LightOJ 1100 - Again Array Queries【巧用数组】求最接近的两个数

1100 - Again Array Queries

PDF (English)StatisticsForum

Time Limit: 3 second(s)Memory Limit: 32 MB

Given an array with n integers, and you are given two indices i and j (i ≠ j) in the array. You have to find two integers in the range whose difference is minimum. You have to print this value. The array is indexed from 0 to n-1.

Input

Input starts with an integer T (≤ 5), denoting the number of test cases.

Each case contains two integers n (2 ≤ n ≤ 10⁵) and q (1 ≤ q ≤ 10000). The next line contains n space separated integers which form the array. These integers range in [1, 1000].

Each of the next q lines contains two integers i and j (0 ≤ i < j < n).

Output

For each test case, print the case number in a line. Then for each query, print the desired result.

Sample Input

Output for Sample Input

2

5 3

10 2 3 12 7

0 2

0 4

2 4

2 1

1 2

0 1

Case 1:

1

1

4

Case 2:

1

Notes

Dataset is huge, use faster I/O methods.

本来自己也是想到这样暴力了，但是没继续实施下去，后来发现一位大神暴力过了，也是醉了，果断暴力过！

比较巧妙的是利用数组的下标进行统计最近的距离，这个结合的桶排序的思想，比较巧妙........

然后就是q次暴力了......

暴力查找距离最短的时候，选一个动态参考点，就可以在1000（区间最大范围）次范围内解决....

大神们用的是什么划分树什么的，膜拜.....

#include<stdio.h>#include<string.h>using namespace std;int x[100005],cnt[1005],n,m;void slove(){int l,r;scanf("%d%d",&l,&r);if(r-l>=1000)//如果多于1000个数的话，绝对有重复的（鸽巢原理）{printf("0\n");return;}memset(cnt,0,sizeof(cnt));for(int i=l;i<=r;++i)//统计每个数出现的个数{++cnt[x[i]];}int kase=-1,dis=10005,tp=0;for(int i=0;i<1005&&tp<=r-l;++i){if(cnt[i]>1)//出现多于一次，绝对是 0 {dis=0;break;}if(cnt[i])//出现过的话{++tp;//计算统计过的个数if(kase!=-1&&i-kase<dis)//更新最短{dis=i-kase;}kase=i;//移动参考位置}}printf("%d\n",dis);}int main(){int t;//freopen("shuju.txt","r",stdin);scanf("%d",&t);for(int i=1;i<=t;++i){scanf("%d%d",&n,&m);for(int j=0;j<n;++j){scanf("%d",x+j);}printf("Case %d:\n",i);while(m--){slove();}}return 0;}