poj 3624 Balanced Lineup

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

此題就是輸入一些數，求區間最大值與最小值的差。

此題用RMQ
dpma[i][j]：表示第j個開始，連續2^i個數的最大值；
dpmi[i][j]：表示第j個開始，連續2^i個數的最小值；
dpma[i][j]=max(dpma[i-1][j],dpma[i-1][j+(1<<(i-1))]);
（因爲，從j開始的2^(i-1)個數的最大值和(j+(1<<(i-1)))開始的2^(i-1)個數的最大值比較，就產生了，從i開始的2^i個數的最大值）
dpmi[i][j]=min(dpmi[i-1][j],dpmi[i-1][j+(1<<(i-1))]);
（同上）
那查找就是
如果查u到v，想要O（1），要用儘可能長的段覆蓋（重複也可以），
就可以求出

int k=(int)(log(v-u+1.0)/log(2.0));

return max(dpma[k][u],dpma[k][v-(1<< k)+1]);

return min(dpmi[k][u],dpmi[k][v-(1<< k)+1]);

---

#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
int n;
int dpma[18][50010];
int dpmi[18][50010];
int a[50010];
void makeMa(){
  for (int i=0;i<n;++i) 
    dpma[0][i]=a[i];
  for (int i=1;(1<<i)<=n;++i)
    for (int j=0;j+(1<<i)-1<n;++j)
    dpma[i][j]=max(dpma[i-1][j],dpma[i-1][j+(1<<(i-1))]);
}
int getMa(int u,int v){
  int k=(int)(log(v-u+1.0)/log(2.0));
  return max(dpma[k][u],dpma[k][v-(1<<k)+1]);
}
void makeMi(){
  for (int i=0;i<n;++i) 
    dpmi[0][i]=a[i];
  for (int i=1;(1<<i)<=n;++i)
    for (int j=0;j+(1<<i)-1<n;++j)
    dpmi[i][j]=min(dpmi[i-1][j],dpmi[i-1][j+(1<<(i-1))]);
}
int getMi(int u,int v){
  int k=(int)(log(v-u+1.0)/log(2.0));
  return min(dpmi[k][u],dpmi[k][v-(1<<k)+1]);
}
int main(){
  int q;
  while (scanf("%d%d",&n,&q)!=EOF){
    for (int i=0;i<n;++i) scanf("%d",&a[i]);
    makeMa();
    makeMi();
    while (q--){
      int u,v;
      scanf("%d%d",&u,&v);
      u--;
      v--;
      int t1=getMa(u,v);
      int t2=getMi(u,v);
      printf("%d\n",t1-t2);
    }
  }
  return 0;
}