Sample Input
5 6
1 2 3 4 5
Q 1 5 //1-5结点的最大值
U 3 6 //将点3的数值换成6
Q 3 4
Q 4 5
U 2 9
Q 1 5
Sample Output
5
6
5
9
1 # include <iostream> 2 # include <cstdio> 3 # include <cstring> 4 # include <algorithm> 5 # include <cmath> 6 # include <queue> 7 # define LL long long 8 using namespace std ; 9 10 const int maxn = 200010; 11 12 int MAX[maxn<<2] ; //结点开4倍 13 14 void PushUP(int rt) //更新到父节点 15 { 16 MAX[rt] = max(MAX[rt * 2] , MAX[rt * 2 + 1] ); //rt 为当前结点 17 } 18 19 void build(int l , int r , int rt) //构建线段树 20 { 21 if (l == r) 22 { 23 scanf("%d" , &MAX[rt]) ; 24 return ; 25 } 26 int m = (l + r) / 2 ; 27 build(l , m , rt * 2) ; 28 build(m + 1 , r , rt * 2 +1) ; 29 PushUP(rt) ; 30 } 31 32 void updata(int p , int add , int l , int r , int rt) //单点替换 33 { 34 if (l == r) 35 { 36 MAX[rt] = add ; 37 return ; 38 } 39 int m = (l + r) / 2 ; 40 if (p <= m) 41 updata(p , add , l , m , rt * 2) ; 42 else 43 updata(p , add , m + 1 , r , rt * 2 + 1) ; 44 PushUP(rt) ; 45 } 46 47 int query(int L , int R , int l , int r , int rt) //区间求最大值 48 { 49 if (L <= l && r <= R) 50 return MAX[rt] ; 51 int m = (l + r) / 2 ; 52 int ret = 0 ; 53 if (L <= m) 54 ret = max(ret , query(L , R , l , m , rt * 2) ) ; 55 if (R > m) 56 ret = max(ret , query(L , R , m + 1 , r , rt * 2 + 1) ); 57 return ret ; 58 } 59 60 int main () 61 { 62 //freopen("in.txt","r",stdin) ; 63 int n , m ; 64 while(scanf("%d %d" , &n , &m) != EOF) 65 { 66 build(1 , n , 1) ; 67 char op[10] ; 68 while(m--) 69 { 70 scanf("%s" , op) ; 71 if (op[0] == 'E') //结束 72 break ; 73 int a , b ; 74 scanf("%d %d" , &a , &b) ; 75 if (op[0] == 'Q') //求a,b区间的和最大值 76 printf("%d ", query(a , b , 1 , n , 1)) ; 77 else if (op[0] == 'U') //将第a个地方的值改为b 78 updata(a , b , 1 , n , 1) ; 79 } 80 } 81 82 return 0 ; 83 }