ZOJ2112 Dynamic Rankings

题意

Dynamic Rankings

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Time Limit: 10 Seconds
    
Memory Limit: 32768 KB

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The Company Dynamic Rankings has developed a new kind of computer that is no
longer satisfied with the query like to simply find the k-th smallest number
of the given N numbers. They have developed a more powerful system such that
for N numbers a[1], a[2], ..., a[N], you can ask it like: what is the k-th smallest
number of a[i], a[i+1], ..., a[j]? (For some i<=j, 0<k<=j+1-i that
you have given to it). More powerful, you can even change the value of some
a[i], and continue to query, all the same.



Your task is to write a program for this computer, which



- Reads N numbers from the input (1 <= N <= 50,000)



- Processes M instructions of the input (1 <= M <= 10,000). These instructions
include querying the k-th smallest number of a[i], a[i+1], ..., a[j] and change
some a[i] to t.

Input



The first line of the input is a single number X (0 < X <= 4), the number
of the test cases of the input. Then X blocks each represent a single test case.



The first line of each block contains two integers N and M, representing N numbers
and M instruction. It is followed by N lines. The (i+1)-th line represents the
number a[i]. Then M lines that is in the following format



Q i j k or

C i t



It represents to query the k-th number of a[i], a[i+1], ..., a[j] and change
some a[i] to t, respectively. It is guaranteed that at any time of the operation.
Any number a[i] is a non-negative integer that is less than 1,000,000,000.



There're NO breakline between two continuous test cases.

Output



For each querying operation, output one integer to represent the result. (i.e.
the k-th smallest number of a[i], a[i+1],..., a[j])



There're NO breakline between two continuous test cases.

Sample Input



2

5 3

3 2 1 4 7

Q 1 4 3

C 2 6

Q 2 5 3

5 3

3 2 1 4 7

Q 1 4 3

C 2 6

Q 2 5 3

Sample Output



3

6

3

6

(adviser)

Site: http://zhuzeyuan.hp.infoseek.co.jp/index.files/our_contest_20040619.htm

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Author: XIN, Tao

Source: Online Contest of Christopher's Adventure

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Status

分析

跟POJ2104 Kth Number差不多，只不过多了一个单点修改。那么把单点修改拆成删除和插入两个操作就好了。

时间复杂度\(O((N+2*M) \log SIZE \log N)\)

代码

#include<bits/stdc++.h>
#define rg register
#define il inline
#define co const
template<class T>il T read(){
    rg T data=0,w=1;rg char ch=getchar();
    while(!isdigit(ch)) {if(ch=='-') w=-1;ch=getchar();}
    while(isdigit(ch)) data=data*10+ch-'0',ch=getchar();
    return data*w;
}
template<class T>il T read(rg T&x) {return x=read<T>();}
typedef long long ll;

co int N=1e5+1,INF=1e9;
struct rec {int op,x,y,z;}q[3*N],lq[3*N],rq[3*N];
int T,n,m,t,p,a[N],c[N],ans[N];
int ask(int x){
    int y=0;
    for(;x;x-=x&-x) y+=c[x];
    return y;
}
void change(int x,int y){
    for(;x<=n;x+=x&-x) c[x]+=y;
}
void solve(int lval,int rval,int st,int ed){
    if(st>ed) return;
    if(lval==rval){
        for(int i=st;i<=ed;++i)
            if(q[i].op>0) ans[q[i].op]=lval;
        return;
    }
    int mid=lval+rval>>1;
    int lt=0,rt=0;
    for(int i=st;i<=ed;++i){
        if(q[i].op<=0){
            if(q[i].y<=mid) change(q[i].x,q[i].z),lq[++lt]=q[i];
            else rq[++rt]=q[i];
        }
        else{
            int cnt=ask(q[i].y)-ask(q[i].x-1);
            if(cnt>=q[i].z) lq[++lt]=q[i];
            else q[i].z-=cnt,rq[++rt]=q[i];
        }
    }
    for(int i=ed;i>=st;--i)
        if(q[i].op<=0&&q[i].y<=mid) change(q[i].x,-q[i].z);
    for(int i=1;i<=lt;++i) q[st+i-1]=lq[i];
    for(int i=1;i<=rt;++i) q[st+lt+i-1]=rq[i];
    solve(lval,mid,st,st+lt-1);
    solve(mid+1,rval,st+lt,ed);
}
int main(){
//  freopen(".in","r",stdin),freopen(".out","w",stdout);
    read(T);
    while(T--){
        read(n),read(m);
        t=p=0;
        for(int i=1;i<=n;++i){
            read(a[i]);
            q[++t].op=0,q[t].x=i,q[t].y=a[i],q[t].z=1;
        }
        for(int i=1;i<=m;++i){
            static char op[2];scanf("%s",op);
            if(op[0]=='Q'){
                static int l,r,k;read(l),read(r),read(k);
                q[++t].op=++p,q[t].x=l,q[t].y=r,q[t].z=k;
            }
            else{
                static int x,y;read(x),read(y);
                q[++t].op=-1,q[t].x=x,q[t].y=a[x],q[t].z=-1;
                q[++t].op=0,q[t].x=x,q[t].y=y,q[t].z=1;
                a[x]=y;
            }
        }
        solve(0,INF,1,t);
        for(int i=1;i<=p;++i) printf("%d\n",ans[i]);
    }
    return 0;
}