POJ 3264 Balanced Lineup

思路：线段树求最大值减最小值，每个结点分别维护最大值和最小值即可。

#include <cstdio>#include <queue>#include <cstring>#include <iostream>#include <cstdlib>#include <algorithm>#include <vector>#include <map>#include <string>#include <set>#include <ctime>#include <cmath>#include <cctype>using namespace std;#define maxn 50010#define LL long long#define lson i<<1,l,m#define rson (i<<1)|1,m+1,rstruct ST{int l,r;int Max,Min;}st[maxn<<2];void pushup(int i){st[i].Max=max(st[i<<1].Max,st[(i<<1)|1].Max);st[i].Min=min(st[i<<1].Min,st[(i<<1)|1].Min);}void build(int i,int l,int r){st[i].l=l;st[i].r=r;if (st[i].l==st[i].r){scanf("%d",&st[i].Max);st[i].Min = st[i].Max;return;}int m = (st[i].l+st[i].r)>>1;build(lson);build(rson);pushup(i);}int query1(int i,int l,int r){if (st[i].l==l && st[i].r==r)return st[i].Max;int m = (st[i].l+st[i].r)>>1;if (r<=m)return query1(i<<1,l,r);else if (l>m)return query1((i<<1)|1,l,r);elsereturn max(query1(lson),query1(rson));}int query2(int i,int l,int r){if (st[i].l==l && st[i].r==r)return st[i].Min;int m = (st[i].l+st[i].r)>>1;if (r<=m)return query2(i<<1,l,r);else if (l>m)return query2((i<<1)|1,l,r);elsereturn min(query2(lson),query2(rson));}int n,q;int cas=1,T;int main(){while (scanf("%d%d",&n,&q)!=EOF){build(1,1,n);while (q--){int l,r;scanf("%d%d",&l,&r);printf("%d\n",query1(1,l,r)-query2(1,l,r));}}//freopen("in","r",stdin);//scanf("%d",&T);//printf("time=%.3lf",(double)clock()/CLOCKS_PER_SEC);return 0;}

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 31734251 54 62 2

Sample Output

630