地址：http://acm.hdu.edu.cn/showproblem.php?pid=1754
题意：给n个数字。m次操作，每次操作更新一个数字或者查询区间最大值。
mark：典型线段树题。不过a的时候学了一下树状数组求区间最值。感觉对树状数组的理解又深刻了一点。不过这个更新不是O(lgn)而是O(lgn*lgn)的，比线段树慢！
代码：
线段树：
1 # include <stdio.h> 2 # include <string.h> 3 4 5 #define max(a,b) (a>b?a:b) 6 int n ; 7 int tr[200010<<2] ; 8 9 10 void u(int a, int b, int l, int r, int rt) 11 { 12 int m = (l+r)/2 ; 13 if (l == r) { 14 tr[rt] = b; 15 return; 16 } 17 if (a <= m) u(a, b, l, m, rt*2); 18 else u(a, b, m+1, r, rt*2+1); 19 tr[rt] = max(tr[rt*2], tr[rt*2+1]); 20 } 21 22 23 int q(int a, int b, int l, int r, int rt) 24 { 25 int m = (l + r) / 2, a1, a2 ; 26 if (a == l && b == r) return tr[rt]; 27 if (b <= m) return q(a, b, l, m, rt*2) ; 28 else if (a > m) return q(a, b, m+1, r, rt*2+1) ; 29 a1 = q(a, m, l, m, rt*2), a2 = q(m+1, b, m+1, r, rt*2+1) ; 30 return max(a1, a2) ; 31 } 32 33 34 int main () 35 { 36 int m, i, a, b ; 37 char op[5] ; 38 39 while (~scanf ("%d%d", &n, &m)) 40 { 41 //memset (tr, 0, sizeof(tr)); 42 for (i = 1 ; i <= n ; i++) 43 scanf ("%d", &a), u(i, a, 1, n, 1); 44 while (m--) 45 { 46 scanf ("%s %d %d", op, &a, &b) ; 47 if (op[0] == 'U') u(a, b, 1, n, 1); 48 else printf ("%d ", q(a, b, 1, n, 1)) ; 49 } 50 } 51 return 0 ; 52 }
树状数组（不加inline优化max函数的话会TLE！！！）
1 # include <stdio.h> 2 # include <string.h> 3 4 5 int n, num[200010], tr[200010]; 6 int lowbit(int x){return x & -x;} 7 inline int max(int a, int b){return a>b?a:b;} 8 9 10 void u(int a, int b){ 11 int i, j; 12 num[a] = b; 13 for (i = a; i <= n; i+=lowbit(i)) 14 { 15 tr[i] = num[i]; 16 for (j = 1; j < lowbit(i); j <<= 1) 17 tr[i] = max(tr[i], tr[i-j]); 18 } 19 } 20 21 22 int q(int a, int b){ 23 int i = b, ans = num[b]; 24 while (i >= a) 25 if (i-lowbit(i)+1 >= a) 26 ans = max(ans, tr[i]), i -= lowbit(i); 27 else 28 ans = max(ans, num[i]), i--; 29 return ans; 30 } 31 32 33 int main () 34 { 35 int a, b, i, m; 36 char op[5]; 37 while (~scanf ("%d%d", &n, &m)) 38 { 39 memset(tr, 0, sizeof(tr)); 40 for (i = 1; i <= n; i++) 41 scanf ("%d", &a), u(i, a); 42 while (m--) 43 { 44 scanf ("%s %d %d", op, &a, &b); 45 if (op[0] == 'U') u(a, b); 46 else printf ("%d ", q(a, b)) ; 47 } 48 } 49 return 0 ; 50 }