POJ3264

基础
Time Limit:5000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u
Submit Status



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

<span style="color:#3366ff;">/*********************************************************************    author    :   Grant Yuan    time      :   2014.7.26    algorithm :   线段树**********************************************************************/#include <iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<algorithm>#define MAX 50003#define INF 999999999using namespace std;int n,m;struct node{int l;int r;int Min;int Max;};int m1,m2;node tree[4*MAX];int h[MAX];void build(int left,int right,int root){     tree[root].l=left;     tree[root].r=right;    if(left==right){     tree[root].Max=h[left];     tree[root].Min=h[left];    return;    }    int mid=(left+right)>>1;    build(left,mid,root*2);    build(mid+1,right,root*2+1);    tree[root].Max=max(tree[root*2].Max,tree[root*2+1].Max);    tree[root].Min=min(tree[root*2].Min,tree[root*2+1].Min);}void findMax(int left,int right,int root){   if(left<=tree[root].l&&right>=tree[root].r)   {m1=max(m1,tree[root].Max);   return;}    int mid=(tree[root].l+tree[root].r)>>1;    if(mid>=right)        findMax(left,right,root*2);    else if(mid<left)        findMax(left,right,root*2+1);    else{        findMax(left,right,2*root);        findMax(left,right,2*root+1);    }}void findMin(int left,int right,int root){    if(left<=tree[root].l&&right>=tree[root].r)      {m2=min(m2,tree[root].Min);      return;}    int mid=(tree[root].l+tree[root].r)>>1;    if(mid>=right)        findMin(left,right,root*2);    else if(mid<left)        findMin(left,right,root*2+1);    else{       findMin(left,right,2*root);       findMin(left,right,2*root+1);    }}int main(){  cin>>n>>m;int a,b;  for(int i=1;i<=n;i++)    scanf("%d",&h[i]);    build(1,n,1);    for(int i=0;i<m;i++)        {scanf("%d%d",&a,&b); m1=0;m2=INF;     findMax(a,b,1);       findMin(a,b,1);        printf("%d\n",m1-m2);        }    return 0;}</span>