题目链接：
http://acm.hdu.edu.cn/showproblem.php?pid=1754
题目大意：
求区间最大值+单点修改
解题思路：
直接套用模板即可
1 #include<bits/stdc++.h>
2 #define MID(l, r) (l + (r - l) / 2)
3 #define lson(o) (o * 2)
4 #define rson(o) (o * 2 + 1)
5 using namespace std;
6 typedef long long ll;
7 const int INF = 1e9 +7;
8 const int maxn = 1e6 + 10;
9 int a[maxn];
10 struct node
11 {
12 int l, r, mmax, mmin, sum;
13 }tree[maxn];
14 void build(int o, int l, int r)
15 {
16 tree[o].l = l, tree[o].r = r;
17 if(l == r)
18 {
19 tree[o].mmax = tree[o].mmin = tree[o].sum = a[l];
20 return;
21 }
22 int m = MID(l, r);
23 int lc = lson(o), rc = rson(o);
24 build(lc, l, m);
25 build(rc, m + 1, r);
26 tree[o].mmax = max(tree[lc].mmax, tree[rc].mmax);
27 tree[o].mmin = min(tree[lc].mmin, tree[rc].mmin);
28 tree[o].sum = tree[lc].sum + tree[rc].sum;
29 }
30 int ql, qr;//查询区间[ql, qr]中的max,min,sum
31 int ans_max, ans_min, ans_sum;
32 void query_init()//查询前，将全局变量初始化
33 {
34 ans_max = -INF;
35 ans_min = INF;
36 ans_sum = 0;
37 }
38 void query(int o)
39 {
40 if(ql <= tree[o].l && qr >= tree[o].r)//[L, R]包含在[ql, qr]区间内，直接用该节点的信息，达到线段树查询快的操作
41 {
42 ans_max = max(ans_max, tree[o].mmax);
43 ans_min = min(ans_min, tree[o].mmin);
44 ans_sum += tree[o].sum;
45 return;
46 }
47 int m = MID(tree[o].l, tree[o].r);
48 if(ql <= m)query(lson(o));
49 if(qr > m)query(rson(o));
50 }
51 //单点更新，a[p] = v;
52 int p, v;
53 void update(int o)
54 {
55 if(tree[o].l == tree[o].r)
56 {
57 tree[o].mmax = v;
58 tree[o].mmin = v;
59 tree[o].sum = v;
60 return;
61 }
62 int m = MID(tree[o].l, tree[o].r);
63 int lc = lson(o), rc = rson(o);
64 if(p <= m)update(lc);
65 else update(rc);
66 tree[o].mmax = max(tree[lc].mmax, tree[rc].mmax);
67 tree[o].mmin = min(tree[lc].mmin, tree[rc].mmin);
68 tree[o].sum = tree[lc].sum + tree[rc].sum;
69 }
70 int main()
71 {
72 int n, m;
73 while(scanf("%d%d", &n, &m) != EOF)
74 {
75 for(int i = 1; i <= n; i++)scanf("%d", &a[i]);
76 build(1, 1, n);
77 char s[5];
78 int x, y;
79 while(m--)
80 {
81 scanf("%s%d%d", s, &x, &y);
82 if(s[0] == 'Q')
83 {
84 ql = x, qr = y;
85 query_init();
86 query(1);
87 printf("%d\n", ans_max);
88 }
89 else if(s[0] == 'U')
90 {
91 p = x, v = y;
92 update(1);
93 }
94 }
95 }
96 return 0;
97 }
来源：https://www.cnblogs.com/fzl194/p/9025005.html