//  cout<<tha<<" "<<thb<<" "<<thc<<" "<<thd<<endl;//  cout<<tha+thb+thc+thd<<endl;printf("%.12f\n",ans);return 0;
}

Problem C. Old Chess Sets

题意

给你两个集合，问你这两个集合加起来，能不能构成国际象棋的32个棋子。

题解

签到题

代码

#include<bits/stdc++.h>
using namespace std;int k1,k2;
int mp[2][35];
int num[35];
int getid(string s){if(s=="black")return 0;if(s=="white")return 1;if(s=="king")return 0;if(s=="queen")return 1;if(s=="bishop")return 2;if(s=="knight")return 3;if(s=="rook")return 4;if(s=="pawn")return 5;
}
vector<string>ans1,ans2;
int main()
{num[0]=1,num[1]=1,num[2]=2,num[3]=2,num[4]=2,num[5]=8;memset(mp,0,sizeof(mp));scanf("%d%d",&k1,&k2);for(int i=0;i<k1;i++){string s1,s2;cin>>s1>>s2;mp[getid(s1)][getid(s2)]++;}int flag = 0;for(int i=0;i<k2;i++){string s1,s2;cin>>s1>>s2;if(mp[getid(s1)][getid(s2)]<num[getid(s2)]){mp[getid(s1)][getid(s2)]++;ans1.push_back(s1);ans2.push_back(s2);}}if(ans1.size()+k1==32){for(int i=0;i<ans1.size();i++)cout<<ans1[i]<<" "<<ans2[i]<<endl;}else{cout<<"impossible"<<endl;}
}

Problem D. Checkmate with Bishop and Knight

题意

给你一个国际象棋的局势，问你应该下哪一步，才能将死对面。

题解

直接暴力枚举就好了

代码

#include<bits/stdc++.h>
using namespace std;pair<int,int> pos(string s){return make_pair(s[0]-'a',s[1]-'1');
}
pair<int,int> bishop,knight,wking,bking;
int mp[10][10];
int dkx[8]={1,-1,1,-1,1,-1,0,0},dky[8]={1,-1,-1,1,0,0,1,-1};
int dbx[4]={1,-1,1,-1},dby[4]={1,-1,-1,1};
int dnx[8]={1,-1,1,-1,2,-2,2,-2},dny[8]={2,-2,-2,2,1,-1,-1,1};
bool test(int x,int y)
{for(int i=0;i<8;i++){int nx=x+dkx[i];int ny=y+dky[i];if(nx<0||nx>=8)continue;if(ny<0||ny>=8)continue;if(mp[nx][ny]==1)return true;}for(int i=0;i<4;i++){for(int j=0;j<8;j++){int nx=x+dbx[i]*(j+1);int ny=y+dby[i]*(j+1);if(nx<0||nx>=8)continue;if(ny<0||ny>=8)continue;if(mp[nx][ny]==2)return true;if(mp[nx][ny])break;}}for(int i=0;i<8;i++){int nx=x+dnx[i];int ny=y+dny[i];if(nx<0||nx>=8)continue;if(ny<0||ny>=8)continue;if(mp[nx][ny]==3)return true;}return false;
}
bool check()
{if(!test(bking.first,bking.second))return 0;for(int i=0;i<8;i++){int nx=bking.first+dkx[i];int ny=bking.second+dky[i];if(nx<0||nx>=8)continue;if(ny<0||ny>=8)continue;if(!test(nx,ny))return 0;}return 1;
}
int main()
{string s;cin>>s;wking=pos(s),mp[wking.first][wking.second]=1;cin>>s;bishop=pos(s),mp[bishop.first][bishop.second]=2;cin>>s;knight=pos(s),mp[knight.first][knight.second]=3;cin>>s;bking=pos(s),mp[bking.first][bking.second]=4;for(int i=0;i<8;i++){int nx=knight.first+dnx[i];int ny=knight.second+dny[i];if(nx<0||nx>=8)continue;if(ny<0||ny>=8)continue;if(mp[nx][ny])continue;mp[nx][ny]=3;mp[knight.first][knight.second]=0;knight=make_pair(nx,ny);if(check()){printf("N%c%c\n",'a'+nx,'1'+ny);return 0;}mp[nx][ny]=0;knight=make_pair(nx-dnx[i],ny-dny[i]);mp[knight.first][knight.second]=3;}for(int i=0;i<8;i++){int nx=wking.first+dkx[i];int ny=wking.second+dky[i];if(nx<0||nx>=8)continue;if(ny<0||ny>=8)continue;if(mp[nx][ny])continue;int ff=0;for(int j=0;j<8;j++){int nnx=nx+dkx[j];int nny=ny+dky[j];if(nnx<0||nnx>=8)continue;if(nny<0||nny>=8)continue;if(mp[nnx][nny]==4)ff=1;}if(ff)continue;mp[nx][ny]=1;mp[wking.first][wking.second]=0;wking=make_pair(nx,ny);if(check()){printf("K%c%c\n",'a'+nx,'1'+ny);return 0;}mp[nx][ny]=0;wking=make_pair(nx-dkx[i],ny-dky[i]);mp[wking.first][wking.second]=1;}for(int i=0;i<4;i++){for(int j=0;j<8;j++){int nx=bishop.first+dbx[i]*(j+1);int ny=bishop.second+dby[i]*(j+1);if(nx<0||nx>=8)continue;if(ny<0||ny>=8)continue;if(mp[nx][ny])break;mp[nx][ny]=2;mp[bishop.first][bishop.second]=0;bishop=make_pair(nx,ny);if(check()){printf("B%c%c\n",'a'+nx,'1'+ny);return 0;}mp[nx][ny]=0;bishop=make_pair(nx-dbx[i]*(j+1),ny-dby[i]*(j+1));mp[bishop.first][bishop.second]=2;}}cout<<"impossible"<<endl;
}

Problem H. Kids’ Play

题意

给你一个有向图图，n点n边，点的度数小于等于2，每个点都在环上

然后Q次查询，每次查询从一个点开始，问这个点跑k步之后在哪个点上。

题解

处理出每个环，然后对于每一个点都倍增处理这个点走2^k步之后在哪儿。

然后暴力去搞一搞就好了

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5+7;
vector<int> E[maxn],cir[maxn];
int vis[maxn],p[maxn][2],n,m,cnt,step[maxn][20],be[maxn],two[maxn];
void dfs(int x)
{if(vis[x])return;vis[x]=1;cir[cnt].push_back(x);be[x]=cnt;for(int i=0;i<E[x].size();i++){step[x][0]=E[x][i];dfs(E[x][i]);}
}
int main()
{two[0]=1;for(int i=1;i<20;i++)two[i]=two[i-1]*2;scanf("%d%d",&n,&m);for(int i=1;i<=n;i++){int a,b;scanf("%d%d",&a,&b);p[a][0]=i,p[b][1]=i;step[i][0]=i;}for(int i=1;i<=n;i++)E[p[i][1]].push_back(p[i][0]);for(int i=1;i<=n;i++)if(!vis[i]){cnt++;dfs(i);}for(int j=1;j<20;j++)for(int i=1;i<=n;i++)step[i][j]=step[step[i][j-1]][j-1];for(int i=1;i<=m;i++){int a,b;scanf("%d%d",&a,&b);b%=cir[be[a]].size();int now = a;for(int j=19;j>=0;j--){if(b>=two[j]){now=step[now][j];b-=two[j];}}cout<<now<<endl;}
}

Problem I. Odd Factor (64 MB ML!)

题意

给你一个无向图，让你删除一些边，使得每个点的度数都是奇数

题解

先找一棵树出来，然后自下而上，如果这个点度数为偶数，就删除这个点与根相连的边就好了

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e6+7;
vector<int> E[maxn];
vector<int> id[maxn];
vector<int>ans1,ans2,ans3;
int vis[maxn];
int V[maxn],d[maxn],fa[maxn],p[maxn];
int n,m,flag=0;
void dfs(int x)
{vis[x]=1;for(int i=0;i<E[x].size();i++){if(vis[E[x][i]])continue;d[E[x][i]]++;d[x]++;ans1.push_back(x);ans2.push_back(E[x][i]);ans3.push_back(1);p[E[x][i]]=ans1.size()-1;dfs(E[x][i]);}if(p[x]==-1){if(d[x]%2==0)flag=1;}else{if(d[x]%2==0){d[x]--;d[ans1[p[x]]]--;ans3[p[x]]=0;}}
}
int main()
{memset(p,-1,sizeof(p));scanf("%d%d",&n,&m);for(int i=0;i<m;i++){int a,b;scanf("%d%d",&a,&b);E[a].push_back(b);E[b].push_back(a);id[a].push_back(i);id[b].push_back(i);}for(int i=1;i<=n;i++)if(!vis[i])dfs(i);if(flag){cout<<"-1"<<endl;return 0;}int tmp = 0;for(int i=0;i<ans1.size();i++)if(ans3[i])tmp++;cout<<tmp<<endl;for(int i=0;i<ans1.size();i++)if(ans3[i]==1)cout<<ans1[i]<<" "<<ans2[i]<<endl;return 0;
}

Problem K. Average Speed

题意

给你每个时间段汽车的速度，多次查询汽车在某个时间一共跑了多远

题解

签到题，水题

代码

#include<bits/stdc++.h>
using namespace std;string s;
double gettime(string s)
{double h = (s[0]-'0')*10+s[1]-'0';double m = (s[3]-'0')*10+s[4]-'0';double ss = (s[6]-'0')*10+s[7]-'0';return h+m/60.0+ss/3600.0;
}
int main()
{double now = 0;double lasttime = 0;double lastspeed = 0;while(getline(cin,s)){if(s.size()>8){double time = gettime(s);now+=(time-lasttime)*lastspeed;lastspeed=0;for(int i=9;i<s.size();i++)lastspeed=lastspeed*10+s[i]-'0';lasttime=time;}else{double time = gettime(s);cout<<s<<" ";printf("%.2f km\n",(time-lasttime)*lastspeed+now);}}
}

Problem L. Subway

题意

给你一个二维平面，上面有地铁，地铁的速度和人的速度给你

现在让你从起点到家，问你怎么走最快。

题解

把直接走路和地铁的边都扔进邻接矩阵里面，然后去跑floyd就好了

代码

#include<bits/stdc++.h>
using namespace std;double x1,y1,x2,y2;
double lx,ly,nx,ny;
int cnt=0;
double X[250],Y[250];
double len[250];
double dp[250][250];
double dis(double x,double y,double xx,double yy){return sqrt((x-xx)*(x-xx)+(y-yy)*(y-yy));
}
int main()
{for(int i=0;i<250;i++)for(int j=0;j<250;j++)if(i==j)dp[i][j]=0;else dp[i][j]=1e9;cin>>x1>>y1>>x2>>y2;cnt++;X[0]=x1,Y[0]=y1;while(cin>>nx>>ny){int st=cnt;len[cnt]=0;X[cnt]=nx,Y[cnt]=ny;for(int i=0;i<cnt;i++){dp[cnt][i]=min(dp[cnt][i],dis(nx,ny,X[i],Y[i])/1000.0*6);dp[i][cnt]=min(dp[i][cnt],dis(nx,ny,X[i],Y[i])/1000.0*6);}cnt++;while(cin>>lx>>ly){if(lx==-1&&ly==-1)break;swap(lx,nx);swap(ly,ny);len[cnt]=dis(lx,ly,nx,ny)+len[cnt-1];X[cnt]=nx,Y[cnt]=ny;for(int i=0;i<cnt;i++){dp[cnt][i]=min(dp[cnt][i],dis(nx,ny,X[i],Y[i])/1000.0*6);dp[i][cnt]=min(dp[i][cnt],dis(nx,ny,X[i],Y[i])/1000.0*6);}for(int i=st;i<cnt;i++){dp[cnt][i]=min(dp[cnt][i],(len[cnt]-len[i])/4000.0*6);dp[i][cnt]=min(dp[i][cnt],(len[cnt]-len[i])/4000.0*6);}cnt++;}}X[cnt]=x2,Y[cnt]=y2;for(int i=0;i<=cnt;i++){for(int j=0;j<=cnt;j++){if(i==j)dp[i][j]=0;else dp[i][j]=min(dp[i][j],dis(X[i],Y[i],X[j],Y[j])/1000.0*6);}}for(int k=0;k<=cnt;k++){for(int j=0;j<=cnt;j++){for(int i=0;i<=cnt;i++)dp[i][j]=min(dp[i][j],dp[i][k]+dp[k][j]);}}printf("%.0f\n",dp[0][cnt]);
}

Problem M. A multiplication game

题意

一开始数是1，两人在玩博弈论，该一个人的时候，就让他乘上2~9中的一个数，如果这个数大于等于p就胜利。

题解

直接dfs搜吧，感觉上状态很小的样子

代码

#include<bits/stdc++.h>
using namespace std;
long long n;
map<long long,int>H;
int dfs(long long o){if(H[o])return H[o]-1;if(o>=n)return 1;int flag = 1;for(int i=2;i<=9;i++)if(dfs(o*1ll*i))flag=0;H[o]=flag+1;return flag;
}
int main()
{while(cin>>n){if(n==1){cout<<"Stan wins."<<endl;continue;}H.clear();int o=dfs(1);if(o==0)cout<<"Stan wins."<<endl;else cout<<"Ollie wins."<<endl;}
}