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：
1 #include <set> 2 #include <map> 3 #include <deque> 4 #include <queue> 5 #include <stack> 6 #include <cmath> 7 #include <ctime> 8 #include <bitset> 9 #include <cstdio> 10 #include <string> 11 #include <vector> 12 #include <cstdlib> 13 #include <cstring> 14 #include <iostream> 15 #include <algorithm> 16 using namespace std; 17 18 typedef long long LL; 19 typedef pair<LL, LL> pLL; 20 typedef pair<LL, int> pLi; 21 typedef pair<int, LL> pil;; 22 typedef pair<int, int> pii; 23 typedef unsigned long long uLL; 24 25 #define lson rt<<1 26 #define rson rt<<1|1 27 #define lowbit(x) x&(-x) 28 #define name2str(name) (#name) 29 #define bug printf("********* ") 30 #define debug(x) cout<<#x"=["<<x<<"]" <<endl 31 #define FIN freopen("D://code//in.txt","r",stdin) 32 #define IO ios::sync_with_stdio(false),cin.tie(0) 33 34 const double eps = 1e-8; 35 const int mod = 1000000007; 36 const int maxn = 5e4 + 7; 37 const double pi = acos(-1); 38 const int inf = 0x3f3f3f3f; 39 const LL INF = 0x3f3f3f3f3f3f3f3fLL; 40 41 char op[5]; 42 int n, q, x, y, tot1, tot2, cnt; 43 int a[maxn], q1[maxn], q2[maxn], root1[maxn], root2[maxn]; 44 vector<int> v; 45 46 struct que { 47 int op, l, r, x; 48 }ask[maxn]; 49 50 struct node { 51 int l, r, sum; 52 }tree[maxn*40]; 53 54 int getid(int x) { 55 return lower_bound(v.begin(), v.end(), x) - v.begin() + 1; 56 } 57 58 void update(int l, int r, int& x, int y, int pos, int val) { 59 tree[++cnt] = tree[y], tree[cnt].sum += val, x = cnt; 60 if(l == r) return; 61 int mid = (l + r) >> 1; 62 if(pos <= mid) update(l, mid, tree[x].l, tree[y].l, pos, val); 63 else update(mid + 1, r, tree[x].r, tree[y].r, pos, val); 64 } 65 66 int query(int l, int r, int x, int y, int k) { 67 if(l == r) return l; 68 int cnt = tree[tree[y].l].sum - tree[tree[x].l].sum; 69 for(int i = 0; i < tot1; ++i) { 70 cnt -= tree[tree[q1[i]].l].sum; 71 } 72 for(int i = 0; i < tot2; ++i) { 73 cnt += tree[tree[q2[i]].l].sum; 74 } 75 int mid = (l + r) >> 1; 76 if(cnt >= k) { 77 for(int i = 0; i < tot1; ++i) { 78 q1[i] = tree[q1[i]].l; 79 } 80 for(int i = 0; i < tot2; ++i) { 81 q2[i] = tree[q2[i]].l; 82 } 83 return query(l, mid, tree[x].l, tree[y].l, k); 84 } else { 85 for(int i = 0; i < tot1; ++i) { 86 q1[i] = tree[q1[i]].r; 87 } 88 for(int i = 0; i < tot2; ++i) { 89 q2[i] = tree[q2[i]].r; 90 } 91 return query(mid + 1, r, tree[x].r, tree[y].r, k - cnt); 92 } 93 } 94 95 int main(){ 96 #ifndef ONLINE_JUDGE 97 FIN; 98 #endif 99 int t; 100 scanf("%d", &t); 101 while(t--) { 102 scanf("%d%d", &n, &q); 103 cnt = tot1 = tot2 = 0; 104 v.clear(); 105 memset(root2, 0, sizeof(root2)); 106 for(int i = 1; i <= n; ++i) { 107 scanf("%d", &a[i]); 108 v.push_back(a[i]); 109 } 110 for(int i = 1; i <= q; ++i) { 111 scanf("%s", op); 112 if(op[0] == 'C') ask[i].op = 1; 113 else ask[i].op = 2; 114 if(ask[i].op == 1) { 115 scanf("%d%d", &ask[i].l, &ask[i].x); 116 v.push_back(ask[i].x); 117 } else { 118 scanf("%d%d%d", &ask[i].l, &ask[i].r, &ask[i].x); 119 } 120 } 121 sort(v.begin(), v.end()); 122 v.erase(unique(v.begin(), v.end()), v.end()); 123 int sz = v.size(); 124 for(int i = 1; i <= n; ++i) { 125 a[i] = getid(a[i]); 126 update(1, sz, root1[i], root1[i-1], a[i], 1); 127 } 128 for(int i = 1; i <= q; ++i) { 129 int op = ask[i].op; 130 if(op == 1) { 131 int pos = ask[i].l, x = ask[i].x, y = a[pos]; 132 x = getid(x); 133 a[pos] = x; 134 while(pos <= n) { 135 update(1, sz, root2[pos], root2[pos], y, -1); 136 update(1, sz, root2[pos], root2[pos], x, 1); 137 pos += lowbit(pos); 138 } 139 } else { 140 int l = ask[i].l, r = ask[i].r, k = ask[i].x; 141 int x = l - 1; 142 tot1 = tot2 = 0; 143 while(x) { 144 q1[tot1++] = root2[x]; 145 x -= lowbit(x); 146 } 147 x = r; 148 while(x) { 149 q2[tot2++] = root2[x]; 150 x -= lowbit(x); 151 } 152 printf("%d ", v[query(1, sz, root1[l-1], root1[r], k)-1]); 153 } 154 } 155 } 156 return 0; 157 }