BZOJ 1699: [Usaco2007 Jan]Balanced Lineup排队

1699: [Usaco2007 Jan]Balanced Lineup排队

Description

每天,农夫 John 的N(1 <= N <= 50,000)头牛总是按同一序列排队. 有一天, John 决定让一些牛们玩一场飞盘比赛. 他准备找一群在对列中为置连续的牛来进行比赛. 但是为了避免水平悬殊,牛的身高不应该相差太大. John 准备了Q (1 <= Q <= 180,000) 个可能的牛的选择和所有牛的身高 (1 <= 身高 <= 1,000,000). 他想知道每一组里面最高和最低的牛的身高差别. 注意: 在最大数据上, 输入和输出将占用大部分运行时间.

Input

* 第一行: N 和 Q. * 第2..N+1行: 第i+1行是第i头牛的身高.

 * 第N+2..N+Q+1行: 两个整数, A 和 B (1 <= A <= B <= N), 表示从A到B的所有牛.

Output

*第1..Q行: 所有询问的回答 (最高和最低的牛的身高差), 每行一个.

Sample Input

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

Sample Output

6
3
0

——分隔符——

传说中的线段树水题，维护最大值和最小值。

代码：

#include<cstdio>using namespace std;struct Btree{int left;int right;int num;int max;int min;};const int Maxn=50000;Btree tree[Maxn*4+1];int height[Maxn+1];int n,q;inline int remax(int a,int b){if (a>b) return a;return b;}inline int remin(int a,int b){if (a<b) return a;return b;}void build(int x,int left,int right){tree[x].left=left;tree[x].right=right;if (left==right){tree[x].min=tree[x].max=tree[x].num=height[left];}else{int mid=(left+right)/2;build(x*2,left,mid);build(x*2+1,mid+1,right);tree[x].max=remax(tree[x*2].max,tree[x*2+1].max);tree[x].min=remin(tree[x*2].min,tree[x*2+1].min);}}int queryMax(int x,int left,int right){if (left<=tree[x].left && tree[x].right<=right){return tree[x].max;}else{int mid=(tree[x].left+tree[x].right)/2;int maxNum=-1;if (left<=mid) maxNum=remax(maxNum,queryMax(x*2,left,right));if (right>=mid+1) maxNum=remax(maxNum,queryMax(x*2+1,left,right));return maxNum; }}int queryMin(int x,int left,int right){if (left<=tree[x].left && tree[x].right<=right){return tree[x].min;}else{int mid=(tree[x].left+tree[x].right)/2;int minNum=2147483647;if (left<=mid) minNum=remin(minNum,queryMin(x*2,left,right));if (right>=mid+1) minNum=remin(minNum,queryMin(x*2+1,left,right));return minNum; }}int main(){scanf("%d%d",&n,&q);for (int i=1;i<=n;i++){scanf("%d",&height[i]);}build(1,1,n);for (;q--;){int l,r;scanf("%d%d",&l,&r);printf("%d\n",queryMax(1,l,r)-queryMin(1,l,r));}return 0;}