HDU 3264 Balanced Lineup（线段树，最值查询）

ACM

题目地址：POJ 3264 Balanced Lineup

题意： 
求区间内最大值-最小值。

分析： 
不用写修改，容易多了。

代码：

/**  Author:      illuz <iilluzen[at]gmail.com>*  Blog:        http://blog.csdn.net/hcbbt*  File:        3264.cpp*  Create Date: 2014-08-05 00:15:22*  Descripton:   */#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;#define repf(i,a,b) for(int i=(a);i<=(b);i++)#define lson(x) ((x) << 1)#define rson(x) ((x) << 1 | 1)typedef long long ll;const int N = 50000;const int ROOT = 1;// below is sement point updatedstruct seg {ll mmax;ll mmin;};struct segment_tree { seg node[N << 2];void update(int pos) {node[pos].mmax = max(node[lson(pos)].mmax, node[rson(pos)].mmax);node[pos].mmin = min(node[lson(pos)].mmin, node[rson(pos)].mmin);}void build(int l, int r, int pos) {if (l == r) {scanf("%lld", &node[pos].mmin);node[pos].mmax = node[pos].mmin;return;}int m = (l + r) >> 1;build(l, m, lson(pos));build(m + 1, r, rson(pos));update(pos);}// query the segment [x, y]ll query(int l, int r, int pos, int x, int y, bool ismax) {if (x <= l && r <= y) {return ismax ? node[pos].mmax : node[pos].mmin;}int m = (l + r) >> 1;ll res = ismax ? 0 : 1e6 + 1;if (x <= m)res = query(l, m, lson(pos), x, y, ismax);if (y > m) {res = ismax ? max(res, query(m + 1, r, rson(pos), x, y, ismax)) :min(res, query(m + 1, r, rson(pos), x, y, ismax));}return res;}} sgm;int n, q, l, r;int main() {while (~scanf("%d%d", &n, &q)) {sgm.build(1, n, ROOT);while (q--) {scanf("%d%d", &l, &r);printf("%lld\n", sgm.query(1, n, ROOT, l, r, 1) - sgm.query(1, n, ROOT, l, r, 0));}}return 0;}