Codeforces Round 501 (Div. 3)
Codeforces Round 501 (Div. 3)
思路：记录每个区间
#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=3e4+5,INF=0x3f3f3f3f,Mod=1e9+7; const double eps=1e-6; void solve(){ int n,m;cin>>n>>m; vector<int>a(m+1,0); for(int i=0,l,r;i<n;++i){ cin>>l>>r; for(int j=l;j<=r;++j)a[j]=1; } int ans=0; for(int i=1;i<=m;++i){ ans+=!a[i]; } cout<<ans<<'\n'; for(int i=1;i<=m;++i){ if(!a[i])cout<<i<<' '; } 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; }
思路：枚举每个字符，使前i个字符都相等
#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=3e4+5,INF=0x3f3f3f3f,Mod=1e9+7; const double eps=1e-6; void solve(){ int n;cin>>n; string a,b;cin>>a>>b; vector<int>aa(26,0),bb(26,0); for(int i=0;i<n;++i){ aa[a[i]-'a']++; bb[b[i]-'a']++; } for(int i=0;i<26;++i){ if(aa[i]!=bb[i]){ cout<<-1;return ; } } vector<int>ans; for(int i=0;i<n;++i){ if(b[i]!=a[i]){ int p=0; for(int j=i+1;j<n;++j){ if(a[j]==b[i]){ p=j;break; } } for(int j=p;j>i;--j){ ans.push_back(j); swap(a[j],a[j-1]); } } } cout<<ans.size()<<'\n'; for(auto v:ans)cout<<v<<' '; 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; }
思路：需要改越少的数，那么把所有差值排序，从差值大的开始改
#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=3e4+5,INF=0x3f3f3f3f,Mod=1e9+7; const double eps=1e-6; void solve(){ int n,m;cin>>n>>m; vector<int>d(n); int all=0; for(int i=0,x,y;i<n;++i){ cin>>x>>y; all+=x; d[i]=x-y; } sort(d.begin(),d.end(),greater<int>()); if(all<=m)cout<<0; else{ all-=m; int ans=-1; for(int i=0;i<d.size();++i){ all-=d[i]; if(all<=0){ ans=i+1;break; } } 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; }
思路：每次最多加n-1，判断下是否能移动k次，以及s不能超过最大增值（n-1）*k；
判断满足条件后，一定是存在答案的，每次加的数就是min（n-1，s-k）（满足的条件：最多加n-1、剩下的需要加的数至少够加满k次）
#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=3e4+5,INF=0x3f3f3f3f,Mod=1e9+7; const double eps=1e-6; void solve(){ int n,k,s;cin>>n>>k>>s; int c=n-1; if(s>c*k||k>s){ cout<<"NO";return ; } cout<<"YES\n"; int now=1; while(k--){ int d=min(s-k,n-1); s-=d; if(now+d<=n)now+=d; else now-=d; cout<<now<<' '; } } signed main(){ ios::sync_with_stdio(0),cin.tie(0),cout.tie(0); int T=1; //init(); //cin>>T; while(T--){ solve(); } return 0; }
E1 - Stars Drawing (Easy Edition)
思路同下题
#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=1e3+5,INF=0x3f3f3f3f,Mod=1e9+7; const double eps=1e-6; int l[N][N],r[N][N],u[N][N],d[N][N]; char g[N][N]; int n,m; void P(int x,int y,int s){ int ll=max(1ll,y-s+1),rr=min(m,y+s-1),uu=max(1ll,x-s+1),dd=min(n,x+s-1); for(int i=ll,j=uu;i<=rr||j<=dd;++i,++j){ if(i>=ll&&i<=rr)g[x][i]='.'; if(j>=uu&&j<=dd)g[j][y]='.'; } } void solve(){ cin>>n>>m; for(int i=1;i<=n;++i) { for(int j=1;j<=m;++j){ cin>>g[i][j]; if(g[i][j]=='*'){ l[i][j]=r[i][j]=u[i][j]=d[i][j]=1; if(g[i][j-1]=='*')l[i][j]+=l[i][j-1]; if(g[i-1][j]=='*')u[i][j]+=u[i-1][j]; } } } for(int i=n;i>=1;--i){ for(int j=m;j>=1;--j){ if(g[i][j]=='*'){ if(g[i][j+1]=='*')r[i][j]+=r[i][j+1]; if(g[i+1][j]=='*')d[i][j]+=d[i+1][j]; } } } vector<pair<PII,int>>ans; for(int i=1;i<=n;++i){ for(int j=1;j<=m;++j){ int s=min(min(l[i][j],r[i][j]),min(u[i][j],d[i][j])); if(s>1){ ans.push_back({{i,j},s-1}); P(i,j,s); } } } for(int i=1;i<=n;++i){ for(int j=1;j<=m;++j){ if(g[i][j]=='*'){ cout<<-1; return ; } } } cout<<ans.size()<<'\n'; for(auto v:ans){ cout<<v.first.first<<' '<<v.first.second<<' '<<v.second<<'\n'; } }; signed main(){ ios::sync_with_stdio(0),cin.tie(0),cout.tie(0); int T=1; //init(); //cin>>T; while(T--){ solve(); } return 0; }
E2 - Stars Drawing (Hard Edition)
思路：维护每个点（i，j）的上下左右连续的‘*’个数；枚举每个点（i，j），若四个方向的连续‘*’数大于1，说明能构成星星
#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=1e3+5,INF=0x3f3f3f3f,Mod=1e9+7; const double eps=1e-6; int l[N][N],r[N][N],u[N][N],d[N][N]; char g[N][N]; int n,m; void P(int x,int y,int s){ int ll=max(1ll,y-s+1),rr=min(m,y+s-1),uu=max(1ll,x-s+1),dd=min(n,x+s-1); for(int i=ll,j=uu;i<=rr||j<=dd;++i,++j){ if(i>=ll&&i<=rr)g[x][i]='.'; if(j>=uu&&j<=dd)g[j][y]='.'; } } void solve(){ cin>>n>>m; for(int i=1;i<=n;++i) { for(int j=1;j<=m;++j){ cin>>g[i][j]; if(g[i][j]=='*'){ l[i][j]=r[i][j]=u[i][j]=d[i][j]=1; if(g[i][j-1]=='*')l[i][j]+=l[i][j-1]; if(g[i-1][j]=='*')u[i][j]+=u[i-1][j]; } } } for(int i=n;i>=1;--i){ for(int j=m;j>=1;--j){ if(g[i][j]=='*'){ if(g[i][j+1]=='*')r[i][j]+=r[i][j+1]; if(g[i+1][j]=='*')d[i][j]+=d[i+1][j]; } } } vector<pair<PII,int>>ans; for(int i=1;i<=n;++i){ for(int j=1;j<=m;++j){ int s=min(min(l[i][j],r[i][j]),min(u[i][j],d[i][j])); if(s>1){ ans.push_back({{i,j},s-1}); P(i,j,s); } } } for(int i=1;i<=n;++i){ for(int j=1;j<=m;++j){ if(g[i][j]=='*'){ cout<<-1; return ; } } } cout<<ans.size()<<'\n'; for(auto v:ans){ cout<<v.first.first<<' '<<v.first.second<<' '<<v.second<<'\n'; } }; signed main(){ ios::sync_with_stdio(0),cin.tie(0),cout.tie(0); int T=1; //init(); //cin>>T; while(T--){ solve(); } return 0; }
