USACO Pearl Pairing

洛谷 P2902 [USACO08MAR]珍珠配对Pearl Pairing

https://www.luogu.org/problem/P2902

JDOJ 2577: USACO 2008 Mar Gold 3.Pearl Pairing

https://neooj.com:8082/oldoj/problem.php?id=2577

Description

At Bessie's recent birthday party, she received N (2 <= N <= 100,000;
N%2 == 0) pearls, each painted one of C different colors (1 <= C
< = N).

Upon observing that the number of pearls N is always even, her
creative juices flowed and she decided to pair the pearls so that
each pair of pearls has two different colors.

Knowing that such a set of pairings is always possible for the
supplied testcases, help Bessie perform such a pairing. If there
are multiple ways of creating a pairing, any solution suffices.

Input

* Line 1: Two space-separated integers: N and C

* Lines 2..C + 1: Line i+1 tells the count of pearls with color i: C_i

Output

* Lines 1..N/2: Line i contains two integers a_i and b_i indicating
        that Bessie can pair two pearls with respective colors a_i and
        b_i.

Sample Input

8 3
2
2
4

Sample Output

1 3
1 3
2 3
3 2

HINT

INPUT DETAILS:

There are 8 pearls and 3 different colors. Two pearls have color I; two
have color II; four have color III.

OUTPUT DETAILS:

Bessie pairs each pearl of color III with one of color I and II.

 

题目大意：

在Bessie最近的生日聚会上，她收到N（2<=N<=100,000; N％2==0）珍珠，每个都涂上C种不同颜色之一（1<=C<=N）。

观察到珍珠N的数量总是均匀的，她的创意来了，决定配对珍珠，使每双珍珠有两种不同的颜色。数据保证存在答案。请帮助Bessie执行这样的配对，如果有多种创建配对的方法，任意输出即可。

SPJ开了。

我是为了刷背包才看到这道题的，想破脑袋也没想出来咋背包，后来用模拟做了一遍，思路是这个样子。

既然数据保证存在答案，我们便开始思考什么样子的数据才会保证存在答案。

然后我们发现每种颜色的数量不可能超过N/2要不然没法玩。

所以我们就想出了这样一种方案：

记录下每个珍珠的颜色，然后分成两组（因为N是偶数）

由于不可能有颜色数量超过N/2，所以我们枚举N/2次，每次取两组的第i位，就可以百分百保证不会有重复。

脑筋急转弯的味道很浓，你信我的就没问题。

其实也不需要再打一遍sort，因为你存数组的时候就是从小到大存的。

代码如下：

#include<cstdio>
using namespace std;
int s[];
int n,c;
int k;
int main()
{
    scanf("%d%d",&n,&c);
    for(int i=;i<=c;i++)
    {
        int x;
        scanf("%d",&x);
        for(int j=;j<=x;j++)
            s[++k]=i;
    }
    for(int i=;i<=n/;i++)
        printf("%d %d\n",s[i],s[i+(n/)]);
    return ;
}

sort与否随你便。