Codeforces Round #251(Div. 2) 439C. Devu and Partitioning of the Array 构造

C. 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.

Examples

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

题意：

出n个数，将n个数分成k个不相交集合，并且其中恰好有p个集合的元素之和为偶数，输出分解方案。

题解：

题目要求p个集合偶数和，那么就有q=k-p个集合奇数和。

奇+奇=偶

奇+偶=奇

偶+偶=偶

所以和为奇数的集合里必须包含奇数个奇数，故第一个无解的情况为：奇数的个数<q。

对于多出来的奇数，可以偶数个奇数组成偶数，以此来不扰乱偶数的组合，但是在组成完成q个奇数集合后，剩下的奇数如果是奇数个，那么就是第二个无解的情况：

(奇数的个数-q)%2!=0.

如果按照上面的方法组成的集合个数小于k，则也是无解。

接下来就是构造，先处理奇数，再处理偶数就好了。

/****************
*PID:439c div2
*Auth:Jonariguez
*****************
*/
#define lson k*2,l,m
#define rson k*2+1,m+1,r
#define rep(i,s,e) for(i=(s);i<=(e);i++)
#define For(j,s,e) For(j=(s);j<(e);j++)
#define sc(x) scanf("%d",&x)
#define In(x) scanf("%I64d",&x)
#define pf(x) printf("%d",x)
#define pfn(x) printf("%d\n",(x))
#define Pf(x) printf("%I64d",(x))
#define Pfn(x) printf("%I64d\n",(x))
#define Pc printf(" ")
#define PY puts("YES")
#define PN puts("NO")
#include <stdio.h>
#include <string.h>
#include <string>
#include <math.h>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
typedef __int64 LL;
typedef int Ll;
Ll quick_pow(Ll a,Ll b,Ll MOD){a%=MOD;Ll res=1;while(b){if(b&1)res=(res*a)%MOD;b/=2;a=(a*a)%MOD;}return res;}

const int maxn=100000+10;
int a[maxn];
vector<int> odd,eve;

int main()
{
int i,j,n,k,p;
while(scanf("%d%d%d",&n,&k,&p)!=EOF){
odd.clear();eve.clear();
int ji=0,ou=0;
for(i=1;i<=n;i++){
scanf("%d",&a[i]);
if(a[i]&1){
ji++;odd.push_back(a[i]);
}else {
ou++;eve.push_back(a[i]);
}
}
int q=k-p;
if(ji<q || ji>=q && (ji-q)&1){
puts("NO");continue;
}
if(q+(ji-q)/2+ou<k){
puts("NO");continue;
}
PY;
if(p==0){
for(i=0;i<k-1;i++)
printf("1 %d\n",odd[i]);
printf("%d",n-k+1);
for(;i<odd.size();i++)
printf(" %d",odd[i]);
for(i=0;i<eve.size();i++)
printf(" %d",eve[i]);
puts("");continue;
}else if(q==0){
if(eve.size()>=k-1){
for(i=0;i<k-1;i++)
printf("1 %d\n",eve[i]);
printf("%d",n-k+1);
for(;i<eve.size();i++)
printf(" %d",eve[i]);
for(i=0;i<odd.size();i++)
printf(" %d",odd[i]);
puts("");continue;
}else {
int cnt=0;
for(i=0;i<eve.size();i++){
printf("1 %d\n",eve[i]);
cnt++;
}
int temp=eve.size();
j=0;
for(i=temp+1;i<=k-1;i++){
printf("2 ");
printf("%d %d\n",odd[j],odd[j+1]);
cnt+=2;j+=2;
}
if(n-cnt>0){
printf("%d",n-cnt);
for(;j<odd.size();j++)
printf(" %d",odd[j]);
puts("");
}
continue;
}
}
int cnt=0;
for(i=0;i<q;i++){
printf("1 %d\n",odd[i]);
cnt++;
}
if(ou<p){
for(i=0;i<eve.size();i++){
printf("1 %d\n",eve[i]);
cnt++;
}
int c=eve.size()+1;
for(i=q;c<p && i<odd.size();i+=2){
printf("2 %d %d\n",odd[i],odd[i+1]);
c++;cnt+=2;
}
printf("%d",n-cnt);
for(;i<odd.size();i++)
printf(" %d",odd[i]);
puts("");
}else {
for(i=0;i<p-1;i++){
printf("1 %d\n",eve[i]);cnt++;
}
printf("%d",n-cnt);
for(;i<eve.size();i++)
printf(" %d",eve[i]);
for(i=q;i<odd.size();i++)
printf(" %d",odd[i]);
puts("");
}
}
return 0;
}

/*
8 4 4
1 1 1 1 1 1 1 1

*/