HDU 1754 线段树入门解题报告

---恢复内容开始---

题意：给定区间,每个人的成绩, Q次询问,求每次询问区间中的最大值

思路：构造线段树

代码：

#include<stdio.h>
#include<algorithm>
#include<string>
#include<iostream>
using namespace std;

int n, m, ans;
string str;

struct Tree
{
    int left, right, maxx;
}tree[200000 * 4];  //线段树开4倍空间 

void build(int left, int right, int k)
{
    tree[k].left = left, tree[k].right = right;
    if(left == right)
    {
        scanf("%d", &tree[k].maxx);
        return ;
    }
    int mid = (left + right) / 2;
    build(left, mid, 2 * k);
    build(mid + 1, right, 2 * k + 1);
    tree[k].maxx = max(tree[2 * k].maxx, tree[2 * k + 1].maxx); //向上回溯最大值记录 
}

void update(int find_node, int up_num, int k)//单点修改 
{
    if(tree[k].left == find_node && tree[k].right == find_node)
    {
        tree[k].maxx = up_num;
        return ;
    }
    int mid = (tree[k].left + tree[k].right) / 2;
    if(find_node <= mid)
        update(find_node, up_num, 2 * k);
    else
        update(find_node, up_num, 2 * k + 1);  //修改过后回溯修改最大值 
    tree[k].maxx = max(tree[2 * k].maxx, tree[2 * k + 1].maxx);
}

void query(int find_left, int find_right, int k)//区间查询取最大值 
{
    if(find_left == tree[k].left && find_right == tree[k].right)
    {
        ans = max(ans, tree[k].maxx);
        return ;
    }
    int mid = (tree[k].left + tree[k].right) / 2;
    if(find_right <= mid)
        query(find_left, find_right, 2 * k);
    else if(find_left > mid)
        query(find_left, find_right, 2 * k + 1);
    else
    {
        query(find_left, mid, 2 * k);
        query(mid + 1, find_right, 2 * k + 1);
    }
}

int main()
{
    int a, b;
    while(scanf("%d%d", &n, &m)!=EOF)
    {
        build(1, n, 1);
        for(int i = 1; i <= m; i ++)
        {
            cin >> str;
            if(str == "U")
            {
                scanf("%d%d", &a, &b);
                update(a, b, 1);
            }
            else
            {
                ans = -1;
                scanf("%d%d", &a, &b);
                query(a, b, 1);
                printf("%d\n", ans);
            }
        }    
    }
    return 0;
}