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 ", 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 }