题意: 中文题
query 区间最大值,保存最大值就好了
点更新
学长博客:https://blog.youkuaiyun.com/sun897949163/article/details/52092178
链接:hdu 1754
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#define INF 0x3f3f3f3f
#define ll long long
using namespace std;
const int inf = 0x3f3f3f3f;
const int maxn = 200005;
const int maxm = 1000005;
int n, m, a[maxn];
struct node {
int l, r;
int v;
}tree[maxn << 2];
void pushup(int rt) {
tree[rt].v = max(tree[rt << 1].v, tree[rt << 1 | 1].v);
}
void build(int rt, int l, int r) {
tree[rt].l = l;
tree[rt].r = r;
if(r == l) {
tree[rt].v = a[l];
return;
}
int mid = (l + r) >> 1;
build(rt << 1, l, mid);
build(rt << 1 | 1, mid + 1, r);
pushup(rt);
}
void update(int rt, int k, int v) {
if(tree[rt].l == k && tree[rt].r == k) {
tree[rt].v = v;
return;
}
int mid = (tree[rt].l + tree[rt].r) >> 1;
if(k <= mid)
update(rt << 1, k, v);
if(k > mid)
update(rt << 1 | 1, k, v);
pushup(rt);
}
int query(int rt, int l, int r) {
if(tree[rt].l == l && tree[rt].r == r)
return tree[rt].v;
int mid = (tree[rt].l + tree[rt].r) >> 1;
if(r <= mid)
return query(rt << 1, l, r);
if(l > mid)
return query(rt << 1 | 1, l, r);
if(l <= mid && r > mid)
return max(query(rt << 1, l, mid), query(rt << 1 | 1, mid + 1, r));
return 0;
}
int main ()
{
while(scanf("%d %d", &n, &m) != EOF) {
for(int i = 1; i <= n; i++) {
scanf("%d", &a[i]);
}
build(1, 1, n);
int b, c;
char str[5];
for(int i = 1; i <= m; i++) {
scanf("%s %d %d", str, &b, &c);
if(str[0] == 'Q') {
printf("%d\n", query(1, b, c));
}
else {
update(1, b, c);
}
}
}
return 0;
}
本文介绍了一个基于线段树实现的区间最大值查询及点更新算法。该算法适用于解决如HDU1754等问题,通过递归构建、更新和查询线段树来高效地处理区间查询和单点更新操作。

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