【POJ】Buy Tickets（线段树）

点更新用法很巧妙的一道题。倒序很容易的找到该人的位置，而update操作中需要不断的变化下标才能合理的插入。看了别人写的代码，学习了。

 

 1 #include <iostream>
 2 #include <cstring>
 3 #include <cstdio>
 4 #include <map>
 5 using namespace std;
 6 
 7 const int maxn = 2e5 + 10;
 8 int sumv[maxn << 2], data[maxn];
 9 
10 void PushUp (int rt) {
11     sumv[rt] = sumv[rt << 1] + sumv[rt << 1 | 1];
12 }
13 
14 void build (int l, int r, int rt) {
15     if (l == r) {
16         sumv[rt] = 1;
17         return ;
18     }
19     int m = (l + r) / 2;
20     build (l, m, rt << 1);
21     build (m + 1, r, rt << 1 | 1);
22     PushUp (rt);
23 }
24 
25 void update (int p, int v, int l, int r, int rt) {
26     if (l == r) {
27         data[l] = v; sumv[rt] = 0;
28         return ;
29     }
30     int m = (l + r) >> 1;
31     if (sumv[rt * 2] >= p) {
32         update (p, v, l, m, rt * 2);
33     } else {
34         update (p - sumv[rt * 2], v, m + 1, r, rt * 2 + 1);
35     }
36     PushUp(rt);
37 }
38 
39 int main () {
40     pair<int, int> p[maxn];
41     int n;
42     while (~scanf ("%d", &n)) {
43         for (int i = 0; i < n; ++ i) {
44             scanf ("%d%d", &p[i].first, &p[i].second);
45         }
46         build (1, n, 1);
47         for (int i = n - 1; i >= 0; -- i) {
48             update (p[i].first + 1, p[i].second, 1, n, 1);
49         }
50         for (int i = 1; i <= n; ++ i) {
51             if (i == 1) {
52                 printf ("%d", data[i]);
53             } else {
54                 printf (" %d", data[i]);
55             }
56         }
57         puts("");
58     }
59     return 0;
60 }

 

转载于:https://www.cnblogs.com/Destiny-Gem/p/3879892.html