[bzoj2288] [POJ Challenge] Birthday present

Use heap maintenance doubly linked list to greedy ...

The data range obviously does not allow O (nm) of the dp>_<. And the DP light state is n*m. Obviously can't optimize.

Probably think of greedy mess ... Initially want to greedy through a few small negative numbers to connect a positive number into a paragraph, but in the end is to join or directly throw away bad judgment

Then I ran to the puzzle ... The solving is very good, even I have read >_<. Address: http://www.cnblogs.com/tuigou/p/4868127.html

Although the choice of positive and negative values is different, the actual operation is to combine the numbers of the two sides together. Also, for negative numbers at the left or right end, deleting them will not reduce the number of segments currently selected.

1#include <cstdio>2#include <iostream>3#include <cstring>4#include <algorithm>5#include <queue>6#include <cstdlib>7#include <cmath>8 #definell Long Long9 using namespacestd;Ten Const intmaxn=100233; One structzs{ A     intID; - }; -Priority_queue <zs>Q; the inta[maxn],cnt,pre[maxn<<1],next[maxn<<1],v[maxn<<1]; - intI,j,n,m,zsnum,ans; - BOOLdel[maxn<<1]; -  + intRA,FH;CharRx; -InlineintRead () { +Rx=getchar (), ra=0, fh=1; A      while((rx<'0'|| Rx>'9') &&rx!='-') rx=GetChar (); at     if(rx=='-') fh=-1, rx=GetChar (); -      while(rx>='0'&&rx<='9') ra*=Ten, ra+=rx- -, Rx=getchar ();returnTalfh; - } - BOOL operator< (ZS A,zs b) {returnABS (V[a.id]) >ABS (V[b.id]);} - intMain () { -N=read (), M=read ();if(!m) {Puts ("0");return 0;} in      for(i=1; i<=n;i++) A[i]=read (); cnt=0; -      for(i=1; i<=n;v[cnt]+=a[i++]) to         if(LL) a[i]* (LL) v[cnt]<0||! CNT) cnt++; +      for(i=1; i<=cnt;i++)if(v[i]>0) zsnum++,ans+=V[i]; -     if(zsnum>m) { the          for(i=1; i<=cnt;i++) Q.push ((ZS) {i}), pre[i]=i-1, next[i]=i+1;//, printf ("%d", V[i]);p UTS (""); *pre[1]=next[cnt]=0; $          for(i=zsnum-m;i;i--){Panax Notoginseng              while(!q.empty () &&del[q.top (). Id]) Q.pop ();if(Q.empty ()) Break; -             intX=q.top (). id,pr=pre[x],nex=Next[x]; the              +Q.pop (), Ans-=abs (V[x]), del[x]=1; A             if(! (pr&&NEX)) { the                 if(v[x]<0) i++,ans+=ABS (V[x]); +                 if(PR) next[pr]=0;if(NEX) pre[nex]=0; -             } $             Else{ $del[pr]=del[nex]=1; -v[++cnt]=v[pr]+v[x]+V[nex]; - Q.push (ZS) {cnt}); the                 if(PRE[PR]) pre[cnt]=pre[pr],next[pre[cnt]]=CNT; -                 if(Next[nex]) next[cnt]=next[nex],pre[next[cnt]]=CNT;Wuyi             } the         } -     } Wuprintf"%d\n", ans); -     return 0; About}

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[bzoj2288] [POJ Challenge] Birthday present