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 = 60000+1000; const int M = 2400000; 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[10010]; 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] = 0; if(l != r) { int mid = (l+r) >> 1; lson[root] = build(l, mid); rson[root] = build(mid+1, r); } return root; } int Insert(int root, int pos, int val) { int newroot = tot++, tmp = newroot; int l = 0, r = m-1; c[newroot] = c[root] + val; while(l < r) { int mid = (l+r)>>1; 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+1; } 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 = 0; while(x > 0) { ret += c[lson[use[x]]]; x -= lowbit(x); } return ret; } int Query(int left, int right, int k) { int left_root = T[left-1], right_root = T[right]; int l = 0, r = m-1; for(int i = left-1; 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) >> 1; int tmp = Sum(right) - Sum(left-1) + c[lson[right_root]] - c[lson[left_root]]; if(tmp >= k) { r = mid; for(int i = left-1; 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 + 1; k -= tmp; for(int i = left-1; 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 = 0; m = 0; for(int i = 1; i <= n; i++) { scanf("%d", &a[i]); t[m++] = a[i]; } char op[10]; for(int i = 0;i < q;i++) { scanf("%s",op); if(op[0] == 'Q') { query[i].kind = 0; scanf("%d%d%d",&query[i].l,&query[i].r,&query[i].k); } else { query[i].kind = 1; scanf("%d%d", &query[i].l, &query[i].r); t[m++] = query[i].r; } } Init_hash(m); T[0] = build(0, m-1); for(int i = 1; i <= n; i++) T[i] = Insert(T[i-1], Hash(a[i]), 1); for(int i = 1; i <= n; i++) S[i] = T[0]; for(int i = 0; i < q; i++) { if(query[i].kind == 0) printf("%d ",t[Query(query[i].l,query[i].r,query[i].k)]); else { add(query[i].l, Hash(a[query[i].l]), -1); add(query[i].l, Hash(query[i].r), 1); a[query[i].l] = query[i].r; } } } return 0; }