hdu I Hate It

 1 #include <stdio.h>
 2 #include <string.h>
 3 #include <math.h>
 4 #include <algorithm>
 5 #include <iostream>
 6 #include <ctype.h>
 7 #include <iomanip>
 8 #include <queue>
 9 #include <stdlib.h>
10 
11 using namespace std;
12 
13 const int maxn = 200000 + 10, INF = -214748364;
14 int a[maxn], b[4*maxn], A, B;        //a为输入数组，maxv为线段树结点数组，存该相应区间的最大值
15 
16 void update(int o, int L, int R)        //线段树更新，把a[A]改为B
17 {
18     if(L == R)     
19         b[o] = B;
20     else
21     {
22         int M = L + (R - L) / 2;
23         if(A <= M)     
24             update(2*o, L, M);
25         else    
26             update(2*o+1, M+1, R);
27         b[o] = max(b[2*o], b[2*o+1]);
28     }
29 }
30 
31 int query(int o, int L, int R)      //线段树查询[A, B]间的最大值
32 {
33     if(A <= L && R <= B)
34        return b[o];
35     int M = L + (R - L) / 2, ans = INF;
36     if(A <= M)   
37         ans = max(ans, query(2*o, L, M));
38     if(B > M)    
39         ans = max(ans, query(2*o+1, M+1, R));
40     return ans;
41 }
42 
43 void build(int o, int L, int R)     //建树
44 {
45 
46     if(L == R)
47     {
48         b[o] = a[L];
49         return;
50     }
51     int M = L + (R - L) / 2;
52     if(L <= M)     
53         build(2*o, L, M);
54     if(R > M)      
55         build(2*o+1, M+1, R);
56     b[o]=max(b[2*o], b[2*o+1]);
57 
58 }
59 int main()
60 {
61     int N, M;
62     char c;
63     while(~scanf("%d%d\n", &N, &M))
64     {
65         for(int i = 1; i <= N; i++) scanf("%d", &a[i]);
66         build(1, 1, N);
67         /*for (int i = 1; i <= 2*N; ++i)
68              {
69                  printf("%d\n",b[i]);    //5 3 5 2 3 4 5 1 2
70              }*/
71         while(M--)
72         {
73             scanf("\n%c%d%d", &c, &A, &B);
74             if(c == ‘U‘){
75              update(1, 1, N);
76              /*for (int i = 1; i <= 2*N; ++i)
77              {
78                  printf("%d\n",b[i]);  //6 6 5 2 6 4 5 1 2
79              }*/
80             }
81             else printf("%d\n", query(1, 1, N));
82         }
83     }
84     return 0;
85 }