hdu 1754

地址：http://acm.hdu.edu.cn/showproblem.php?pid=1754

题意：给n个数字。m次操作，每次操作更新一个数字或者查询区间最大值。

mark：典型线段树题。不过a的时候学了一下树状数组求区间最值。感觉对树状数组的理解又深刻了一点。不过这个更新不是O(lgn)而是O(lgn*lgn)的，比线段树慢！

代码：

线段树：

# include <stdio.h>
# include <string.h>


#define max(a,b) (a>b?a:b)
int n ;
int tr[200010<<2] ;


void u(int a, int b, int l, int r, int rt)
{    
    int m = (l+r)/2 ;
    if (l == r) {
        tr[rt] = b;
        return;
    }
    if (a <= m) u(a, b, l, m, rt*2);
    else u(a, b, m+1, r, rt*2+1);
    tr[rt] = max(tr[rt*2], tr[rt*2+1]);
}


int q(int a, int b, int l, int r, int rt)
{
    int m = (l + r) / 2, a1, a2 ;
    if (a == l && b == r) return tr[rt];
    if (b <= m) return q(a, b, l, m, rt*2) ;
    else if (a > m) return q(a, b, m+1, r, rt*2+1) ;
    a1 = q(a, m, l, m, rt*2), a2 = q(m+1, b, m+1, r, rt*2+1) ;
    return max(a1, a2) ;
}


int main ()
{
    int m, i, a, b ;
    char op[5] ;
    
    while (~scanf ("%d%d", &n, &m))
    {
        //memset (tr, 0, sizeof(tr));
        for (i = 1 ; i <= n ; i++)
            scanf ("%d", &a), u(i, a, 1, n, 1);
        while (m--)
        {
            scanf ("%s %d %d", op, &a, &b) ;
            if (op[0] == 'U') u(a, b, 1, n, 1);
            else printf ("%d
", q(a, b, 1, n, 1)) ;
        }
    }
    return 0 ;
}

树状数组（不加inline优化max函数的话会TLE！！！）

# include <stdio.h>
# include <string.h>


int n, num[200010], tr[200010];
int lowbit(int x){return x & -x;}
inline int max(int a, int b){return a>b?a:b;}


void u(int a, int b){
    int i, j;
    num[a] = b;
    for (i = a; i <= n; i+=lowbit(i))
    {
        tr[i] = num[i];
        for (j = 1; j < lowbit(i); j <<= 1)
            tr[i] = max(tr[i], tr[i-j]);
    }
}


int q(int a, int b){
    int i = b, ans = num[b];
    while (i >= a)
        if (i-lowbit(i)+1 >= a)
            ans = max(ans, tr[i]), i -= lowbit(i);
        else
            ans = max(ans, num[i]), i--;
    return ans;
}

    
int main ()
{
    int a, b, i, m;
    char op[5];
    while (~scanf ("%d%d", &n, &m))
    {
        memset(tr, 0, sizeof(tr));
        for (i = 1; i <= n; i++)
            scanf ("%d", &a), u(i, a);
        while (m--)
        {
            scanf ("%s %d %d", op, &a, &b);
            if (op[0] == 'U') u(a, b);
            else printf ("%d
", q(a, b)) ;
        }
    }
    return 0 ;
}