AtCoder Beginner Contest 311
AtCoder Beginner Contest 311
First ABC
思路：找到第一个a，b，c都出现的位置
#include<bits/stdc++.h> using namespace std; #define int long long //#define int __int128 typedef pair<int,int>PII; typedef pair<string,int>PSI; typedef pair<string,string>PSS; const int N=1e6+5,INF=0x3f3f3f3f,Mod=1e9+7; const double eps=1e-6; void solve(){ int n;cin>>n; string s; cin>>s; int a=0,b=0,c=0; for(int i=0;i<n;++i){ if(s[i]=='A')a=1; if(s[i]=='B')b=1; if(s[i]=='C')c=1; if(a&&b&&c){ cout<<i+1; return ; } } } signed main(){ ios::sync_with_stdio(0),cin.tie(0),cout.tie(0); int t=1; //init(); //cin>>t; while(t--){ solve(); } return 0; }
Vacation Together
思路：找到最大的连续'o'的列长度
#include<bits/stdc++.h> using namespace std; #define int long long //#define int __int128 typedef pair<int,int>PII; typedef pair<string,int>PSI; typedef pair<string,string>PSS; const int N=1e6+5,INF=0x3f3f3f3f,Mod=1e9+7; const double eps=1e-6; void solve(){ int n,d,ans=0;cin>>n>>d; vector<string>ve(n); for(int i=0;i<n;++i)cin>>ve[i]; char pre='o'; for(int i=0,c=0;i<d;++i){ bool ok=true; for(int j=1;j<n;++j)if(ve[j][i]!=ve[j-1][i]){ ok=false;break; } if(ok&&ve[0][i]==pre)c++; else c=0; ans=max(ans,c); } cout<<ans; } signed main(){ ios::sync_with_stdio(0),cin.tie(0),cout.tie(0); int t=1; //init(); //cin>>t; while(t--){ solve(); } return 0; }
Find it!
思路：每个点都指向一点，那么一定有环，从一点开始搜，搜到搜过的点即找到环
#include<bits/stdc++.h> using namespace std; #define int long long //#define int __int128 typedef pair<int,int>PII; typedef pair<string,int>PSI; typedef pair<string,string>PSS; const int N=2e5+5,INF=0x3f3f3f3f,Mod=1e9+7; const double eps=1e-6; int n; vector<int>ve(N); vector<int>st(N); int dfs(int u){ if(st[u]){ return u; } st[u]=1; return dfs(ve[u]); } void solve(){ cin>>n; for(int i=1;i<=n;++i){ cin>>ve[i]; } st[1]=1; int res=dfs(ve[1]); vector<int>ans; int p=ve[res]; ans.push_back(res); while(p!=res){ ans.push_back(p); p=ve[p]; } cout<<ans.size()<<'\n'; for(auto v:ans)cout<<v<<' '; } signed main(){ //ios::sync_with_stdio(0),cin.tie(0),cout.tie(0); int t=1; //init(); //cin>>t; while(t--){ solve(); } return 0; }
Grid Ice Floor
思路：从（1，1）开始bfs，一个方向要搜到底，标记搜过的点，最后统计
#include<bits/stdc++.h> using namespace std; #define int long long //#define int __int128 typedef pair<int,int>PII; typedef pair<string,int>PSI; typedef pair<string,string>PSS; const int N=2e5+5,INF=0x3f3f3f3f,Mod=1e9+7; const double eps=1e-6; int n,m,ans=0; vector<string>s(205); int st[205][205],vis[205][205]; int dx[4]={-1,0,1,0}; int dy[4]={0,1,0,-1}; void bfs(int x,int y){ queue<PII>q; q.push({x,y}); st[1][1]=vis[1][1]=1; while(q.size()){ auto [u,v]=q.front();q.pop(); for(int i=0;i<4;++i){ int xx=u,yy=v; while(s[xx+dx[i]][yy+dy[i]]!='#'){ xx+=dx[i],yy+=dy[i]; vis[xx][yy]=1; } if(!st[xx][yy]){ q.push({xx,yy}); st[xx][yy]=1; } } } for(int i=0;i<n;++i) for(int j=0;j<m;++j)ans+=vis[i][j]; } void solve(){ cin>>n>>m; for(int i=0;i<n;++i){ cin>>s[i]; } bfs(1,1); cout<<ans; } signed main(){ ios::sync_with_stdio(0),cin.tie(0),cout.tie(0); int t=1; //init(); //cin>>t; while(t--){ solve(); } return 0; }
Defect-free Squares
思路：把洞的地方标为1，用前缀和判断是否为正方形。枚举每个点，二分找到以该点为右下角的最大正方形
#include<bits/stdc++.h> using namespace std; #define int long long //#define int __int128 typedef pair<int,int>PII; typedef pair<string,int>PSI; typedef pair<string,string>PSS; const int N=3e3+5,INF=0x3f3f3f3f,Mod=1e9+7,mod=998244353; const int inf=0x3f3f3f3f3f3f; const double eps=1e-6; const int dx[4]={-1,0,1,0}; const int dy[4]={0,1,0,-1}; int h,w,n; int a[N][N]; void solve(){ cin>>h>>w>>n; for(int i=0,u,v;i<n;++i){ cin>>u>>v; a[u][v]=1; } int res=0; for(int i=1;i<=h;++i) for(int j=1;j<=w;++j){ if(!a[i][j]){ a[i][j]+=(a[i-1][j]+a[i][j-1]-a[i-1][j-1]); int l=0,r=min(i,j)-1,k; while(l<=r){ int mid=l+r>>1; auto check=[&](int x){ int ii=i-x,jj=j-x; int s=a[i][j]-a[i][jj-1]-a[ii-1][j]+a[ii-1][jj-1]; return !s; }; if(check(mid))k=mid,l=mid+1; else r=mid-1; } //cout<<i<<' '<<j<<':'<<k<<'\n'; res+=(k+1); } else a[i][j]+=(a[i-1][j]+a[i][j-1]-a[i-1][j-1]); } cout<<res; } signed main(){ ios::sync_with_stdio(0),cin.tie(0),cout.tie(0); int T=1; //init(); //cin>>T; while(T--){ solve(); } return 0; }
F - Yet Another Grid Task
思路：dp[i][j]表示前i列，第j行以及以下的行都为黑的数量。
dp[i][j]状态转移： 1. s[i][j]自己变为黑：dp[i][j] = dp[i-1][j~m]
2.s[i][j]受s[i-1][j-1]的影响而变黑 dp[i][j] = dp[i-1][j-1]
即dp[i][j] = dp[i-1][ （j-1）~ m]，可直接维护前一列的行的后缀和；
这里dp[0][n+1] = 1；
dp[i][j]只和前一列有关系，可优化为一维
#include<bits/stdc++.h> using namespace std; #define int long long //#define int __int128 typedef pair<int,int>PII; typedef pair<string,int>PSI; typedef pair<string,string>PSS; const int N=3e3+5,INF=0x3f3f3f3f,Mod=1e9+7,mod=998244353; const int inf=0x3f3f3f3f3f3f; const double eps=1e-6; const int dx[4]={-1,0,1,0}; const int dy[4]={0,1,0,-1}; void solve(){ int n,m;cin>>n>>m; vector<string>s(n); for(int i=0;i<n;++i)cin>>s[i]; for(int i=0;i<n;++i){ for(int j=0;j<m;++j){ if(s[i][j]=='#'&&i+1<n){ s[i+1][j]='#'; if(j+1<m)s[i+1][j+1]='#'; } } } vector<int>dp(n+1,0); dp[n]=1; for(int i=0;i<m;++i){ vector<int>f(n+1,0); int pre=0; for(int j=n;j>=0;--j){ pre=(pre+dp[j])%mod; if(j&&s[j-1][i]=='#')continue; f[j]=pre; if(j)f[j]=(f[j]+dp[j-1])%mod; } dp=f; } int ans=0; for(int i=0;i<=n;++i)ans=(ans+dp[i])%mod; cout<<ans; return ; } signed main(){ ios::sync_with_stdio(0),cin.tie(0),cout.tie(0); int T=1; //init(); //cin>>T; while(T--){ solve(); } return 0; }
