poj3264 Balanced Lineup（树状数组）
Balanced Lineup
| Time Limit: 5000MS | Memory Limit: 65536K | |
| Total Submissions: 64655 | Accepted: 30135 | |
| Case Time Limit: 2000MS | ||
Description
For the daily milking, Farmer John's N cows (1 ≤ N ≤ 50,000) always line up in the same order. One day Farmer John decides to organize a game of Ultimate Frisbee with some of the cows. To keep things simple, he will take a contiguous range of cows from the milking lineup to play the game. However, for all the cows to have fun they should not differ too much in height.
Farmer John has made a list of Q (1 ≤ Q ≤ 200,000) potential groups of cows and their heights (1 ≤ height ≤ 1,000,000). For each group, he wants your help to determine the difference in height between the shortest and the tallest cow in the group.
Input
Line 1: Two space-separated integers, N and Q.
Lines 2..N+1: Line i+1 contains a single integer that is the height of cow i
Lines N+2..N+Q+1: Two integers A and B (1 ≤ A ≤ B ≤ N), representing the range of cows from A to B inclusive.
Lines 2..N+1: Line i+1 contains a single integer that is the height of cow i
Lines N+2..N+Q+1: Two integers A and B (1 ≤ A ≤ B ≤ N), representing the range of cows from A to B inclusive.
Output
Lines 1..Q: Each line contains a single integer that is a response to a reply and indicates the difference in height between the tallest and shortest cow in the range.
Sample Input
6 3 1 7 3 4 2 5 1 5 4 6 2 2
Sample Output
6 3 0
Source
题意：就是求区间[l,r]的最大值与最小值之差
首先，介绍一下树状数组，树状数组的复杂度为O(logn),之所以这么快，它是用到了分块优化的思想
，比如：1-5就分为1、1-2、3、1-2-3-4、5；而树状数组最基本实现的就是求前缀和和区间单点更新；在本题中，我们也能效仿这样吗，当然。单点更新极值的话跟之前的一样；重点是求区间的极值，我们不能像求前缀和一样sum[r]-sum[l]，那应该怎么办呢？
其实这里还是用到分块优化的思想，但是我们不能直接用Max[r]，因为这里表示区间可能超出了我们要求的区间，但是我们可以利用一些包含在[l,r]这里面的块，从而降低复杂度。
假设我们暴力的话，就是从右往左遍历一遍a数组然后取最值，那么在这之中，如果有一些分块在[l,r]之中，就可以直接使用这些块了，就不用去遍历他们，但是遇到有的块超出了[l,r],我们就只能直接比较a[r]了。
代码：
#include<iostream>
#include<string.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
#define mod 1000000007
#define INF 0x3f3f3f3f
#define MAX 50005
int Max[MAX],Min[MAX];
int a[MAX];
int n,q;
inline void read(int &x){
char ch;
bool flag=false;
for (ch=getchar();!isdigit(ch);ch=getchar())if (ch=='-') flag=true;
for (x=0;isdigit(ch);x=x*10+ch-'0',ch=getchar());
x=flag?-x:x;
}
inline void write(int x){
static const int maxlen=100;
static char s[maxlen];
if (x<0) { putchar('-'); x=-x;}
if(!x){ putchar('0'); return; }
int len=0; for(;x;x/=10) s[len++]=x % 10+'0';
for(int i=len-1;i>=0;--i) putchar(s[i]);
}
int lowbit(int x)
{
return x&-x;
}
void updata(int i,int val)
{
while(i<=n)
{
Min[i]=min(Min[i],val);
Max[i]=max(Max[i],val);
i+=lowbit(i);
}
}
int query(int l,int r)
{
int maxn=a[l],minn=a[r];
while(1)
{
maxn=max(maxn,a[r]),minn=min(minn,a[r]);
if(l==r) break;
for(r-=1;r-l>=lowbit(r);r-=lowbit(r))
maxn=max(Max[r],maxn),minn=min(minn,Min[r]);
}
return maxn-minn;
}
int main()
{
memset(Min,INF,sizeof(Min));
read(n);read(q);
for(int i=1;i<=n;i++){
read(a[i]);
updata(i,a[i]);
}
while(q--)
{
int a,b;
read(a),read(b);
write(query(a,b));
putchar('\n');
}
return 0;
}
时间复杂度近似为log2(n)。
