zoj 2112 （主席树，树状数组套线段树）

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

题意：查找区间第K大值，值可修改

分析：

每一棵线段树是维护每一个序列前缀的值在任意区间的个数，
如果还是按照静态的来做的话，那么每一次修改都要遍历O(n)棵树，
时间就是O(2*M*nlogn)->TLE
考虑到前缀和，我们通过树状数组来优化，即树状数组套主席树，
每个节点都对应一棵主席树，那么修改操作就只要修改logn棵树，
o(nlognlogn+Mlognlogn)时间是可以的，
但是直接建树要nlogn*logn（10^7）会MLE
我们发现对于静态的建树我们只要nlogn个节点就可以了，
而且对于修改操作，只是修改M次，每次改变俩个值（减去原先的，加上现在的）
也就是说如果把所有初值都插入到树状数组里是不值得的，
所以我们分两部分来做，所有初值按照静态来建，内存O(nlogn)，
而修改部分保存在树状数组中，每次修改logn棵树，每次插入增加logn个节点

代码：

//#pragma comment(linker,"/STACK:102400000,102400000")#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#include <stack>#include <map>#include <set>#include <vector>#include <queue>#define mem(p,k) memset(p,k,sizeof(p));#define pb push_back//#define lson l,m,rt<<1//#define rson m+1,r,rt<<1|1#define inf 0x3f3f3f3f#define ll long longusing namespace std;const int N=50010 ;struct SD{    int flag,l,r,k;}que[10010];int T,n,q,len,tot;int num[N],head1[N],head2[N];int tree[N*50],lson[N*50],rson[N*50],use[N];char s[2];vector<int> vec;int Hash(int k){    //cout<<lower_bound(vec.begin(),vec.end(),k)-vec.begin()+1;    return lower_bound(vec.begin(),vec.end(),k)-vec.begin()+1;}void update(int pre,int &now,int k,int val,int l,int r){    now=tot++;    tree[now]=tree[pre]+val;    lson[now]=lson[pre];rson[now]=rson[pre];    if(l==r)return;    int m=(l+r)>>1;    if(k<=m)update(lson[pre],lson[now],k,val,l,m);    else update(rson[pre],rson[now],k,val,m+1,r);}int lowbit(int i){ return -i&i; }void add(int k,int f,int val){    for(int i=k;i<=n;i+=lowbit(i)){        update(head2[i],head2[i],f,val,1,len);    }}int sum(int k){    int s=0;    for(int i=k;i>0;i-=lowbit(i)){        s+=tree[lson[use[i]]];    }    return s;}int Query(int L,int R,int k){    int s1=head1[L-1],s2=head1[R],l=1,r=len;    for(int i=L-1;i>0;i-=lowbit(i))use[i]=head2[i];    for(int i=R;i>0;i-=lowbit(i))use[i]=head2[i];    //cout<<L<<R<<"==="<<endl;    while(l<r){        int tmp=sum(R)-sum(L-1)+tree[lson[s2]]-tree[lson[s1]];        int m=(l+r)>>1;//cout<<tmp<<"  "<<k<<endl;        if(k<=tmp){            r=m;            s1=lson[s1];s2=lson[s2];            for(int i=L-1;i>0;i-=lowbit(i))use[i]=lson[use[i]];            for(int i=R;i>0;i-=lowbit(i))use[i]=lson[use[i]];        }        else{            k-=tmp;            l=m+1;            s1=rson[s1],s2=rson[s2];            for(int i=L-1;i>0;i-=lowbit(i))use[i]=rson[use[i]];            for(int i=R;i>0;i-=lowbit(i))use[i]=rson[use[i]];        }//cout<<l<<r<<endl;    }    //cout<<"  ~~~~~~ "<<endl;    return vec[l-1];}int main(){    cin>>T;    while(T--){        cin>>n>>q;        tot=1;        vec.clear();        mem(tree,0);        mem(lson,0);        mem(rson,0);        for(int i=1;i<=n;i++)scanf("%d",num+i),vec.pb(num[i]);        for(int i=1;i<=q;i++){            scanf("%s",s);            if(s[0]=='Q'){                que[i].flag=0;                scanf("%d%d%d",&que[i].l,&que[i].r,&que[i].k);            }            else{                que[i].flag=1;                scanf("%d%d",&que[i].l,&que[i].k);                vec.pb(que[i].k);            }        }        sort(vec.begin(),vec.end());        vec.erase(unique(vec.begin(),vec.end()),vec.end());        len=vec.size();        //for(int i=0;i<vec.size();i++)cout<<vec[i]<<endl;        head1[0]=0;        for(int i=1;i<=n;i++){            update(head1[i-1],head1[i],Hash(num[i]),1,1,len);        }        for(int i=1;i<=n;i++){            head2[i]=0;        }        for(int i=1;i<=q;i++){            if(que[i].flag){                add(que[i].l,Hash(num[que[i].l]),-1);                add(que[i].l,Hash(que[i].k),1);                num[que[i].l]=que[i].k;            }            else{                printf("%d\n",Query(que[i].l,que[i].r,que[i].k));            }        }    }    return 0;}