Codeforces Round #374 (Div. 2) D. Maxim and Array 贪心+ 最小堆

D. Maxim and Array
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
Recently Maxim has found an array of n integers, needed by no one. He immediately come up with idea of changing it: he invented positive integer x and decided to add or subtract it from arbitrary array elements. Formally, by applying single operation Maxim chooses integer i (1 ≤ i ≤ n) and replaces the i-th element of array ai either with ai + x or with ai - x. Please note that the operation may be applied more than once to the same position.
Maxim is a curious minimalis, thus he wants to know what is the minimum value that the product of all array elements (i.e. ) can reach, if Maxim would apply no more than k operations to it. Please help him in that.
Input
The first line of the input contains three integers n, k and x (1 ≤ n, k ≤ 200 000, 1 ≤ x ≤ 109) — the number of elements in the array, the maximum number of operations and the number invented by Maxim, respectively.
The second line contains n integers a1, a2, ..., an () — the elements of the array found by Maxim.
Output
Print n integers b1, b2, ..., bn in the only line — the array elements after applying no more than k operations to the array. In particular,  should stay true for every 1 ≤ i ≤ n, but the product of all array elements should be minimum possible.
If there are multiple answers, print any of them.
Examples
input
5 3 1
5 4 3 5 2

output

5 4 3 5 -1 

input

5 3 1
5 4 3 5 5

output

5 4 0 5 5 

input

5 3 1
5 4 4 5 5

output

5 1 4 5 5 

input

3 2 7
5 4 2

output

5 11 -5 

贪心，一开始的思路错了，一开始想先找出负数的个数，如果为偶数个则找到绝对值最小的那个将它变成异号，否则就不变。

然后再找到绝对值最小的那个，将剩余的k全部加或者减在上面，就是这个地方有了问题，就比如 

3 2 3

-2 4 6

如果照着我的想法，就是 -8 4 6

但是 -5 7 6明显更好，所以我想到了应该k的变化应该一次次进行，每次找到绝对值最小的那个进行变化。

#include<iostream>
#include<cstring>
#include<cstdio>
#include<vector>
#include<queue>
#include<algorithm>
#include<cmath>
#include<stack>
const int inf=0x3f3f3f3f;
typedef long long LL;
using namespace std;
const int MAXN=2e5+10;
LL a[MAXN];
LL b[MAXN];
struct Node
{
    int id;
    LL value;
    Node(int i,LL v)
    {
        id=i;
        value=v;
    }
};
struct cmp
{
    bool operator()(Node a,Node b){
        return a.value>b.value;
    }
};
int main()
{
    int n,k,x;
    while(cin>>n>>k>>x)
    {
        int sum=0;
        for(int i=1;i<=n;i++)
        {
            scanf("%I64d",&a[i]);
            b[i]=a[i];
            if(a[i]<0)
            {
                sum++;
                b[i]*=-1LL;
            }
        }
        int h=0;
        if(sum%2==0)
        {
            LL MIN=b[1];
            int index=1;
            for(int i=2;i<=n;i++)
            {
                if(MIN>b[i])
                {
                    MIN=b[i];
                    index=i;
                }
            }
            int t=k;
            if(MIN/x+1<=k)
            {
                b[index]-=LL(MIN/x+1)*LL(x);
                k-=(MIN/x+1);
            }
            else
            {
                b[index]-=LL(k*x);
                k=0;
            }
            if(a[index]<0)
            {
                a[index]+=LL(t-k)*LL(x);
            }
            else
            {
                a[index]-=LL(t-k)*LL(x);
            }
        }
        if(k)
        {
            for(int i=1;i<=n;i++)
            {
                b[i]=a[i];
                if(b[i]<0)
                {
                    b[i]*=(-1LL);
                }
            }
            priority_queue<Node,vector<Node>,cmp> Q;
            for(int i=1;i<=n;i++)
            {
                Q.push(Node(i,b[i]));
            }
            while(k)
            {
                Node u=Q.top();
                Q.pop();
                b[u.id]+=LL(x);
                Q.push(Node(u.id,b[u.id]));
                k--;
            }
            for(int i=1;i<=n;i++)
            {
                if(a[i]<0)
                {
                    a[i]=b[i]*-1LL;
                }
                else
                {
                    a[i]=b[i];
                }
            }
        }
        printf("%I64d",a[1]);
        for(int i=2;i<=n;i++)
        {
            printf(" %I64d",a[i]);
        }
        cout<<endl;

    }
}