poj 3264 Balanced Lineup 题解

Balanced Lineup

Submit Status

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.

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

USACO 2007 January Silver

Submit Status

——————————————————————我是分割线——————————————————————————————————

【解法】

　　很容易想到用线段树解，但是代码复杂度太高，懒得打了。

　　所以就用刚学的ST算法解决。

【代码】

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<cmath>
 5 #include<algorithm>
 6 #include<queue>
 7 #include<cstdlib>
 8 #include<iomanip>
 9 #include<cassert>
10 #include<climits>
11 #define maxn 100001
12 #define F(i,j,k) for(int i=j;i<=k;i++)
13 #define M(a,b) memset(a,b,sizeof(a))
14 #define FF(i,j,k) for(int i=j;i>=k;i--)
15 #define inf 0x7fffffff
16 #define maxm 21
17 using namespace std;
18 int read(){
19     int x=0,f=1;char ch=getchar();
20     while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
21     while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
22     return x*f;
23 }
24 int fm[maxn][maxm],fi[maxn][maxm],p[maxn];
25 int n,q;
26 inline int init()
27 {
28     cin>>n>>q;
29     F(i,1,n){
30         cin>>p[i];
31     }
32     F(i,1,n){
33         fm[i][0]=fi[i][0]=p[i];
34     }
35     int m=floor((int)(log10((double)n)/log10((double)2)));
36     F(j,1,m)F(i,1,n){
37         fm[i][j]=max(fm[i+(1<<(j-1))][j-1],fm[i][j-1]);
38         fi[i][j]=min(fi[i+(1<<(j-1))][j-1],fi[i][j-1]);
39     }
40 }
41 inline int stmax(int a,int b)
42 {
43     int m=floor((int)(log10((double)(b-a+1))/log10((double)2)));
44     return max(fm[a][m],fm[b-(1<<m)+1][m]);
45 }
46 inline int stmin(int a,int b)
47 {
48     int m=floor((int)(log10((double)(b-a+1))/log10((double)2)));
49     return min(fi[a][m],fi[b-(1<<m)+1][m]);
50 }
51 int main()
52 {
53     std::ios::sync_with_stdio(false);//cout<<setiosflags(ios::fixed)<<setprecision(1)<<y;
54 //  freopen("data.in","r",stdin);
55 //  freopen("data.out","w",stdout);
56     init();int c;
57     while(q--)
58     {
59         int a,b;
60         cin>>a>>b;
61         if(a>b) swap(a,b);
62         c=stmax(a,b)-stmin(a,b);
63         cout<<c<<endl;
64     }
65     return 0;
66 }

poj 3264