Dynamic Rankings（动态第k大+树套树）

题目链接：

　　http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=1112

题目：

思路：

　　树套树板子题。

代码实现如下：

 #include <set>
 #include <map>
 #include <deque>
 #include <queue>
 #include <stack>
 #include <cmath>
 #include <ctime>
 #include <bitset>
 #include <cstdio>
 #include <string>
 #include <vector>
 #include <cstdlib>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 using namespace std;

 typedef long long LL;
 typedef pair<LL, LL> pLL;
 typedef pair<LL, int> pLi;
 typedef pair<int, LL> pil;;
 typedef pair<int, int> pii;
 typedef unsigned long long uLL;

 #define lson rt<<1
 #define rson rt<<1|1
 #define lowbit(x) x&(-x)
 #define name2str(name) (#name)
 #define bug printf("*********\n")
 #define debug(x) cout<<#x"=["<<x<<"]" <<endl
 #define FIN freopen("D://code//in.txt","r",stdin)
 #define IO ios::sync_with_stdio(false),cin.tie(0)

 const double eps = 1e-;
 const int mod = ;
 const int maxn = 5e4 + ;
 const double pi = acos(-);
 const int inf = 0x3f3f3f3f;
 const LL INF = 0x3f3f3f3f3f3f3f3fLL;

 char op[];
 int n, q, x, y, tot1, tot2, cnt;
 int a[maxn], q1[maxn], q2[maxn], root1[maxn], root2[maxn];
 vector<int> v;

 struct que {
     int op, l, r, x;
 }ask[maxn];

 struct node {
     int l, r, sum;
 }tree[maxn*];

 int getid(int x) {
     return lower_bound(v.begin(), v.end(), x) - v.begin() + ;
 }

 void update(int l, int r, int& x, int y, int pos, int val) {
     tree[++cnt] = tree[y], tree[cnt].sum += val, x = cnt;
     if(l == r) return;
     int mid = (l + r) >> ;
     if(pos <= mid) update(l, mid, tree[x].l, tree[y].l, pos, val);
     else update(mid + , r, tree[x].r, tree[y].r, pos, val);
 }

 int query(int l, int r, int x, int y, int k) {
     if(l == r) return l;
     int cnt = tree[tree[y].l].sum - tree[tree[x].l].sum;
     for(int i = ; i < tot1; ++i) {
         cnt -= tree[tree[q1[i]].l].sum;
     }
     for(int i = ; i < tot2; ++i) {
         cnt += tree[tree[q2[i]].l].sum;
     }
     int mid = (l + r) >> ;
     if(cnt >= k) {
         for(int i = ; i < tot1; ++i) {
             q1[i] = tree[q1[i]].l;
         }
         for(int i = ; i < tot2; ++i) {
             q2[i] = tree[q2[i]].l;
         }
         return query(l, mid, tree[x].l, tree[y].l, k);
     } else {
         for(int i = ; i < tot1; ++i) {
             q1[i] = tree[q1[i]].r;
         }
         for(int i = ; i < tot2; ++i) {
             q2[i] = tree[q2[i]].r;
         }
         return query(mid + , r, tree[x].r, tree[y].r, k - cnt);
     }
 }

 int main(){
 #ifndef ONLINE_JUDGE
     FIN;
 #endif
     int t;
     scanf("%d", &t);
     while(t--) {
         scanf("%d%d", &n, &q);
         cnt = tot1 = tot2 = ;
         v.clear();
         memset(root2, , sizeof(root2));
         for(int i = ; i <= n; ++i) {
             scanf("%d", &a[i]);
             v.push_back(a[i]);
         }
         for(int i = ; i <= q; ++i) {
             scanf("%s", op);
             if(op[] == 'C') ask[i].op = ;
             else ask[i].op = ;
             if(ask[i].op == ) {
                 scanf("%d%d", &ask[i].l, &ask[i].x);
                 v.push_back(ask[i].x);
             } else {
                 scanf("%d%d%d", &ask[i].l, &ask[i].r, &ask[i].x);
             }
         }
         sort(v.begin(), v.end());
         v.erase(unique(v.begin(), v.end()), v.end());
         int sz = v.size();
         for(int i = ; i <= n; ++i) {
             a[i] = getid(a[i]);
             update(, sz, root1[i], root1[i-], a[i], );
         }
         for(int i = ; i <= q; ++i) {
             int op = ask[i].op;
             if(op == ) {
                 int pos = ask[i].l, x = ask[i].x, y = a[pos];
                 x = getid(x);
                 a[pos] = x;
                 while(pos <= n) {
                     update(, sz, root2[pos], root2[pos], y, -);
                     update(, sz, root2[pos], root2[pos], x, );
                     pos += lowbit(pos);
                 }
             } else {
                 int l = ask[i].l, r = ask[i].r, k = ask[i].x;
                 int x = l - ;
                 tot1 = tot2 = ;
                 while(x) {
                     q1[tot1++] = root2[x];
                     x -= lowbit(x);
                 }
                 x = r;
                 while(x) {
                     q2[tot2++] = root2[x];
                     x -= lowbit(x);
                 }
                 printf("%d\n", v[query(, sz, root1[l-], root1[r], k)-]);
             }
         }
     }
     return ;
 }