POJ-3264 (Balanced Lineup)

最新推荐文章于 2022-04-04 13:58:27 发布

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最新推荐文章于 2022-04-04 13:58:27 发布

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原文链接：http://www.cnblogs.com/wangtao971115/p/10358342.html

本文介绍了一种使用线段树进行区间查询的方法，通过构建线段树并利用递归方式查询指定区间的最大值和最小值，从而快速计算出该区间内数值的极差。

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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

Sample Output

6
3
0

询问a~b的极差

用两个query分别求最大值和最小值,然后相减即可

#include<cstdio>
#include<algorithm>
#define maxn 100007
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
using namespace std;
int mini[maxn<<2],maxi[maxn<<2];
int a[maxn],n;

void pushup(int rt)
{
   mini[rt]=min(mini[rt<<1],mini[rt<<1|1]);
   maxi[rt]=max(maxi[rt<<1],maxi[rt<<1|1]);
}

void build(int l,int r,int rt) {
    if (l == r) {
        mini[rt]=maxi[rt]=a[l];
        return ;
    }
    int m = (l + r) >> 1;
    build(lson);
    build(rson);
    pushup(rt);
}

int query(int L,int R,int l,int r,int rt) {
    if (L <= l && r <= R) {
        return maxi[rt];
    }
    int m = (l + r) >> 1;
    int ret = 0;
    if (L <= m) ret =query(L , R , lson);
    if (R > m) ret =max(ret,query(L , R , rson));
    return ret;
}

int query2(int L,int R,int l,int r,int rt) {
    if (L <= l && r <= R) {
        return mini[rt];
    }
    int m = (l + r) >> 1;
    int ret = 9999999;///注意这里求最小值,ret要重置为极大值
    if (L <= m) ret =query2(L , R , lson);
    if (R > m) ret =min(ret,query2(L , R , rson));
    return ret;
}

int main()
{
    int n,q,x,y;
    scanf("%d%d",&n,&q);
    for(int i=1;i<=n;i++)
        scanf("%d",&a[i]);
    build(1,n,1);
    for(int i=0;i<q;i++)
    {
        scanf("%d%d",&x,&y);
        printf("%d\n",query(x,y,1,n,1)-query2(x,y,1,n,1));
    }
    return 0;
}

转载于:https://www.cnblogs.com/wangtao971115/p/10358342.html