题目传送门
1 /*
2 结点存储下面有几个空位
3 每次从根结点往下找找到该插入的位置，
4 同时更新每个节点的值
5 */
6 #include <cstdio>
7 #define lson l, m, rt << 1
8 #define rson m+1, r, rt << 1 | 1
9
10 const int MAX_N = 200000 + 10;
11 int pos[MAX_N], val[MAX_N];
12 int sum[MAX_N << 2];
13 int que[MAX_N];
14 int id;
15
16 void build(int l, int r, int rt)
17 {
18 sum[rt] = r - l + 1; //节点有多少个空位置
19 if (l == r) return ;
20 int m = (l + r) >> 1;
21 build (lson);
22 build (rson);
23 }
24
25 void update(int p, int l, int r, int rt)
26 {
27 sum[rt]--; //插入一个减少一个空位置
28 if (l == r) //找到了传递该位置
29 {
30 id = l;
31 return ;
32 }
33 int m = (l + r) >> 1;
34 if (p <= sum[rt<<1])
35 {
36 update (p, lson);
37 }
38 else
39 {
40 p -= sum[rt<<1]; //减去
41 update (p, rson);
42 }
43 }
44
45 int main(void) //POJ 2828 Buy tickets
46 {
47 //freopen ("inE.txt", "r", stdin);
48 int n;
49
50 while (~scanf ("%d", &n))
51 {
52 build (1, n, 1);
53 for (int i=1; i<=n; ++i)
54 {
55 scanf ("%d%d", &pos[i], &val[i]);
56 }
57 for (int i=n; i>=1; --i) //最后插入的位置不变
58 {
59 update (pos[i]+1, 1, n, 1);
60 que[id] = val[i];
61 }
62 printf ("%d", que[1]);
63 for (int i=2; i<=n; ++i)
64 {
65 printf (" %d", que[i]);
66 }
67 puts ("");
68 }
69
70 return 0;
71 }