CodeForce 439C Devu and Partitioning of the Array（模拟）

最新推荐文章于 2021-02-09 09:31:27 发布

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最新推荐文章于 2021-02-09 09:31:27 发布

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探讨如何将包含不同整数的数组划分成k个不相交的非空子集，使p个子集的元素之和为偶数，剩余的k-p个子集之和为奇数。提供一种有效的算法解决方案。

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Devu and Partitioning of the Array

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Devu being a small kid, likes to play a lot, but he only likes to play with arrays. While playing he came up with an interesting question which he could not solve, can you please solve it for him?

Given an array consisting of distinct integers. Is it possible to partition the whole array into k disjoint non-empty parts such that p of the parts have even sum (each of them must have even sum) and remaining k - p have odd sum? (note that parts need not to be continuous).

If it is possible to partition the array, also give any possible way of valid partitioning.

Input

The first line will contain three space separated integers n, k, p (1 ≤ k ≤ n ≤ 105; 0 ≤ p ≤ k). The next line will contain n space-separated distinct integers representing the content of array a: a1, a2, ..., an (1 ≤ ai ≤ 109).

Output

In the first line print "YES" (without the quotes) if it is possible to partition the array in the required way. Otherwise print "NO" (without the quotes).

If the required partition exists, print k lines after the first line. The ith of them should contain the content of the ith part. Print the content of the part in the line in the following way: firstly print the number of elements of the part, then print all the elements of the part in arbitrary order. There must be exactly p parts with even sum, each of the remaining k - p parts must have odd sum.

As there can be multiple partitions, you are allowed to print any valid partition.

                             Sample test(s)
                           

                               input
                             
5 5 3
2 6 10 5 9

                               output
                             
YES
1 9
1 5
1 10
1 6
1 2

                               input
                             
5 5 3
7 14 2 9 5

                               output
                             
NO

                               input
                             
5 3 1
1 2 3 7 5

                               output
                             
YES
3 5 1 3
1 7
1 2

题目链接：http://codeforces.com/problemset/problem/439/C

题意：给出n个数，要分成k份，每份有若干个数，但是只需要关注该份的和为奇数还是偶数，要求偶数堆的个数为p。输出方案。

分析：先输出k-p-1组奇数，然后输出p-1组偶数；如果 p!=0&&(k-p)!=0,再输出一个奇数，这时奇数的已经构造完成，最后把剩下的全部输出为一组即可。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
using namespace std;
const int MAXN = 1e5 + 100;
int a[MAXN], b[MAXN];
int main()
{
    int n, p, k, i, odd = 0, even = 0, c;
    scanf("%d%d%d",&n,&k,&p);
    for(i = 0; i < n; i++)
    {
        scanf("%d",&c);
        if(c&1)
            a[odd++] = c;
        else
            b[even++] = c;
    }
    if(odd < k-p || (odd - k + p) % 2 == 1 || even + (odd - k + p) / 2 < p)
        printf("NO\n");
    else
    {
        printf("YES\n");
        int tmp = k - p;
        for(i = 0; i < tmp - 1; i++)
            printf("1 %d\n",a[--odd]);
        for(i = 0; i < p - 1; i++)
        {
            if(even)
                printf("1 %d\n", b[--even]);
            else
            {
                printf("2 %d %d\n", a[odd-1], a[odd-2]);
                odd -= 2;
            }
        }
        if(tmp && p)
            printf("1 %d\n", a[--odd]);
        printf("%d", odd+even);
        while(odd)
            printf(" %d",a[--odd]);
        while(even)
            printf(" %d\n", b[--even]);
        printf("\n");
    }
    return 0;
}