题目：poj.org/problem?id=3264
题意：求一段区间内最大值与最小值的差。
看到区间最值首先想到RMQ--ST算法。但本题出现在了kuangbin专题的线段树里。
用线段树也无思维难点，但有两个坑：
1. 查询函数中，区间不包含时的返回值。
2.用cin,cout会TLE。用c的输入方式。
1 #include<iostream> 2 #include<cstring> 3 #include<cstdio> 4 #include<string> 5 #include<vector> 6 #include<list> 7 #include<cmath> 8 #include<set> 9 #include<utility> 10 #include<algorithm> 11 using namespace std; 12 13 14 #define maxn 50010 15 int N, Q; 16 int cow[maxn]; 17 struct node 18 { 19 int minn, maxx; 20 }tree[maxn*4]; 21 22 void build_tree(int l,int r,int i) 23 { 24 if (l == r) 25 { 26 tree[i].maxx = cow[l]; 27 tree[i].minn = cow[l]; 28 return; 29 } 30 int mid = (l + r) / 2; 31 build_tree(l, mid, i + i); 32 build_tree(mid + 1, r, i + i + 1); 33 tree[i].maxx = max(tree[i + i].maxx, tree[i + i + 1].maxx); 34 tree[i].minn = min(tree[i + i].minn, tree[i + i + 1].minn); 35 } 36 int querymax(int tl, int tr, int l, int r, int i) 37 { 38 if (tl > r || tr < l)return 0; 39 if (tl <= l&&r <= tr) 40 { 41 return tree[i].maxx; 42 } 43 int mid = (l + r) / 2; 44 return max(querymax(tl, tr, l, mid, i + i) , querymax(tl, tr, mid + 1, r, i + i + 1)); 45 } 46 int querymin(int tl, int tr, int l, int r, int i) 47 { 48 if (tl > r || tr < l) 49 return INT_MAX;//这里return返回的值要注意 50 if (tl <= l&&r <= tr) 51 { 52 return tree[i].minn; 53 } 54 int mid = (l + r) / 2; 55 return min(querymin(tl, tr, l, mid, i + i), querymin(tl, tr, mid + 1, r, i + i + 1)); 56 } 57 int main() 58 { 59 60 scanf("%d%d", &N, &Q); 61 for (int i = 1; i <= N; i++) 62 scanf("%d", &cow[i]); 63 build_tree(1, N, 1); 64 int a, b; 65 for (int i = 1; i <= Q; i++) 66 { 67 scanf("%d%d", &a, &b); 68 if (a == b) 69 { 70 cout << "0" << endl; continue; 71 } 72 int qjmax = querymax(a, b, 1, N, 1); 73 int qjmin = querymin(a, b, 1, N, 1); 74 printf("%d ", qjmax - qjmin); 75 } 76 return 0; 77 }