I Hate It

Time Limit: 9000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 33125    Accepted Submission(s): 13059

5 6
1 2 3 4 5
Q 1 5
U 3 6
Q 3 4
Q 4 5
U 2 9
Q 1 5

5
6
5
9

Hint
Huge input,the C function scanf() will work better than cin
 

解题思路：

基本的线段树单点更新，查询某一区间的最大值。

#include <iostream>
#include <string.h>
#include <stdio.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
const int maxn=200005;
int grade[maxn<<2];
int n,m;
int a,b;
char c;

void PushUp(int rt)//从下自上更新最大值
{
    if(grade[rt<<1]>=grade[rt<<1|1])
        grade[rt]=grade[rt<<1];
    else
        grade[rt]=grade[rt<<1|1];
}
void built(int l,int r,int rt)
{
    if(l==r)
    {
        scanf("%d",&grade[rt]);
        return;
    }
    int m=(l+r)>>1;
    built(lson);
    built(rson);
    PushUp(rt);
}
void change(int p,int num,int l,int r,int rt)
{
    if(l==r)
    {
        grade[rt]=num;
        return;
    }
    int m=(l+r)>>1;
    if(p<=m)
        change(p,num,lson);
    else
        change(p,num,rson);
    PushUp(rt);
}

int query(int L,int R,int l,int r,int rt)
{
    if(L<=l&&r<=R)
    {
        return grade[rt];
    }
    int ans=0;
    int m=(l+r)>>1;
    if(L<=m) ans=max(ans,query(L,R,lson));//取最大值
    if(R>m) ans=max(ans,query(L,R,rson));
    return ans;
}
int main()
{
    while(scanf("%d%d",&n,&m)==2)//加上==2就不超时了，或者写成~scanf("%d%d",&n,&m)也可以
    {
        built(1,n,1);
        for(int i=1;i<=m;i++)
        {
            scanf("%s",&c);
            if(c==‘Q‘)
            {
                scanf("%d%d",&a,&b);
                printf("%d\n",query(a,b,1,n,1));
            }
            else
            {
                scanf("%d%d",&a,&b);
                change(a,b,1,n,1);
            }
        }
    }
    return 0;
}

[ACM] hdu 1754 I Hate It （线段树，单点更新）,布布扣,bubuko.com

[ACM] hdu 1754 I Hate It （线段树，单点更新）

原文：http://blog.csdn.net/sr_19930829/article/details/23869157