Bzoj2453: Maintenance queue

2453: Maintenance queue time limit: 10 sec memory limit: 128 MB
Submit: 183 solved: 89
[Submit] [Status] Description: Have you ever played marbles when you were a child? Child A has some marbles. A prefers to queue them, numbered from left to right as 1 to n. In order to make the entire queue colorful and beautiful, the children want to know the number of marbles of different colors in a row. Of course, a sometimes replaces the color of a bullet ball in the queue based on personal preferences. However, a hasn't learned programming yet, and he thinks brainstorming is a waste of mental power. So he asks you for help. The first line of the input file contains two integers, N and M. N integers in the second line, indicating the color of the marbles in the initial queue. Next m rows, each row is in the form of "q l r" or "r x C ", "q l r" indicates how many different colors of marbles A wants to know from the first marbles in the queue to the r marbles, "R x C" indicates that a changes the marbles at position X to the C color. For each Q operation, output a line indicates the query result. Sample Input
2 3
1 2
Q 1 2
R 1 2
Q 1 2
Sample output2
1 hint

 

For 100% of the data, there are 1 ≤ n ≤ 10000, 1 ≤ m ≤ 10000, Child A will not be modified more than 1000 times, all colors are represented by an integer ranging from 1 to 10 ^ 6.

 

Source

2011 Fujian Training

Question:

Use pre [I] to record the position of the previous ball of the same color as I.

When you ask about l to R, if pre [I] <r indicates that the ball from L to I is useless, then ans ++

We can use this idea... Multipart

Each part is sorted by Pre [I], and the question is the same as that of the tutorial.

----- Hzwer

Code:

 1 #include<cstdio> 2 #include<cstdlib> 3 #include<cmath> 4 #include<cstring> 5 #include<algorithm> 6 #include<iostream> 7 #include<vector> 8 #include<map> 9 #include<set>10 #include<queue>11 #include<string>12 #define inf 100000000013 #define maxn 10000+100014 #define maxm 1000000+100015 #define eps 1e-1016 #define ll long long17 #define pa pair<int,int>18 using namespace std;19 inline int read()20 {21     int x=0,f=1;char ch=getchar();22     while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}23     while(ch>=‘0‘&&ch<=‘9‘){x=10*x+ch-‘0‘;ch=getchar();}24     return x*f;25 }26 int n,m,block,b[maxn],c[maxn],pre[maxn],pos[maxn],last[maxm];27 void reset(int x)28 {29     int l=(x-1)*block+1,r=min(x*block,n);30     for(int i=l;i<=r;i++)pre[i]=b[i];31     sort(pre+l,pre+r+1);32 }33 int find(int x,int y)34 {35     int l=(x-1)*block+1,r=x*block,mid;36     while(l<=r)37     {38         mid=(l+r)>>1;39         if(pre[mid]>=y)r=mid-1;else l=mid+1;40     }41     return l-(x-1)*block-1;42 }43 int query(int x,int y)44 {45     int sum=0,bx=pos[x],by=pos[y];46     if(by-bx<=1)47     {48         for(int i=x;i<=y;i++)if(b[i]<x)sum++;49     }50     else51     {52         for(int i=x;i<=bx*block;i++)if(b[i]<x)sum++;53         for(int i=(by-1)*block+1;i<=y;i++)if(b[i]<x)sum++;54     }55     for(int i=bx+1;i<by;i++)sum+=find(i,x);56     return sum;57 }58 void change(int x,int y)59 {60     for(int i=1;i<=n;i++)last[c[i]]=0;61     c[x]=y;62     for(int i=1;i<=n;i++)63     {64         int t=b[i];65         b[i]=last[c[i]];66         last[c[i]]=i;67         if(t!=b[i])reset(pos[i]);68     }69 }70 int main()71 {72     freopen("input.txt","r",stdin);73     freopen("output.txt","w",stdout);74     n=read();m=read();75     block=floor(sqrt(n));76     for(int i=1;i<=n;i++)77      {78          c[i]=read();79          pos[i]=(i-1)/block+1;80          b[i]=last[c[i]];81          last[c[i]]=i;82      }83     for(int i=1;i<=pos[n];i++)reset(i);84     char ch;int x,y;85     while(m--)86     {87         ch=‘ ‘;88         while(ch!=‘Q‘&&ch!=‘R‘)ch=getchar();x=read();y=read();89         if(ch==‘R‘)change(x,y);else printf("%d\n",query(x,y));90     } 91     return 0;92 }

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Bzoj2453: Maintenance queue