uva 11898 - Killer Problem

最新推荐文章于 2019-08-05 22:21:11 发布

原创最新推荐文章于 2019-08-05 22:21:11 发布 · 860 阅读

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解决一个数组查询问题，需在给定区间内找到最小绝对差。利用计数排序优化算法，实现10^8级别的查询效率。

Killer Problem

You are given an array of N integers and Q queries. Each query is a closed interval [l, r]. You should find the minimum absolute difference between all pairs in that interval.

First line contains an integer T (T $\le$ 10). T sets follow. Each set begins with an integer N ( N $\le$ 200000). In the next line there are N integers a_i ( 1 $\le$ a_i $\le$ 10⁴), the number in the i-th cell of the array. Next line will contain Q ( Q $\le$ 10⁴). Q lines follow, each containing two integers l_i, r_i ( 1 $\le$ l_i, r_i $\le$ N, l_i < r_i) describing the beginning and ending of of i-th range. Total number of queries will be less than 15000.

Output

For the i-th query of each test output the minimum | a_j–a_k| for l_i $\le$ j, k $\le$ r_i (j $\ne$ k) a single line.

Sample Input

1
10
1 2 4 7 11 10 8 5 1 10000
4
1 10
1 2
3 5
8 10

Sample Output

大力出奇迹啊！10^8的算法也过了。主要是a的范围很小，可以用计数排序，然后找相邻两个间的差值，注意如果有某个数计数超过两次，则答案为0，然后停止循环，加上这个优化就过了。。。（后来认识到，区间范围大于10^4时，一定有某个数出现2次，所以复杂度是10^8）

#include <cstdio>
#include <algorithm>
#include <vector>
#include <map>
#include <queue>
#include <iostream>
#include <stack>
#include <set>
#include <cstring>
#include <stdlib.h>
#include <cmath>
using namespace std;
typedef long long LL;
typedef pair<int, int> P;
const int maxn = 200000 + 5;
const int maxk = 10000 + 5;
const int INF = 1000000000;

int a[maxn];
int countn[maxk];

int main(){
    int t;
    scanf("%d", &t);
    while(t--){
        int n;
        scanf("%d", &n);
        int Max = 0;
        for(int i = 1;i <= n;i++){
            scanf("%d", &a[i]);
            Max = max(Max, a[i]);
        }
        int q;
        scanf("%d", &q);
        while(q--){
            int l, r;
            scanf("%d%d", &l, &r);
            memset(countn, 0, sizeof(countn));
            int ans = INF;
            for(int i = l;i <= r;i++){
                countn[a[i]]++;
                if(countn[a[i]]>1){
                    ans = 0;
                    break;
                }
            }
            if(ans != 0){
                int last = -1;
                for(int i = 1;i <= Max;i++){
                    if(countn[i] == 0) continue;
                    if(countn[i] > 1){
                        ans = 0;
                        break;
                    }
                    if(last == -1){
                        last = i;
                        continue;
                    }
                    ans = min(ans, i-last);
                    last = i;
                }
            }
            printf("%d\n", ans);
        }
    }
    return 0;
}