Codeforces Round #673 (Div. 1)
A
模拟一下？
#include<iostream> #include<cstring> #include<cstdio> #include<cstring> #include<vector> #include<queue> #include<algorithm> #define pii pair<int,int> #define pb push_back #define mp make_pair #define fi first #define se second using namespace std; inline int read(){ int f=1,ans=0;char c=getchar(); while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();} while(c>='0'&&c<='9'){ans=ans*10+c-'0';c=getchar();} return f*ans; } const int MAXN=3e5+11; int T,N,Ans[MAXN];vector<int> vec[MAXN]; int main(){ //freopen("1.in","r",stdin); T=read(); while(T--){ memset(Ans,127/3,sizeof(Ans)); N=read();for(int i=1;i<=N;i++) vec[i].clear(),vec[i].pb(0);for(int i=1;i<=N;i++) vec[read()].pb(i); for(int i=1;i<=N;i++) vec[i].pb(N+1); for(int i=1;i<=N;i++){ int Maxn=0; for(int j=1;j<vec[i].size();j++) Maxn=max(Maxn,vec[i][j]-vec[i][j-1]); Ans[Maxn]=min(Ans[Maxn],i); } for(int i=1;i<=N;i++) Ans[i]=min(Ans[i],Ans[i-1]); for(int i=1;i<=N;i++){ if(Ans[i]>N) printf("-1 "); else printf("%d ",Ans[i]); }printf("\n"); }return 0; }
B
由于 $1$ 的特性，先通过 $2n$ 次操作将所以数移到 $1$，再通过 $2n$ 次操作，移回去即可。
#include<iostream> #include<cstring> #include<cstdio> #include<cstring> #include<vector> #include<queue> #include<algorithm> #include<climits> #include<bitset> #define pii pair<int,int> #define pb push_back #define mp make_pair #define fi first #define se second using namespace std; inline int read(){ int f=1,ans=0;char c=getchar(); while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();} while(c>='0'&&c<='9'){ans=ans*10+c-'0';c=getchar();} return f*ans; } const int MAXN=1e4+11; int T,N,A[MAXN],sum,tot; pair<pii,int> Ans[MAXN<<2]; int main(){ //freopen("B.in","r",stdin); T=read(); while(T--){ N=read(),sum=0; for(int i=1;i<=N;i++) A[i]=read(),sum+=A[i];tot=0; if((sum%N)){printf("-1\n");continue;}sum/=N; for(int i=2;i<=N;i++){ int x=(i-(A[i]%i))%i; ++tot;Ans[tot].fi.fi=1,Ans[tot].fi.se=i,Ans[tot].se=x; ++tot;Ans[tot].fi.fi=i,Ans[tot].fi.se=1,Ans[tot].se=(x+A[i])/i; } for(int i=2;i<=N;i++){ ++tot;Ans[tot].fi.fi=1,Ans[tot].fi.se=i,Ans[tot].se=sum; }printf("%d\n",tot); for(int i=1;i<=tot;i++) printf("%d %d %d\n",Ans[i].fi.fi,Ans[i].fi.se,Ans[i].se); }return 0; }
C
按位枚举。
#include<iostream> #include<cstring> #include<cstdio> #include<cstring> #include<vector> #include<queue> #include<algorithm> #include<climits> #include<bitset> #define int long long #define pii pair<int,int> #define pb push_back #define mp make_pair #define fi first #define se second using namespace std; inline int read(){ int f=1,ans=0;char c=getchar(); while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();} while(c>='0'&&c<='9'){ans=ans*10+c-'0';c=getchar();} return f*ans; } const int MAXN=3e5+11; int N,A[MAXN],cur,now;pii B[MAXN]; vector<int> vec[MAXN]; bool cmp(int a,int b){return a>b;} signed main(){ // /freopen("C.in","r",stdin); N=read();for(int i=1;i<=N;i++) A[i]=B[i].fi=read(),B[i].se=i;sort(B+1,B+N+1); for(int i=31;i>=1;i--){ for(int j=1;j<=N;j++) vec[j].clear(); int cnt=0;vec[++cnt].pb(B[1].se); for(int j=2;j<=N;j++){ int pre=(B[j-1].fi>>i),now=(B[j].fi>>i); if(now==pre) vec[cnt].pb(B[j].se); else{++cnt;vec[cnt].pb(B[j].se);} } int S0=0,S1=0,Sum=0; for(int j=1;j<=cnt;j++){ sort(vec[j].begin(),vec[j].end(),cmp); int t0=0,t1=0,siz=vec[j].size(); Sum+=(siz*(siz-1))/2; for(auto p:vec[j]){ bool ww=(A[p]>>(i-1))&1; if(ww) S0+=t0,t1++; else t0++,S1+=t1; } } if(S0<=S1) now+=S0; else cur|=(1ll<<(i-1)),now+=S1; //cerr<<"cnt:"<<cnt<<" now:"<<now<<" cur:"<<cur<<" i:"<<i<<" "<<S0<<" "<<S1<<endl; }printf("%lld %lld\n",now,cur);return 0; }
D
由于删除不太好做考虑变为增加，而由于操作有后效性考虑建 $kruskal$ 重构树，保证当时间为 $t$ 时当前子树里的点联通。线段树维护。
#include<iostream> #include<cstring> #include<cstdio> #include<cstring> #include<vector> #include<queue> #include<algorithm> #include<climits> #include<bitset> #define pii pair<int,int> #define pb push_back #define mp make_pair #define fi first #define se second using namespace std; inline int read(){ int f=1,ans=0;char c=getchar(); while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();} while(c>='0'&&c<='9'){ans=ans*10+c-'0';c=getchar();} return f*ans; } const int MAXN=4e5+11; int N,M,T[MAXN];vector<int> vec[MAXN]; struct Segment{ int Maxn[MAXN<<2]; void Modify(int k,int l,int r,int ps,int w){ if(l==r){Maxn[k]=w;return;} int mid=(l+r)>>1;if(ps<=mid) Modify(k<<1,l,mid,ps,w);else Modify(k<<1|1,mid+1,r,ps,w); Maxn[k]=max(Maxn[k<<1],Maxn[k<<1|1]);return; } int Query(int k,int l,int r,int x,int y){ if(x<=l&&r<=y) return Maxn[k];int mid=(l+r)>>1,res=0; if(x<=mid) res=max(res,Query(k<<1,l,mid,x,y)); if(mid<y) res=max(res,Query(k<<1|1,mid+1,r,x,y)); return res; } }S; struct Edge{int u,v,tim;}E[MAXN]; bool cmp(Edge x1,Edge x2){return x1.tim>x2.tim;} struct Union{ int f[MAXN];void init(){for(int i=1;i<=2*N;i++) f[i]=i;return;} int find(int x){return f[x]==x?x:f[x]=find(f[x]);} }U;pii Que[MAXN]; int dfn[MAXN],siz[MAXN],INF=INT_MAX,MM[MAXN],Val[MAXN],dep[MAXN],tot,fa[MAXN][21],cnt,Q; void dfs(int u,int fath){ dep[u]=dep[fath]+1;fa[u][0]=fath;if(u<=N) siz[u]=1,dfn[u]=++dfn[0]; for(int i=1;(1<<i)<=dep[u];i++) fa[u][i]=fa[fa[u][i-1]][i-1]; for(auto v:vec[u]) if(v!=fath) dfs(v,u),siz[u]+=siz[v],dfn[u]=min(dfn[u],dfn[v]);return; } int Qmax(int u,int t){ for(int i=20;i>=0;i--) if(T[fa[u][i]]>=t&&fa[u][i]) u=fa[u][i]; return u; } int main(){ //freopen("D.in","r",stdin); N=read(),M=read(),Q=read();U.init();tot=N; for(int i=1;i<=N;i++) Val[i]=read(),MM[Val[i]]=i; for(int i=1;i<=M;i++) E[i].u=read(),E[i].v=read(),E[i].tim=INF;int ps=0; for(int i=1;i<=Q;i++){int opt=read(),u=read();if(opt==1) Que[++cnt].fi=u,Que[cnt].se=ps;else E[u].tim=ps=i;}sort(E+1,E+M+1,cmp); for(int i=1;i<=M;i++){ int u=E[i].u,v=E[i].v,tim=E[i].tim; if(U.find(u)==U.find(v)) continue; u=U.find(u),v=U.find(v); ++tot;U.f[u]=U.f[v]=tot;T[tot]=tim; vec[tot].pb(u),vec[tot].pb(v); }memset(dfn,127/3,sizeof(dfn));dfn[0]=0; for(int i=1;i<=tot;i++) if(U.find(i)==i) dfs(i,0); for(int i=1;i<=N;i++) S.Modify(1,1,N,dfn[i],Val[i]); for(int i=1;i<=cnt;i++){ int u=Que[i].fi,tim=Que[i].se+1; int G=Qmax(u,tim),L=dfn[G],R=dfn[G]+siz[G]-1,WW; printf("%d\n",WW=S.Query(1,1,N,L,R));if(!WW) continue; S.Modify(1,1,N,dfn[MM[WW]],0); }return 0; }
