POJ 3264: Balanced Lineup

2019独角兽企业重金招聘Python工程师标准>>>

题目在此

解题思路：查询区间最大值/最小值之差，最基础的线段树应用。

代码：


#include <cstdio>// 线段树为完全二叉树，因此可用 root * 2 和 root * 2 + 1
// 来求得左右孩子节点的位移
#define L(root) ((root) << 1)
#define R(root) (((root) << 1) + 1)
#define MAX(a, b) ((a) > (b) ? (a) : (b))
#define MIN(a, b) ((a) < (b) ? (a) : (b))const int MAXN = 50001;
int cows[MAXN];
int maxn, minn;struct st {// 左右区间int left, right;// 区间最大、最小值int max, min;
} st[MAXN * 4];// 建树
void build(int root, int l, int r) {// 区间为叶子节点时，初始化各值后返回st[root].left = l, st[root].right = r;if (l == r) {st[root].max = st[root].min = cows[l];return;}// 否则，递归构建左右节点int m = l + ((r - l) >> 1);build(L(root), l, m);build(R(root), m + 1, r);st[root].max = MAX(st[L(root)].max, st[R(root)].max);st[root].min = MIN(st[L(root)].min, st[R(root)].min);
}// 查询
void query(int root, int l, int r) {// 查询范围与节点区间范围恰好重合if (st[root].left == l && st[root].right == r) {maxn = MAX(maxn, st[root].max);minn = MIN(minn, st[root].min);return;}// 否则，将区间一分为二int m = st[root].left + ((st[root].right - st[root].left) >> 1);// 查询范围“落”在左子节点区间if (r <= m) {query(L(root), l, r);// 查询范围“落”在右子节点区间} else if (l > m) {query(R(root), l, r);// 左右子节点区间各有一部分被查询范围覆盖：// 将查询范围一分为二，分别查询左右子节点} else {query(L(root), l, m);query(R(root), m + 1, r);}
}int main() {int N, Q;while (scanf("%d%d", &N, &Q) != EOF) {for (int i = 1; i <= N; ++i) {scanf("%d", &cows[i]);}build(1, 1, N);int l, r;while (Q--) {scanf("%d%d", &l, &r);maxn = 0, minn = 0x7FFFFFFF;query(1, l, r);printf("%d\n", maxn - minn);}}return 0;
}

转载于:https://my.oschina.net/Alexanderzhou/blog/204056

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