79 lines
2.3 KiB
C++
79 lines
2.3 KiB
C++
#include <algorithm>
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#include <cstdio>
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#include <cstring>
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using namespace std;
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const int N = 2e5 + 1000;
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// 快读
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int read() {
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int x = 0, f = 1;
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char ch = getchar();
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while (ch < '0' || ch > '9') {
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if (ch == '-') f = -1;
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ch = getchar();
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}
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while (ch >= '0' && ch <= '9') {
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x = (x << 3) + (x << 1) + (ch ^ 48);
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ch = getchar();
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}
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return x * f;
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}
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// zkw线段树
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int d[N << 2];
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int n; // n个叶子节点
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int q; // q个操作,区间查询或单点修改
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int m; // m是指所有父节点们的个数
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//单点修改并psuhup
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void modify(int x, int c) {
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d[m + x] = c;
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for (int i = (m + x) >> 1; i; i >>= 1) d[i] = max(d[i << 1], d[i << 1 | 1]);
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}
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//构建重口味树
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void built() {
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memset(d, 0, sizeof d);
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for (m = 1; m <= n + 1;) m <<= 1; // 强行开够（大于n）方便二进制访问叶节点
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for (int i = 1; i <= n; i++) d[m + i] = read(); //初始值
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for (int i = m; i; i--) d[i] = max(d[i << 1], d[i << 1 | 1]); //待所有叶子节点有值后，统一执行一次向上数据统计信息汇总
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}
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//区间查询
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int query(int l, int r) {
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int ans = -1;
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// 闭区间改为开区间：l=l+m-1 : 将查询区间改为l-1, r=r+m+1 : 将查询区间改为r+1
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// l^r^1 : 相当于判断l与r是否是兄弟节点
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for (l = l + m - 1, r = r + m + 1; l ^ r ^ 1; l >>= 1, r >>= 1) {
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if (~l & 1) ans = max(ans, d[l ^ 1]); // l % 2 == 0 左端点 是左儿子 max(右子树)，想像一下题解中0号端点（哨兵），肯定是%2=0
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if (r & 1) ans = max(ans, d[r ^ 1]); // r % 2 == 1 右端点 是右儿子 max(左子树)，想像一下题解中1号端点，肯定是%2=1
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}
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return ans;
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}
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//方法重载，单点查询 [本题没有用到，用来做模板的]
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int query(int x) {
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return d[m + x]; //直接返回结果的zkw线段树节点值
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}
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int main() {
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//文件输入输出
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#ifndef ONLINE_JUDGE
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freopen("HDU1754.in", "r", stdin);
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#endif
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char op[2];
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while (~scanf("%d%d", &n, &q)) {
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//构建zkw树
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built();
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while (q--) {
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scanf("%s", op);
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int l = read(), r = read();
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if (op[0] == 'Q')
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printf("%d\n", query(l, r));
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else
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modify(l, r);
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}
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}
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return 0;
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} |