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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.
6 3 1 7 3 4 2 5 1 5 4 6 2 2
6 3 0
给定一个区间，求最大值与最小值之差
ST表
#include<iostream> #include<cstdio> #include<cstring> #define N 50010 using namespace std; int ff[1<<20]; int a[N],rmq[20][N],rmx[20][N]; int n,m; void rmq_init() { for (int i=1;i<=n;i++) { rmq[0][i] = a[i]; rmx[0][i] = a[i]; } for (int j=1;j<20;j++) for (int i=1;i+(1<<j)-1<=n;i++){ rmq[j][i] = min(rmq[j-1][i],rmq[j-1][i+(1<<j>>1)]); rmx[j][i] = max(rmx[j-1][i],rmx[j-1][i+(1<<j>>1)]); } memset(ff,-1,sizeof(ff)); for (int i=0;i<20;i++) ff[1<<i] = i; for (int i=0;i<N;i++) if (ff[i] == -1) ff[i] = ff[i-1]; } pair<int,int> rmq_find(int x,int y) { int t=ff[y-x+1]; return make_pair(min(rmq[t][x],rmq[t][y-(1<<t)+1]), max(rmx[t][x],rmx[t][y-(1<<t)+1])); } int main() { int l,r; scanf("%d%d",&n,&m); for (int i=1;i<=n;i++) scanf("%d",&a[i]); rmq_init(); for (int i=1;i<=m;i++) { scanf("%d%d",&l,&r); printf("%d\n",rmq_find(l,r).second-rmq_find(l,r).first); } }
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原文地址：http://www.cnblogs.com/liumengyue/p/5467713.html
