ZOJ2112--Dynamic Rankings (动态区间第k大)

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

主席树动态第k大基本可以手写，，但是感觉理解还不是很深。另外定义两个数组，一个用来新建一颗主席树，所有修改的结果都在这个上面进行，而另一个是记录中间值，相当于temp的效果，不改变原始数组。。（挖个坑）

附主席树专题链接：http://acm.hust.edu.cn/vjudge/contest/view.action?cid=63941#overview

 #include <cstdio>
 #include <cstdlib>
 #include <cstring>
 #include <algorithm>
 #include <iostream>
 using namespace std;
 typedef long long ll;
 const int maxn = 5e4+;
 int n,m,tot,idx;
 ll a[maxn],vec[maxn*];
 struct
 {
     int x,y,k,flag,idx;
 } Q[maxn];

 //   主席树
 int lson[maxn*],rson[maxn*],c[maxn*],root[maxn]; //依次为左儿子 右儿子 线段树  根节点
 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 update(int root,int pos,int val)
 {
     int new_root = tot++;
     int tmp = new_root;
     int l = ,r = idx;
     c[new_root] = c[root] + val;
     while (l < r)
     {
         int mid = (l + r) >> ;
         if (pos <= mid)
         {
             rson[new_root] = rson[root];
             root = lson[root];
             lson[new_root] = tot++;
             new_root = lson[new_root];
             r = mid;
         }
         else
         {
             lson[new_root] = lson[root];
             root = rson[root];
             rson[new_root] = tot++;
             new_root = rson[new_root];
             l = mid + ;
         }
         c[new_root] = c[root] + val;
     }
     return tmp;
 }
 //  树状数组维护
 int s[maxn],use[maxn];
 inline int lowbit (int x)
 {
     return x & -x;
 }
 void add(int k,int pos,int d)
 {
     while (k <= n)
     {
         s[k] = update(s[k],pos,d);
         k += lowbit(k);
     }
 }
 int sum(int pos)
 {
     int res = ;
     while (pos)
     {
         res += c[lson[use[pos]]];
         pos -= lowbit(pos);
     }
     return res;
 }
 int query(int left,int right,int k)
 {
     int l_root = root[left-];
     int r_root = root[right];
     for (int i = left-; i > ; i -= lowbit(i))
         use[i] = s[i];
     for (int i = right; i > ; i -= lowbit(i))
         use[i] =s[i];
     int l = ,r = idx;
     while (l < r)
     {
         int t = sum(right) - sum(left-) + c[lson[r_root]] - c[lson[l_root]];
         int mid = (l + r) >> ;
         if (t >= k)
         {
             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]];
             l_root = lson[l_root];
             r_root = lson[r_root];
             r = mid;
         }
         else
         {
             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]];
             l_root = rson[l_root];
             r_root = rson[r_root];
             k -= t;
             l = mid + ;
         }
     }
     return l;
 }

 void read()
 {
     scanf ("%d%d",&n,&m);
     for (int i = ; i <= n; i++)
     {
         scanf ("%lld",a+i);
         vec[idx++] = a[i];
     }
     for (int i = ; i < m; i++)
     {
         char op[];
         scanf ("%s",op);
         if (op[] == 'C')
         {
             scanf ("%d%d",&Q[i].x,&Q[i].y);
             Q[i].flag = ;
             vec[idx++] = Q[i].y;
         }
         if (op[] == 'Q')
         {
             scanf ("%d%d%d",&Q[i].x,&Q[i].y,&Q[i].k);
             Q[i].flag = ;
         }
     }
 }
 int main(void)
 {
 #ifndef ONLINE_JUDGE
     freopen("in.txt","r",stdin);
 #endif
     int T;
     scanf ("%d",&T);
     while (T--)
     {
         idx = tot = ;
         read();
         sort(vec,vec+idx);                   //离散化
         idx = unique(vec,vec+idx) - vec;
         root[] = build(,idx);
         for (int i = ; i <= n; i++)
         {
             int tmp = lower_bound(vec,vec+idx,a[i]) - vec ;
             root[i] = update(root[i-],tmp,);
         }
         for (int i = ; i <= n; i++)
             s[i] = root[];
         for (int i = ; i < m; i++)
         {
             if (Q[i].flag == )
             {
                 int tmp1 = lower_bound(vec,vec+idx,a[Q[i].x]) - vec ;
                 int tmp2 = lower_bound(vec,vec+idx,Q[i].y) - vec ;
                 add(Q[i].x,tmp1,-);
                 add(Q[i].x,tmp2,);
                 a[Q[i].x] = Q[i].y;
             }
             if (Q[i].flag == )
             {
                 printf("%lld\n",vec[query(Q[i].x,Q[i].y,Q[i].k)]);
             }
         }
     }
     return ;
 }