Bzoj 1071 Team

Title Link: http://www.lydsy.com/JudgeOnline/problem.php?id=1071

The topic is very good, incredibly wrote for a long time, the problem finding really much;

There are two main ways of doing this:

O (N^2LGN), through priority heap maintenance, first equation transformation:a*height+b*speed-c<=a*minheight+b*minspeed;

Add A[i].val=a*height+b*speed-c:

Sort a by height;

Then enumerate I take a[i].s as min

1 /* ***********************************************2 author:forgot933 Created time:2014/12/23 Tuesday morning 9:00:414 File Name:5 ************************************************ */6 7#include <stdio.h>8#include <string.h>9#include <iostream>Ten#include <algorithm> One#include <vector> A#include <queue> -#include <Set> -#include <map> the#include <string> -#include <stdlib.h> -#include <time.h> - using namespacestd; +  - #defineN 5555 +typedefLong Longll; APriority_queue<ll>Q; at  - structnode - { - ll H,s; - ll Val; -     BOOL operator< (ConstNode &b)Const{ in     returnH>b.h; -     } to }a[n]; +  -  the intMain () * { $     intN;Panax Notoginseng ll A,b,c; -Cin>>n>>a>>b>>C; the      for(intI=1; i<=n;i++){ +Cin>>a[i].h>>A[i].s; Aa[i].val=a*a[i].h+b*a[i].s-C; the     } +ll ans=1; -Sort (A +1, a+n+1); $      $      for(intI=1; i<=n;i++) -     { -ll minh=a[i].h; thell mins=A[i].s; -          while(!q.empty ()) Q.pop ();Wuyi Q.push (a[i].val); the          for(intj=1; j<=n;j++) -         if(j!=i&&a[j].s>=mins) Wu         { -Minh=min (minh,a[j].h); Aboutll tmp=b*mins+a*Minh; $             if(a[i].val>tmp) Break; -              while(!q.empty () &&q.top () >tmp) Q.pop (); -             if(a[j].val<=tmp) -             { A Q.push (a[j].val); +ans=Max (ans, (LL) q.size ()); the             } -         } $     } thecout<<ans<<Endl; the     return 0; the}

View Code

Speed

then temporarily minheight=a[i].h;

A[i].h are sorted from large to small;

Next Maintenance heap, we enumerate J for J!=i and A[j].s>=mins,

Update MinHeight at the same time;

Then the Val satisfies the pressure into the heap;

To Q.top () >val q.pop ();

Because the mins is fixed, the Minh is monotonically decreasing, so the back of the front is satisfied (please consider it carefully);

Time is 1100ms;

The second type is O (n*n);

Time is 848ms;

Keywords: monotonous;

1 /* ***********************************************2 author:forgot933 Created time:2014/12/23ðçæú¶þïâîç2:46:364 File name:c.cpp5 ************************************************ */6 7#include <stdio.h>8#include <string.h>9#include <iostream>Ten#include <algorithm> One#include <vector> A#include <queue> -#include <Set> -#include <map> the#include <string> -#include <math.h> -#include <stdlib.h> -#include <time.h> + using namespacestd; -  +typedefLong Longll; A #defineN 5555 at structnode - { -    inth,v; - ll Val; - }h[n],v[n],a[n],r[n]; -  in intcmp1 (node X,node y) - { to     if(X.h==y.h)returnx.v<y.v; +     returnx.h<Y.h; - } the intcmp2 (node X,node y) * { $     if(X.V==Y.V)returnx.h<Y.h;Panax Notoginseng     returnx.v<y.v; - } the  + intCmp3 (node X,node y) A { the     returnx.val<Y.val; + } -  $  $ intMain () - { -     intN; the ll A,b,c; -Cin>>n>>a>>b>>C;Wuyi      for(intI=0; i<n;i++){ theCin>>a[i].h>>a[i].v; -a[i].val=a*a[i].h+b*a[i].v-C; Wuh[i]=v[i]=A[i]; -     } AboutSort (a,a+N,cmp3); $Sort (v,v+n,cmp2); -Sort (h,h+n,cmp1); -     intans=0; -      for(intI=0; i<n;i++) A     { +         intminh=h[i].h,p=0, cnt=0, tot=0; the          for(intj=0; j<n;j++) -         if(v[j].h>=minh&&v[j].v<=h[i].v) $r[tot++]=V[j]; the          for(intj=0; j<tot;j++) the         { the             intminv=r[j].v; thell res=a*minh+b*MINV; -              while(p<n&&a[p].val<=Res) in             { the                 if(a[p].h<minh| | A[P].V&LT;MINV) cnt++; thep++; About             } theAns=max (P-Cnt,ans); the             if(res>=a*r[j].h+b*r[j].v-c) cnt++; the             if(p==n) Break; +         } -     } theprintf"%d\n", ans);Bayi     return 0; the}

View Code

First sort by some keywords.

For I minh=a[i].h;

Then enumerate J to find mins,mins<a[i],s;

then the monotone queue;

There is such a property: when we enumerate pointers in the order of Val from small to large, we enumerate the mins is small to large, so:

(here) The elements of the front must satisfy the back, how to understand?

The mins2 of the enumeration satisfies all elements of the mins1, so the pointer p does not have to be 0.

So it will O (n^2);

Bzoj 1071 Team