ZOJ 2112 Dynamic Rankings （动态第 K 大）（树状数组套主席树）

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

表示不是很懂，先插个眼。。

#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#include<vector>
#include<map>
#include<queue>
#include<stack>
#include<string>
#include<map>
#include<set>
#include<ctime>
#define eps 1e-6
#define LL long long
#define pii (pair<int, int>)
//#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;  

//zoj 2104 动态主席树修改+静态主席树
const int maxn = +;
const int M = ;
int n, q, m, tot;
int a[maxn], t[maxn];
int T[maxn], lson[M], rson[M], c[M];
int S[maxn];
struct Query {
    int kind;
    int l, r, k;
} query[];  

void Init_hash(int k) {
    sort(t, t+k);
    m = unique(t, t+k) - t;
}  

int Hash(int x) {
    return lower_bound(t, t+m, x) - t;
}  

int build(int l, int r) {
    int root = tot++;
    c[root] = ;
    if(l != r) {
        int mid = (l+r) >> ;
        lson[root] = build(l, mid);
        rson[root] = build(mid+, r);
    }
    return root;
}  

int Insert(int root, int pos, int val) {
    int newroot = tot++, tmp = newroot;
    int l = , r = m-;
    c[newroot] = c[root] + val;
    while(l < r) {
        int mid = (l+r)>>;
        if(pos <= mid) {
            lson[newroot] = tot++; rson[newroot] = rson[root];
            newroot = lson[newroot]; root = lson[root];
            r = mid;
        }
        else {
            rson[newroot] = tot++; lson[newroot] = lson[root];
            newroot = rson[newroot]; root = rson[root];
            l = mid+;
        }
        c[newroot] = c[root] + val;
    }
    return tmp;
}  

int lowbit(int x) {
    return x&(-x);
}
int use[maxn];
void add(int x, int pos, int d) {
    while(x <= n) {
        S[x] = Insert(S[x], pos, d);
        x += lowbit(x);
    }
}
int Sum(int x) {
    int ret = ;
    while(x > ) {
        ret += c[lson[use[x]]];
        x -= lowbit(x);
    }
    return ret;
}
int Query(int left, int right, int k) {
    int left_root = T[left-], right_root = T[right];
    int l = , r = m-;
    for(int i = left-; i; i -= lowbit(i)) use[i] = S[i];
    for(int i = right; i; i -= lowbit(i)) use[i] = S[i];
    while(l < r) {
        int mid = (l+r) >> ;
        int tmp = Sum(right) - Sum(left-) + c[lson[right_root]] - c[lson[left_root]];
        if(tmp >= k) {
            r = mid;
            for(int i = left-; i; i -= lowbit(i)) use[i] = lson[use[i]];
            for(int i = right; i; i -= lowbit(i)) use[i] = lson[use[i]];
            left_root = lson[left_root];
            right_root = lson[right_root];
        }
        else {
            l = mid + ;
            k -= tmp;
            for(int i = left-; i; i -= lowbit(i)) use[i] = rson[use[i]];
            for(int i = right; i; i -= lowbit(i)) use[i] = rson[use[i]];
            left_root = rson[left_root];
            right_root = rson[right_root];
        }
    }
    return l;
}  

int main() {
    //freopen("input.txt", "r", stdin);
    int Tcase; cin >> Tcase;
    while(Tcase--) {
        scanf("%d%d", &n, &q);
        tot = ;
        m = ;
        for(int i = ; i <= n; i++) {
            scanf("%d", &a[i]);
            t[m++] = a[i];
        }
        char op[];
        for(int i = ;i < q;i++) {
            scanf("%s",op);
            if(op[] == 'Q') {
                query[i].kind = ;
                scanf("%d%d%d",&query[i].l,&query[i].r,&query[i].k);
            }
            else {
                query[i].kind = ;
                scanf("%d%d", &query[i].l, &query[i].r);
                t[m++] = query[i].r;
            }
        }
        Init_hash(m);
        T[] = build(, m-);
        for(int i = ; i <= n; i++) T[i] = Insert(T[i-], Hash(a[i]), );
        for(int i = ; i <= n; i++) S[i] = T[];
        for(int i = ; i < q; i++) {
            if(query[i].kind == ) printf("%d\n",t[Query(query[i].l,query[i].r,query[i].k)]);
            else {
                add(query[i].l, Hash(a[query[i].l]), -);
                add(query[i].l, Hash(query[i].r), );
                a[query[i].l] = query[i].r;
            }
        }
    }
    return ;
}