【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 }