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题目链接：http://poj.org/problem?id=3264
Balanced Lineup
| Time Limit: 5000MS | Memory Limit: 65536K | |
| Total Submissions: 47515 | Accepted: 22314 | |
| 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
Output
Sample Input
6 3 1 7 3 4 2 5 1 5 4 6 2 2
Sample Output
6 3 0
题目大意：给定一个数组，然后给一个区间[a,b]，输出从a到b中的最大值和最小值的差。
RMQ 和 线段树 均可
RMQ代码：
1 #include <stdio.h> 2 #define MAX(a,b) (a>b ? a:b) 3 #define MIN(a,b) (a>b ? b:a) 4 #define N 50010 5 int a[N],ma[N][25],mi[N][25]; 6 7 void ST(int n) 8 { 9 int i,j; 10 for(i=1;i<=n;i++) 11 mi[i][0]=ma[i][0]=a[i]; 12 13 for (j = 1; (1<<j) <= n; j ++) 14 for (i = 1; i + (1<<j)-1 <= n; i ++) 15 { 16 ma[i][j]=MAX(ma[i][j-1],ma[i+(1<<(j-1))][j-1]); 17 mi[i][j]=MIN(mi[i][j-1],mi[i+(1<<(j-1))][j-1]); 18 } 19 } 20 21 int rmq(int a,int b) 22 { 23 int k = 0; 24 while((1<<(k+1)) <= b-a+1) k++; 25 return MAX(ma[a][k],ma[b-(1<<k)+1][k])-MIN(mi[a][k],mi[b-(1<<k)+1][k]); 26 } 27 28 int main() 29 { 30 int n,i,q,x,y; 31 while(scanf("%d %d",&n,&q)!=-1) 32 { 33 for(i=1;i<=n;i++) 34 scanf("%d",a+i); 35 ST(n); 36 for(i=1;i<=q;i++) 37 { 38 scanf("%d %d",&x,&y); 39 printf("%d\n",rmq(x,y)); 40 } 41 } 42 return 0; 43 }
POJ 3264 区间最大最小值Sparse_Table算法
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原文地址：http://www.cnblogs.com/yoke/p/5858054.html
