國家集訓(xùn)隊作業(yè)sequence partitioning解題報告

Sequence【題目簡述Sequence【題目簡述n對數(shù)(ai,bi)，要求按順序?qū)⑺鼈儎澐殖扇舾啥?l1r1),(l2r2(lp,rp)對于任何兩對數(shù)(apbp)，(aqbq)pTp段，qTqTpTqbpkaMk直接作動態(tài)規(guī)劃的時間復(fù)雜度為maxMax[jmax{ajaj+2ai}iMax[j直接作動態(tài)規(guī)劃的時間復(fù)雜度為maxMax[jmax{ajaj+2ai}iMax[j]iMax[j]的性質(zhì)：1．Max[1]≥Max[2]..≥jMax[j]2i→i+1k~iMaxa1a2aiMax值：假設(shè)這個序列中最大的akMax[1]Max[kakk+1~iapMax[k+1]Max[papMax[1]..Max[i]只需要模擬一個棧即找出所有kMax[k]=ak．如右邊這幅圖中，假設(shè)棧v個元素：k[1],k[2],..k[v]k[0]=0Max[k[p]+1ak[p+1]ak[p]>Maxi→i+1ak[p]>ak[p+1]a值>ai+1qqq2Max[q2+1]Max[qaq，現(xiàn)在將[q2q]Maxai+1-MaxO(log2n)的時間雜度內(nèi)完成而用線段樹也同時可 出一段區(qū)間內(nèi)f[j]+Max[j]的最小值1．Sumb[i]–Sumb[j]≤Curr．2．Minb[j]>max{aj+1,aj+2,..,ai}．Lefti1jMinb[jMax[j]j．雖然不滿足條件2的決策j不一定是連續(xù)的，但是Minb[j]是定值，Max[j]是不斷增大的，當(dāng)j2jMinb[jMax[j]C[j]C除一次Minb[jMax[j]C[j]C除一次，因此這一步總的時間復(fù)雜度還是O(nlog2n)的O(nlog2nlog2R)，實現(xiàn)的Fori←1toWhile(a值≤p將[q1p]EndMaxai-While(C[Root]>jC[j]>jEndf[if[jMax[j]的最小值)EndFor【總結(jié)POJj．【英文原題SequenceGivenasequenceofNorderedpairsegers(Ai,Bi),etooseveralcontiguousparts.Letpbethenumberoftheseparts,whoseare(l1,r1),GivenasequenceofNorderedpairsegers(Ai,Bi),etooseveralcontiguousparts.Letpbethenumberoftheseparts,whoseare(l1,r1),(l2,r2),...,(lp,rp),whichsatisfyli=ri?1+1,li≤ri,l1=1,rp=n.Thepartsthemselvesalsosatisfythefollowingrestrictions:Foranytwopairs(Ap,Bp),(Aq,Bq),where(Ap,Bp)isbelongstotheTpthpartand(Aq,Bq)theTqthpart.IfTp<Tq,thenBp>Aq.LetMibeponentofheithpart,Mi=max{Ali,Ali+1,...,Ari},1≤i≤itistwhereLimitisaLetSibethesumofminimizethevalueponentsofheithpart.NowIwantmax{Si:1≤i≤CouldlmetheTheontainsexactlyonetestcase.lineofontainsegersN(N≤50000),Limit(Limit≤231-1).ThenfollowNachcontainsegerspair(A,B).It'salwaystmax{A1,A2,...,An}≤OutputtheSample443223554Sample9HAnavailableassignment43223554Sample9HAnavailableassignmentistwopairsareopartandlasttwopairsareothesecondpart.ThenB1>