洛谷 P2880 【[USACO07JAN]平衡的陣容Balanced Lineup】

P2880 [USACO07JAN]平衡的陣容Balanced Lineup

大體題意

給你一個區間，然後讓你求得這個區間內最大值和最小值的差.

思路：

這不就是一個SB線段樹維護區間和和區間差嘛，都不用待修改，多簡單

好我們來看一下線段樹維護區間最大值:

很顯然啊，如果你會線段樹的話很容易就明白了，不會的話這裏

int query_max(int rt, int l, int r, int L, int R) {
	if (L <= l && r <= R) return tree[rt].max;//如果在區間內，那麼直接返回區間最大值
	int mid = (l + r) >> 1, maxn = -1;//因爲要取ma所以搞成-1
	if (L <= mid) maxn = max(maxn, query_max(lson, l, mid, L, R));//對兩邊區間都要取max
	if (R > mid) maxn = max(maxn, query_max(rson, mid + 1, r, L, R));
	return maxn;
}

這樣的話區間查詢最小值就很容易了.

int query_min(int rt, int l, int r, int L, int R) {
	if (L <= l && r <= R) return tree[rt].min;
	int mid = (l + r) >> 1, minn = 1e9 + 7;
	if (L <= mid) minn = min(minn, query_min(lson, l, mid, L, R));
	if (R > mid) minn = min(minn, query_min(rson, mid + 1, r, L, R));
	return minn;
}

code

#include <bits/stdc++.h>

#define N 100010
#define M 1010
#define lson rt << 1
#define rson rt << 1 | 1

using namespace std;
int n, q;
struct node {
	int max, min;
}tree[N << 1];

int read() {
	int s = 0, f = 0; char ch = getchar();
	while (!isdigit(ch)) f |= (ch == '-'), ch = getchar();
	while (isdigit(ch)) s = s * 10 + (ch ^ 48), ch = getchar();
	return f ? -s : s;
}

void push_up(int rt) {
	tree[rt].max = max(tree[lson].max, tree[rson].max);
	tree[rt].min = min(tree[lson].min, tree[rson].min);
}

void build(int rt, int l, int r) {
	if (l == r) {
		tree[rt].max = tree[rt].min = read();
		return;
	}
	int mid = (l + r) >> 1;
	build(lson, l, mid);
	build(rson, mid + 1, r);
	push_up(rt);
}

int query_max(int rt, int l, int r, int L, int R) {
	if (L <= l && r <= R) return tree[rt].max;
	int mid = (l + r) >> 1, maxn = -1;
	if (L <= mid) maxn = max(maxn, query_max(lson, l, mid, L, R));
	if (R > mid) maxn = max(maxn, query_max(rson, mid + 1, r, L, R));
	return maxn;
}

int query_min(int rt, int l, int r, int L, int R) {
	if (L <= l && r <= R) return tree[rt].min;
	int mid = (l + r) >> 1, minn = 1e9 + 7;
	if (L <= mid) minn = min(minn, query_min(lson, l, mid, L, R));
	if (R > mid) minn = min(minn, query_min(rson, mid + 1, r, L, R));
	return minn;
}

int main() {
	n = read(), q = read();
	build(1, 1, n);
	for (int i = 1, x, y; i <= q; i++) {
		x = read(), y = read();
		int ma = query_max(1, 1, n, x, y);
		int mi = query_min(1, 1, n, x, y);
		printf("%d\n", ma - mi);
	}
	return 0;
}