HDU 1754 I Hate It（线段树单点更改、区间查找最大值）

#include <cstdio>
#include <iostream>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
#include <map>
using namespace std;

#define ll long long
#define eps 1e-9

const int inf = 0x3f3f3f3f;
const int mod = 1e9+7;

int n, m, ans, a[2000000+8], x, y;
char c;

struct node
{
    int l, r, sum;
}tree[4*2000000+8];

void build(int i, int l, int r)///建树
{
    tree[i].sum = 0;
    tree[i].l = l;
    tree[i].r = r;
    if(l == r)///如果是叶子节点
    {
        tree[i].sum = a[l];///则让它的值存进去
        return ;
    }
    int mid = (tree[i].r+tree[i].l)/2;///如果不是叶子节点
    build(i*2, l, mid);///搜索左儿子的区间
    build(i*2+1, mid+1, r);///搜索右儿子的区间
    tree[i].sum = max(tree[i*2].sum , tree[i*2+1].sum);///使该区间最大的值往上传
}

void pls(int i, int pos, int k)///单点改变
{
    if(tree[i].r == pos && tree[i].l == tree[i].r)///如果时所要寻找的叶子节点
    {
        tree[i].sum = k;///改变该点的值
        return ;
    }
    int mid = (tree[i].l+tree[i].r)/2;
    if(pos <= mid)///如果位置在左边，就往左边找
        pls(i*2, pos, k);
    else///否则往右边找
        pls(i*2+1, pos, k);
    tree[i].sum = max(tree[i*2].sum , tree[i*2+1].sum);///不断更新父亲节点的最大值
}

void search(int i, int l, int r)
{
    if(tree[i].l >= l && tree[i].r <= r)
    {
        ans = max(ans, tree[i].sum);
        return;
    }
    int mid = (tree[i].l+tree[i].r)/2;
    if(l <= mid)
        search(i*2, l, r);///区间不能写成search(i*2, pos, k)，因为这样区间会改变，但是题目要求要查的是[l, r]之间的最大值
    if(mid < r)
        search(i*2+1, l, r);
}

int main()
{
    while(~scanf("%d%d", &n, &m) && (n+m))
    {
        for(int i = 1; i <= n; i++)
            scanf("%d", &a[i]);
        build(1, 1, n);
        for(int i = 1; i <= m; i++)
        {
            ans = -1;
            getchar();
            scanf("%c %d %d", &c, &x, &y);
            if(c == ‘Q‘)
            {
                search(1, x, y);
                printf("%d\n", ans);
            }
            else if(c == ‘U‘)
            {
                pls(1, x, y);
            }
        }
    }
    return 0;
}