Poj 3264 Balanced Lineup ( 线段树

Balanced Lineup

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

6 3
1
7
3
4
2
5
1 5
4 6
2 2

Hint

630

题意

求区间的最大最小值

题解：

板子 好像写的太麻烦了唉

AC代码

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int N = 200000*4+10;struct node{    int left, right;    int sum, num;}tree[N];int arr[N];void build(int l,int r,int step){    tree[step].left = l;    tree[step].right = r;    tree[step].sum = tree[step].num = 0;    if(l == r) {        tree[step].sum = tree[step].num = arr[l];        return ;    }    int mid = (l+r) >> 1;    build(l,mid,step<<1);    build(mid+1,r,step<<1|1);    tree[step].sum = max(tree[step<<1].sum,tree[step<<1|1].sum);    tree[step].num = min(tree[step<<1].num,tree[step<<1|1].num);}int query_1(int l, int r,int step){    if(l==tree[step].left && r==tree[step].right) return tree[step].sum;    int mid = (tree[step].left+tree[step].right) >> 1;    if(r <= mid) return query_1(l,r,step<<1);    if(l > mid) return query_1(l,r,step<<1|1);    else        return max(query_1(l,mid,step<<1),query_1(mid+1,r,step<<1|1));}int query_2(int l, int r,int step){    if(l==tree[step].left && r==tree[step].right) return tree[step].num;    int mid = (tree[step].left+tree[step].right) >> 1;    if(r <= mid) return query_2(l,r,step<<1);    if(l > mid) return query_2(l,r,step<<1|1);    else        return min(query_2(l,mid,step<<1),query_2(mid+1,r,step<<1|1));}int main(){    int n, m;    char ch;    while(~scanf("%d%d",&n,&m)) {        for(int i = 1;i <= n; i++) {            scanf("%d",&arr[i]);        }        build(1,n,1);        int u, v;        while(m--) {            scanf("%d%d",&u,&v);            printf("%d\n",query_1(u,v,1)-query_2(u,v,1));        }    }return 0;}