POJ 3663 Costume Party (快速排序)

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最新推荐文章于 2021-02-22 05:56:31 发布

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原文链接：http://www.cnblogs.com/lidaojian/archive/2012/05/06/2486155.html

此题为美国计算机奥林匹克2008年1月青铜级别赛题，目标是在限定长度下找出所有能共同穿上一套服装的两头奶牛组合。通过快速排序和遍历策略，有效地解决了这一问题。

Costume Party

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9132		Accepted: 3552

Description

It's Halloween! Farmer John is taking the cows to a costume party, but unfortunately he only has one costume. The costume fits precisely two cows with a length of S (1 ≤ S ≤ 1,000,000). FJ has N cows (2 ≤ N ≤ 20,000) conveniently numbered 1..N; cow i has length L_i (1 ≤ L_i ≤ 1,000,000). Two cows can fit into the costume if the sum of their lengths is no greater than the length of the costume. FJ wants to know how many pairs of two distinct cows will fit into the costume.

Input

* Line 1: Two space-separated integers: N and S
* Lines 2..N+1: Line i+1 contains a single integer: L_i

Output

* Line 1: A single integer representing the number of pairs of cows FJ can choose. Note that the order of the two cows does not matter.

Sample Input

4 6

Sample Output

Source

USACO 2008 January Bronze

解题报告：这道题就是让求两牛之间的距离小于等于s的个数，直接遍历找的话TEL，先利用快排从小到大排序，再遍历同时缩小遍历的范围；

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
using namespace std;
const int MAX = 20010;
int a[MAX], n, s, ans;
int main()
{
    int i, j;
    while (scanf("%d%d", &n, &s) != EOF)
    {
        for (i = 0; i < n; ++i)
        {
            scanf("%d", &a[i]);
        }
        ans = 0;
        sort(a, a + n);
        int temp = n;
        for (i = 0; i < temp; ++i)
        {
            for (j = i + 1; j < temp; ++j)
            {
                if (a[i] + a[j] <= s)
                {
                    ans++;
                }
                else
                {
                    temp = j;//缩短遍历的范围
                    break;
                }
            }
        }
        printf("%d\n", ans);
    }
    return 0;
}