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
题意：给出n个数，m个操作，每次操作求x-y区间内的最大差值
思路：线段树水题
AC代码：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<cmath>
#include<utility>
#include<set>
#include<vector>
#include<map>
#include<queue>
#include<stack>
#define maxn 50005
#define INF 0x3f3f3f3f
#define LL long long
#define ULL unsigned long long
#define E 1e-8
#define mod 1000000007
#define P pair<int,int>
#define MID(l,r) (l+(r-l)/2)
#define lson(o) (o<<1) //o*2
#define rson(o) (o<<1|1) //o*2+1
using namespace std;
int a[maxn];
int ql,qr; //查询区间[ql,qr]
int ans_max,ans_min;
struct node
{
int l,r,mmin,mmax;
}tree[maxn<<2];
void build(int o,int l,int r)
{
tree[o].l = l;
tree[o].r = r;
if(l==r){
tree[o].mmax = tree[o].mmin = a[l];
return ;
}
int m = MID(l,r);
int lc = lson(o),rc = rson(o);
build(lc,l,m);
build(rc,m+1,r);
tree[o].mmax = max(tree[lc].mmax,tree[rc].mmax);
tree[o].mmin = min(tree[lc].mmin,tree[rc].mmin);
}
void query_init()
{
ans_max = -INF;
ans_min = INF;
}
void query(int o)
{
if(ql<=tree[o].l&& qr>=tree[o].r){ //[L, R]包含在[ql, qr]区间内
ans_max = max(ans_max,tree[o].mmax);
ans_min = min(ans_min,tree[o].mmin);
return ;
}
int m =MID(tree[o].l,tree[o].r);
if(ql<=m) query(lson(o));
if(qr>m) query(rson(o));
}
int main()
{
int n,m,x,y;
while(scanf("%d %d",&n,&m)!=EOF){
for(int i=1;i<=n;++i) scanf("%d",&a[i]);
build(1,1,n);
while(m--){
scanf("%d %d",&x,&y);
ql = x;
qr = y;
query_init();
query(1);
printf("%d\n",ans_max-ans_min);
}
}
return 0;
}