Bzoj 2453: Maintenance queue && Bzoj 2120: Number of color tiles, bitset

2453: Maintenance Queue time limit:10 Sec Memory limit:128 MB
submit:578 solved:247
[Submit] [Status] [Discuss] Description did you play marbles when you were a kid? Children A have some marbles, a like to queue them, from left to right numbered 1 to N. For the whole queue to be bright and beautiful, the children want to know how many different colored marbles are in a continuous marble. Of course, a sometimes replaces the color of one of the marbles in the queue, depending on your preference. But a has not learned to program, and think brainstorming is too wasteful of brain power, so ask you for help. The first line of the input file contains two integers n and M. The second row n integers representing the color of the initial queue shot. Next m line, each line in the form of "Q L R" or "R x C", "Q L R" indicates that a want to know from the queue L to the first marbles, there are a total number of different colors of the marbles, "R x C" means a to the X position of the marbles replaced by the C color. Output for each q operation, a row of outputs indicates the query result. Sample Input
2 3
1 2
Q 1 2
R 1 2
Q 1 2
Sample Output2
1HINT

For 100% of the data, there are 1≤n≤10000, 1≤m≤10000, children A will not be modified more than 1000 times, all colors are 1 to 10^6 integer representation.

Source

2011 Fujian Training

1#include <bits/stdc++.h>2 using namespacestd;3 #defineMAXN 100104 #defineMAXM 10000105 intN,A[MAXN],LAST[MAXN],PRE[MAXM],BLOCK,POS[MAXN],H[MAXN];6 intRead ()7 {8     ints=0, fh=1;CharCh=GetChar ();9      while(ch<'0'|| Ch>'9'){if(ch=='-') fh=-1; ch=GetChar ();}Ten      while(ch>='0'&&ch<='9') {s=s*Ten+ (ch-'0'); ch=GetChar ();} One     returns*fh; A } - voidClintk) - { the     intL= (K-1) *block+1, R=min (n,k*block), I; -      for(i=l;i<=r;i++) a[i]=Last[i]; -Sort (a+l,a+r+1); - } + voidChange (intXintc) - { +     inti,t; A      for(i=1; i<=n;i++) pre[h[i]]=0; ath[x]=C; -      for(i=1; i<=n;i++) -     { -t=Last[i]; -last[i]=Pre[h[i]]; -         if(t!=Last[i]) cl (pos[i]); inpre[h[i]]=i; -     } to } + intFind (intKintll) - { the     intL= (K-1) *block+1, R=min (k*block,n), mid,t=0; *      while(l<=R) $     {Panax NotoginsengMid= (L+R)/2; -         if(A[MID]&GT;=LL) r=mid-1; the         Else if(A[MID]&LT;LL) t=mid,l=mid+1; +     } A     if(t==0)return 0; the     Else returnT (K-1) *block+1)+1; + } - intQuery (intLintR) $ { $     intans=0, I; -     if(pos[l]==Pos[r]) -     { the          for(i=l;i<=r;i++)if(last[i]<l) ans++; -     }Wuyi     Else the     { -          for(i=l;i<=pos[l]*block;i++)if(last[i]<l) ans++; Wu          for(I= (pos[r]-1) *block+1; i<=r;i++)if(last[i]<l) ans++; -          for(i=pos[l]+1; i<=pos[r]-1; i++) ans+=Find (i,l); About     } $     returnans; - } - intMain () - { A     intm,i,m,s1,s2; +     Charfh[2]; theN=read (); m=read (); -Block= (int) sqrt (n); $Memset (Last,0,sizeof(last)); thememset (PRE,0,sizeof(pre)); the      for(i=1; i<=n;i++) the     { theh[i]=read (); -last[i]=Pre[h[i]]; inpre[h[i]]=i; thepos[i]= (I-1)/block+1; the     } About     if(block*block==n) m=n/Block; the     Elsem=n/block+1; the      for(i=1; i<=m;i++) cl (i); the      for(i=1; i<=m;i++) +     { -scanf"\n%s", FH); S1=read (); s2=read (); the         if(fh[0]=='Q')Bayi         { theprintf"%d\n", Query (S1,S2)); the         } -         ElseChange (S1,S2); -     } the     return 0; the}

View Code

1#include <bits/stdc++.h>2 using namespacestd;3 inta[10010];4bitset<1000010>Vis;5 intMain ()6 {7     intn,m,i,l,r,sum,j;8     Charfh;9scanf"%d%d",&n,&m);Ten      for(i=1; i<=n;i++) scanf ("%d",&a[i]); One      for(i=1; i<=m;i++) A     { -scanf"\n%c%d%d",&fh,&l,&R); -         if(fh=='Q') the         { - Vis.reset (); -sum=0; -              for(j=l;j<=r;j++)if(vis[a[j]]==0) {sum++;vis[a[j]]=1;} +printf"%d\n", sum); -         } +         Else A         { ata[l]=R; -         } -     } -     return 0; -}

View Code

Bzoj 2453: Maintenance queue && Bzoj 2120: Number of color tiles, bitset