NOI2011 Noi Carnival

http://www.lydsy.com/JudgeOnline/problem.php?id=2436

First discretization, the time range after discretization is [1,cnt].

H[I][J], indicates how many times the time range is in [I,j], and can be resolved within n^2 time.

Assume that the site is a and B, respectively.

It is easy to know that the activities of venue A and site B must be:

The arrangement of their activities must have been so spaced out.

We beg F[i][j], indicating that when the time <=i, a site has a J activity, b venue, the maximum number of activities.

We do not know that the last activity is in the site A or B, but it must be the maximum value of the 2 cases, that is to say: F[i][j] "When the time <=i, a site has J activities and the last activity in a site, b site The maximum number of activities" and when the time <=i, A site has a J activity and the last activity on the B site, b site up to how many activities "the maximum value of both cases."

From small to large enumeration I.

First to reverse the loop J, with F[i][j+1] update F[i][j]:f[i][j]=max (f[i][j],f[i][j+1]).

We enumerate J, push backwards, enumerate K,

We can move to F[k][j]:f[k][j]=max (f[k][j],f[i][j]+h[i+1][k) if the next [i+1,k] time zone is the B site in action.

We can move to F[k][j+h[i+1][k]]:f[k][j+h[i+1][k]]=max (F[k][j+h[i+1][k]],f[i][j]) if the next [i+1,k] time interval is a site activity.

This can be solved in n^3 time.

Similarly, we beg G[i][j], said when the time >=i, a site has J activities, B site up to how many activities, is also n^3 time to resolve.

We then ask T[I][J] to indicate that all activities in [i,j] are at the site A, the maximum number of activities of the carnival with relatively little activity.

A very straightforward idea is:

$T [I][j]=max\{min (X+y+h[i][j],f[i-1][x]+g[j+1][y]) \} (0\leqslant x\leqslant n,0\leqslant y\leqslant N) $

But it must have been n^4.

We find that when x gets bigger, f[i-1][x] becomes smaller:

$min (X↑+y+h[i][j],f[i-1][x]↓+g[j+1][y]) $

If y also becomes larger, G[j+1][y] becomes smaller:

$min (x↑+y↑+h[i][j],f[i-1][x]↓+g[j+1][y]↓) $

As we take the min, there is no egg to use.

So y can only get smaller, g[j+1][y] and then grow bigger:

$min (x↑+y↓+h[i][j],f[i-1][x]↓+g[j+1][y]↑) $

So with the increment of x, y decrements.

And easy to know, this must be a single-peak.

So can n^3 find out ans[i][j].

The output of line 1th is $max\{min (I,f[cnt][i]) \} (0\leqslant i\leqslant N) $

The first i+1 output is $max (T[i][j]) (1\leqslant i\leqslant l,r\leqslant j\leqslant CNT) $

#include <cstdio>#include<cstdlib>#include<iostream>#include<fstream>#include<algorithm>#include<cstring>#include<string>#include<cmath>#include<queue>#include<stack>#include<map>#include<utility>#include<Set>#include<bitset>#include<vector>#include<functional>#include<deque>#include<cctype>#include<climits>#include<complex>//#include <bits/stdc++.h> for Cf,uoj, but not for POJ using namespaceStd;typedefLong LongLl;typedefDoubleDb;typedef pair<int,int>Pii;typedef Complex<DB>CP;#defineMmSt (a,v) memset (A,v,sizeof (a))#defineMmcy (A, B) memcpy (A,b,sizeof (a))#defineRe (I,A,B) for (i=a;i<=b;i++)#defineRed (i,a,b) for (i=a;i>=b;i--)#defineFi first#defineSe Second#defineM_p (A, B) Make_pair (A, B)#defineSF scanf#definePF printf#defineBoth (k) (1<< (k))Template<classT>inline T Sqr (t x) {returnx*x;} Template<classT>inlinevoidUpmin (T &t,t tmp) {if(t>tmp) t=tmp;} Template<classT>inlinevoidUpmax (T &t,t tmp) {if(t<tmp) t=tmp;}ConstDB eps=1e-9; inlineintSGN (DB x) {if(ABS (x) <eps)return 0;return(x>0)?1:-1;}ConstDB Pi=acos (-1.0); inlineintGint () {intres=0;BOOLneg=0;CharZ;  for(Z=getchar (); z!=eof && z!='-'&&!isdigit (z); z=GetChar ()); if(z==eof)return 0; if(z=='-') {neg=1; z=GetChar ();}  for(; z!=eof && isdigit (z); res=res*Ten+z-'0', z=GetChar ()); return(neg)?-Res:res; }inline ll Gll () {ll res=0;BOOLneg=0;CharZ;  for(Z=getchar (); z!=eof && z!='-'&&!isdigit (z); z=GetChar ()); if(z==eof)return 0; if(z=='-') {neg=1; z=GetChar ();}  for(; z!=eof && isdigit (z); res=res*Ten+z-'0', z=GetChar ()); return(neg)?-Res:res; }Const intmaxn= $;Const intMaxcnt=2*MAXN;intN;structtdata{intL,r,id;} a[maxn+Ten];intcnt,bak[2*maxn+ -];intt[maxcnt+Ten];inth[maxcnt+Ten][maxcnt+Ten];inlineBOOLCMPR (tdata X,tdata y) {returnx.r<Y.R;}intf[maxcnt+Ten][maxn+Ten],g[maxcnt+Ten][maxn+Ten];intt[maxcnt+Ten][maxcnt+Ten];intans,res[maxn+Ten];intMain () {/*freopen ("show.in", "R", stdin); Freopen ("Show.out", "w", stdout);*/      inti,j,k; N=Gint (); Re (i,1, N) A[i].l=gint (), A[i].r=a[i].l+gint ()-1, a[i].id=i; Re (i,1, N) bak[++cnt]=a[i].l,bak[++cnt]=A[I].R; Sort (Bak+1, bak+cnt+1); CNT=unique (bak+1, bak+cnt+1)-bak-1; Re (i,1, N) A[i].l=lower_bound (bak+1, bak+cnt+1, A[I].L)-bak,a[i].r=lower_bound (bak+1, bak+cnt+1, A[I].R)-Bak; Sort (a+1, a+n+1, CMPR); intHead=1; Re (i,1, CNT) {             while(head<=n && a[head].r<=i) t[a[head].l]++,head++; intsum=0; Red (J,i,1) sum+=t[j],h[j][i]=sum; } mmst (F,-1); f[0][0]=0; Re (i,0, cnt-1) {Red (J,n-1,0) Upmax (f[i][j],f[i][j+1]); Re (J,0, N) {                if(f[i][j]!=-1) Re (k,i+1, CNT) Upmax (f[k][j],f[i][j]+h[i+1][k]); if(f[i][j]!=-1) Re (k,i+1, CNT) Upmax (f[k][j+h[i+1][k]],f[i][j]); }} mmst (G,-1); G[cnt+1][0]=0; Red (i,cnt+1,2) {Red (J,n-1,0) Upmax (g[i][j],g[i][j+1]); Re (J,0, N) {                if(g[i][j]!=-1) Re (k,1, I-1) Upmax (g[k][j],g[i][j]+h[k][i-1]); if(g[i][j]!=-1) Re (k,1, I-1) Upmax (g[k][j+h[k][i-1]],g[i][j]); }} ans=0; Re (J,0, N) Upmax (Ans,min (j,f[cnt][j)); cout<<ans<<Endl; Re (i,1, CNT) Re (j,i,cnt) {T[i][j]=0; /*int x, y; Re (x,0,n) Re (y,0,n) if (f[i-1][x]!=-1 && g[j+1][y]!=-1) Upmax (T[i][j],min (x+y+h[i][j],f[ I-1][x]+g[j+1][y]));*/            intx,y=N;  while(Y1>=0&& g[j+1][y]==-1) y--; Re (x,0, N) {                  if(f[i-1][x]==-1)Continue;  while(Y1>=0&& min (x+y+h[i][j],f[i-1][x]+g[j+1][y]) <=min (x+y-1+h[i][j],f[i-1][x]+g[j+1][y-1])) y--; Upmax (t[i][j],min (x+y+h[i][j],f[i-1][x]+g[j+1][y])); }} re (I,1, N) {            intL=a[i].l,r=a[i].r,id=a[i].id; Res[id]=0; Re (J,1, L) Re (k,r,cnt) Upmax (Res[id],t[j][k]); } Re (I,1, N) cout<<res[i]<<Endl; return 0; }

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NOI2011 Noi Carnival