Bzoj 3236: [Ahoi2013] Jobs

Topics 3236: [Ahoi2013] Job time limit:100 Sec Memory limit:512 MB
submit:732 solved:271
Description

Input

Output

Sample Input3 4
1 2 2
1 2 1 3
1 2 1 1
1 3 1 3
2 3 2 3
Sample Output2 2
1 1
3 2
2 1HINT

n=100000,m=1000000

Source

by wangyisong1996 strengthening data

Solving

A good problem of card evaluation! Using the legendary MO team algorithm can water over, that is, we regard each inquiry x0,y0 as a point on the plane, then the price of violent transfer is the distance between two points, we divide the x-axis, so the total cost of transfer is O (sqrt (n) *n). Since this problem requires a tree-like array to maintain, the complexity of this problem is O (nsqrt (n) logn).

Code

1 /*Author:wnjxyk*/2#include <cstdio>3#include <algorithm>4 using namespacestd;5 6 Const intmaxm=1000000;7 structquestionnode{8     intL,r;9     intb;Ten     intindex; One };  AQuestionnode quest[maxm+Ten]; - structansnode{ -     intAns1,ans2; the }; -Ansnode ans[maxm+Ten]; -  - Const intmaxn=100000; + intcnt[maxn+Ten]; -  + intsiz; A intn,m; at intnum[maxn+Ten]; -  - intsum1[maxn+Ten],sum2[maxn+Ten],cot[maxn+Ten];  -  -InlineintRemin (intAintb) { -     if(A&LT;B)returnA; in     returnb; - } to  +InlineBOOLCMP (Questionnode A,questionnode b) { -     if(CNT[A.L]&LT;CNT[B.L])return true; the     if(CNT[A.L]==CNT[B.L] && A.R&LT;B.R)return true; *     return false; $ }Panax Notoginseng  -InlineintLowbit (intx) { the     returnx&-x; + } A  theInlinevoidAddintC[],intXintnum) { +      for(intI=x;i<=maxn;i+=lowbit (i)) c[i]+=num; - } $  $Inlineint Get(intC[],intx) { -     intres=0; -      for(intI=x;i;i-=lowbit (i)) res+=C[i]; the     returnRes; - }Wuyi  the intMain () { -scanf"%d%d",&n,&m); Wu     //getcnt -      while(siz*siz<n) siz++; About      for(intI=1; i<=siz;i++){ $         intst=siz* (I-1)+1, Ed= (Remin (siz*i,n)); -          for(intj=st;j<=ed;j++){ -cnt[j]=i; -         } A     }     +     //Getnum the      for(intI=1; i<=n;i++) scanf ("%d",&num[i]); -     //getquest $      for(intI=1; i<=m;i++) scanf ("%d%d%d%d",&quest[i].l,&quest[i].r,&quest[i].a,&quest[i].b); the      for(intI=1; i<=m;i++) quest[i].index=i; the     //sortquest theSort (quest+1, quest+m+1, CMP); the     //Getans -     intleft=1, right=0; in      for(intI=1; i<=m;i++){ the         intL=quest[i].l,r=QUEST[I].R; the         inta=quest[i].a,b=quest[i].b; About         intindex=Quest[i].index; the          while(left<l) { thecot[num[left]]--; theAdd (sum1,num[left],-1); +             if(cot[num[left]]==0) Add (sum2,num[left],-1); -left++; the         }Bayi          while(l<Left ) { theleft--; thecot[num[left]]++; -Add (Sum1,num[left],1); -             if(cot[num[left]]==1) Add (Sum2,num[left],1); the         } the          while(right<S) { theright++; thecot[num[right]]++; -Add (Sum1,num[right],1); the             if(cot[num[right]]==1) Add (Sum2,num[right],1); the         } the          while(r<Right ) {94cot[num[right]]--; theAdd (sum1,num[right],-1); the             if(cot[num[right]]==0) Add (sum2,num[right],-1); theright--;98         } AboutAns[index]. ans1=Get(SUM1,B)-Get(sum1,a-1); -Ans[index]. Ans2=Get(SUM2,B)-Get(sum2,a-1);101     }102     //Print103      for(intI=1; i<=m;i++) printf ("%d%d\n", Ans[i]. Ans1,ans[i]. ANS2);104     return 0; the}

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Bzoj 3236: [Ahoi2013] Jobs