poj-3264-Balanced Lineup

poj   3264  Balanced Lineup

link: http://poj.org/problem?id=3264

 

 

　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　Balanced Lineup

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 48747		Accepted: 22833
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.

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

 

 

题解：

快速找到一个区间[a, b] 之间的最大值和最小值的差；

经典的RMQ问题。 利用Sparse Table算法， 动态规划求解。 

 

#include <iostream> 
#include <cstdio> 
#include <cstring> 
#include <algorithm>
#include <cmath> 
using namespace std; 
const int maxn = 50005; 

int n, m,  num[maxn], dp1[maxn][18], dp2[maxn][18]; 

void BuildIndex(){
	for(int i=0; i<n;++i){
		dp1[i][0] = i; 
		dp2[i][0] = i; 
	} 
	for(int i=1; (1<<i)<=n; ++i){
		for(int j=0; j+(1<<i)-1<n; ++j){
			// find max 
			if(num[dp1[j][i-1]] > num[dp1[j+(1<<(i-1))][i-1]]){
				dp1[j][i] = dp1[j][i-1]; 
			}else{
				dp1[j][i] = dp1[j+(1<<(i-1))][i-1]; 
			}

			// find min
			if(num[dp2[j][i-1]] < num[dp2[j+(1<<(i-1))][i-1]]){
				dp2[j][i] = dp2[j][i-1]; 
			}else{
				dp2[j][i] = dp2[j+(1<<(i-1))][i-1]; 
			}
		}
	}
}

int FindMaxIndex(int start, int end){
	int k = (int)((log((end - start + 1)*1.0))/log(2.0)); 
	if(num[dp1[start][k]] > num[dp1[end-(1<<k)+1][k]]){
		return dp1[start][k]; 
	}else{
		return dp1[end-(1<<k)+1][k]; 
	}
}
int FindMinIndex(int start, int end){
	int k = (int)((log((end - start + 1)*1.0))/log(2.0)); 
	if(num[dp2[start][k]] > num[dp2[end-(1<<k)+1][k]]){
		return dp2[end-(1<<k)+1][k]; 
	}else{
		return dp2[start][k]; 
	}
}

int main(){
	freopen("in.txt", "r", stdin); 

	int ans1, ans2, x, y; 
	while(scanf("%d %d", &n, &m) != EOF){
		for(int i=0; i<n; ++i){
			scanf("%d", &num[i]); 
		}
		BuildIndex(); 
		while(m--){
			scanf("%d %d", &x, &y); 
			if(x > y){ swap(x, y); } 
			ans1 = FindMinIndex(x-1, y-1); 
			ans2 = FindMaxIndex(x-1, y-1); 
			printf("%d\n", (num[ans2] - num[ans1]) );
		}
	}
	return 0; 
}