Or virtual good.
Ten days before the beginning of fever, burned for several days, then give me a false burned. Cold, fever, headache, toothache, leg pain.
Exams even explosive, and has a get out of the room.
……
The exam is also quite interesting.
T1 violence will not.
T3 violence will not.
T2 will only cheat Mo team points.
The results T2 just want to lie to 60 A of the program?
Data real water.
I really rubbish.
T1
Violence is a $ \ Theta (N ^ 3) $ sucker violence, however, when the examination did not expect.
Monotone Positive Solutions stack + dp.
Large simulation can also be filled pit A, also fast ......
In fact, quite the water.
#include<cstdio> #include<cmath> #define ll long long using namespace std; int const N=1e6+5; inline int read(){ int ss=0;char bb=getchar(); while(bb<48||bb>57)bb=getchar(); while(bb>=48&&bb<=57)ss=(ss<<1)+(ss<<3)+(bb^48),bb=getchar(); return ss; } int n,top=1; int h[N],stack[N]; ll C,w[N],wf[N],dp[N]; inline int _abs(int x){ return x<0?-x:x; } inline ll max(ll x,ll y){ return x>y?x:y; } inline ll min(ll x,ll y){ return x<y?x:y; } ll find(ll x,ll l,ll r){ ll a=r-l-1,b=-(w[r-1]-w[l]<<1),c=wf[r-1]-wf[l]; b-=(l?C:0)+(r!=n?C:0),c+=C*((l?h[l]:0)+(r!=n?h[r]:0)); x=min(max((ll)(double)b/(-2.0*a)+0.5,x),min(h[l],h[r])); return a*x*x+b*x+c; } signed main(){ n=read(),C=read(); for(register int i=1;i<=n;++i) w[i]=(h[i]=read())+w[i-1],wf[i]=wf[i-1]+1ll*h[i]*h[i]; h[0]=h[++n]=1e9+7; for(register int i=1;i<=n;++i){ dp[i]=(i==1||i==n)?dp[i-1]:(dp[i-1]+C*_abs(h[i]-h[i-1])); while(top^1 && h[stack[top]]<=h[i]) if(top--!=1)dp[i]=min(dp[i],dp[stack[top]]+find(h[stack[top+1]],stack[top],i)); stack[++top]=i; } printf("%lld",dp[n]); return 0; }
T2
Mo water with team had really nothing to say.
The title is a four-dimensional partial order, CDQ can be nested, you can also make a multi-dimensional array tree.
%%% skyh, %%% yxs, the Titans are playing positive solutions.
#include<cstdio> #include<algorithm> using namespace std; int const N=202; inline int read(){ int ss=0;char bb=getchar(); while(bb<48||bb>57)bb=getchar(); while(bb>=48&&bb<=57)ss=(ss<<1)+(ss<<3)+(bb^48),bb=getchar(); return ss; } inline intmin ( int x, int y) { return x <y? x: y; } Inline int max ( int x, int y) { return x> y? x: y; } Inline int ml ( int x) { return x * x- (x- 1 ) * (x- 1 ); } Inline int dl ( int x) { return (x + 1 ) * (x + 1 ) -x * x; } Int r, c, qs, tot, CNT = 1 ; int bl[N][N],a[N][N]; int v[N*N],bar[N*N],now; int as[100005]; struct node{ int xo,yo,xi,yi,id; }q[100005]; struct ljj{ int w,xp,yp; }sj[N*N]; int main(){ r=read(),c=read(),qs=read(); int rs=1,cs=1; for(register int i=1;i<=r;++i) for(register int j=1;j<=c;++j) sj[++tot].xp=i,sj[tot].yp=j,sj[tot].w=read(); sort(sj+1,sj+tot+1,[](ljj skyh,ljj yxs){ return skyh.w<yxs.w; }); a[sj[1].xp][sj[1].yp]=1; for(register int i=2;i<=tot;++i) a[sj[i].xp][sj[i].yp]=cnt=cnt+(sj[i].w!=sj[i-1].w); int xlit=ceil((double)c/cs),ylit=ceil((double)r/rs); for(register int i=1,xl,xr=0,ct=0;i<=xlit;++i){ xl=xr+1,xr=min(xr+cs,c); for(register int j=1,yl,yr=0;j<=ylit;++j){ yl=yr+1,yr=min(yr+rs,r),++ct; for(register int k=yl;k<=yr;++k) for(register int u=xl;u<=xr;++u) bl[k][u]=ct; } } for(register int i=1;i<=qs;++i) q[i].id=i,q[i].xo=read(),q[i].yo=read(),q[i].xi=read(),q[i].yi=read(); sort(q+1,q+qs+1,[](node skyh,node yxs){ return (bl[skyh.xo][skyh.yo]^bl[yxs.xo][yxs.yo])?(bl[skyh.xo][skyh.yo]<bl[yxs.xo][yxs.yo]):(bl[skyh.xi][skyh.yi]>bl[yxs.xi][yxs.yi]); }); int nxo(0),nyo(0),nxi(0),nyi(0),ans(0); for(register int i=1;i<=qs;++i){ int jx=min(q[i].xi,nxi)-max(q[i].xo,nxo)+1,jy=min(q[i].yi,nyi)-max(q[i].yo,nyo)+1; if(jx<=0 || jy<=0 || (nxi-nxo+1)*(nyi-nyo+1)-(jx*jy<<1)>=0){ nxo=q[i].xo,nyo=q[i].yo,nxi=q[i].xi,nyi=q[i].yi; ++now,ans=0; for(register int j=nxo;j<=nxi;++j) for(register int k=nyo;k<=nyi;++k) if(v[a[j][k]]!=now)v[a[j][k]]=now,bar[a[j][k]]=1,++ans; else ans+=ml(++bar[a[j][k]]); as[q[i].id]=ans; continue; } while(nxo<q[i].xo){ for(register int j=nyo;j<=nyi;++j){ ans-=dl(--bar[a[nxo][j]]); if(!bar[a[nxo][j]])v[a[nxo][j]]=now-1; } ++nxo; } while(nyo<q[i].yo){ for(register int j=nxo;j<=nxi;++j){ ans-=dl(--bar[a[j][nyo]]); if(!bar[a[j][nyo]])v[a[j][nyo]]=now-1; } ++nyo; } while(nxi>q[i].xi){ for(register int j=nyo;j<=nyi;++j){ ans-=dl(--bar[a[nxi][j]]); if(!bar[a[nxi][j]])v[a[nxi][j]]=now-1; --Nxi; } While (sex> q [you] .yi) { for (register Corinthians j = nxo; j <= nxi; ++ j) { ans - = DL (- bar [a [j] [sex]]); if (! bar [a [j] [sex]]) v [a [j] [sex]] = now- 1 ; } - sex; } While (nxo> q [you] .xo) { - nxo; for (register Corinthians j = filter; j <= sex; ++ j) if (v [a [nxo] [j]]! = now) ++ ans, v [a [nxo] [j]] = now, Bar [a [nxo] [j]] = 1 ; else ans+=ml(++bar[a[nxo][j]]); } while(nyo>q[i].yo){ --nyo; for(register int j=nxo;j<=nxi;++j) if(v[a[j][nyo]]!=now)++ans,v[a[j][nyo]]=now,bar[a[j][nyo]]=1; else ans+=ml(++bar[a[j][nyo]]); } while(nxi<q[i].xi){ ++nxi; for(register int j=nyo;j<=nyi;++j) if(v[a[nxi][j]]!=now)++ans,v[a[nxi][j]]=now,bar[a[nxi][j]]=1; else ans+=ml(++bar[a[nxi][j]]); } while(nyi<q[i].yi){ ++nyi; for(register int j=nxo;j<=nxi;++j) if(v[a[j][nyi]]!=now)++ans,v[a[j][nyi]]=now,bar[a[j][nyi]]=1; else ans+=ml(++bar[a[j][nyi]]); } as[q[i].id]=ans; } for(register int i=1;i<=qs;++i)printf("%d\n",as[i]); return 0; }
T3
Two diameters of the $ L_1 $ and $ L_2 $ Unicom block merger minimum diameter $ max (\ lfloor \ frac {l_1} {2} \ rfloor + \ lfloor \ frac {l_2} {2} \ rfloor + 1, max (l_1, l_2)) $
This conclusion is well know to do.
To find out the original tree diameter, the diameter of each dot is determined for both ends of the sub-tree formed by the dot diameter.
Then enumeration side, $ \ Theta (1) $ value of minimum diameter size after the merger, updated the answer.
For even edges embodiment, if the original tree ans diameter equal to disconnect and reconnect directly.
Connecting the midpoints of the two diameters can Unicom blocks.
#include<cstdio> #include<cstring> using namespace std; int const N=3e5+5; inline int read(){ int ss=0;char bb=getchar(); while(bb<48||bb>57)bb=getchar(); while(bb>=48&&bb<=57)ss=(ss<<1)+(ss<<3)+(bb^48),bb=getchar(); return ss; } inline int max(int x,int y){ return x>y?x:y; } int n,mp=-1,st,ed,ans=1e9+7,kx,ms,ct=0; int head[N],Next[N<<1],to[N<<1],t=1; int q[N],qj[N]; int pre[N]; char vis[N<<1]; int stl[N],edl[N]; inline void add(int x,int y){ to[++t]=y; Next[t]=head[x],head[x]=t; return ; } void dfs(int x,int y,int ds){ if(ds>mp)mp=ds,st=x; ++ds; for(int i=head[x];i;i=Next[i]) if(to[i]^y && i^ct && i^1^ct)dfs(to[i],x,ds); return ; } int bfs(int x){ memset(pre,0,sizeof(pre)); int u(0),v(1); q[1]=x,pre[x]=-1; while(u^v && v^n){ x=q[++u]; for(register int i=head[x];i;i=Next[i]) if(!pre[to[i]] && i^ct && i^1^ct)pre[q[++v]=to[i]]=i^1; } return v; } int get(int x,int y,bool p){ int mad(0),mal(0); for(int i=head[x];i;i=Next[i]) if(to[i]^y){ int z=get(to[i],x,p)+1; mal=max(max(mal,(p?stl[to[i]]:edl[to[i]])),mad+z); mad=max(mad,z); } (p?stl[x]:edl[x])=mal; return mad; } inline void find(int x){ printf("\n%d %d ",to[x],to[x^1]); ct=x,mp=-1; dfs(to[x],0,0); q[ms=1]=q[bfs(st)]; for(register int i=pre[q[1]];~i;i=pre[to[i]]) q[++ms]=to[i]; printf("%d ",q[ms+1>>1]); mp=-1; dfs(to[x^1],0,0); q[ms=1]=q[bfs(st)]; for(register int i=pre[q[1]];~i;i=pre[to[i]]) q[++ms]=to[i]; printf("%d ",q[ms+1>>1]); return ; } int main(){ n=read(); for(register int i=1,ff,tt;i<n;++i)ff=read(),tt=read(),add(ff,tt),add(tt,ff); dfs(1,0,0),bfs(st); for(register int i=pre[ed=q[n]];~i;i=pre[to[i]]) vis[i]=1,++ms; get(st,0,1),get(ed,0,0); for(register int i=2;i<=t;i+=2) if(vis[i]){ int x=edl[to[i]],y=stl[to[i^1]],now=max((x+1>>1)+(y+1>>1)+1,max(x,y)); if(ans>now)ans=now,qj[kx=1]=i>>1; else if(ans==now)qj[++kx]=i>>1; } else if(vis[i^1]){ int x=stl[to[i]],y=edl[to[i^1]],now=max((x+1>>1)+(y+1>>1)+1,max(x,y)); if(ans>now)ans=now,qj[kx=1]=i>>1; else if(ans==now)qj[++kx]=i>>1; } else if(ans>ms)ans=ms,qj[kx=1]=i>>1; else if(ans==ms)qj[++kx]=i>>1; printf("%d\n%d ",ans,kx); for(register int i=1;i<=kx;++i)printf("%d ",qj[i]); find(qj[1]<<1); return 0; }