E. Painting The Fence 优先队列+贪心

E. Painting The Fence

time limit per test

1.0 s

memory limit per test

256 MB

input

standard input

output

standard output

The fence consists of n

planks, arranged from left to right. Monocarp has m different types of paint, and the number of planks that can be painted in color i is ai (there's not enough paint to color any more planks). The total amount of paint is just enough to paint exactly n planks — in other words, the sum of all ai is equal to n

. Each plank should be painted into exactly one color.

Monocarp has to paint the fence in such a way that the length of every contiguous segment consisting of planks of the same color is not greater than k

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.

Find a suitable way to paint the fence, or say that it is impossible.

Input

The first line contains three integers n

, m and k (1≤n≤2⋅105,1≤m,k≤n)

— the number of planks in the fence, the number of different colors of paint and the maximum length of a contiguous segment of planks with the same color, respectively.

The second line contains a sequence of integers a1,a2,…,am

(1≤ai≤n), where ai is the number of planks that can be painted with color i. The sum of all values of ai is equal to n

.

Output

If it is impossible to paint the fence in such a way that the length of every contiguous segment consisting of planks of the same color is not greater than k

, print −1

.

Otherwise print n

integers, i-th of them should be equal to the index of the color of the i

-th plank. If there are multiple possible answers, print any of them.

Examples

Input

Copy

5 2 1
2 3

Output

Copy

2 1 2 1 2 

Input

Copy

8 2 3
1 7

Output

Copy

-1

Input

Copy

10 3 2
5 2 3

Output

Copy

1 1 3 1 1 2 3 1 2 3 

Note

In the first example the first, third and fifth planks should have color 2

, and all other planks should have color 1

.

In the second example it is impossible to paint the fence in the required way, so the answer is −1

.解题思路：贪心每次找数目最大的数字SS.id，取k个进行排队，然后找第二多的数字S2.id，选一个排在后面， 取出的SS.v,S2.v对数目进行修改，再放入队列中，

但是每次选最多数目时，可能会出现和上次第二多数目S2.id为相同id,则需分类讨论，

贪心保证每次排队都不超过k,且尽可能连续得多。

结构体类型优先队列 运算符必须重载

#include<bits/stdc++.h>
using namespace std;
int n,m,k;
struct node
{
   	int id,v;
	bool operator < (const node & S) const//优先队列 比较符号《必须》重载 
    	{
    		return v<S.v;//默认大值优先级高 
    	}
}a[200005];
priority_queue<node> que;
vector<int > ans;
int main()
{
   	scanf("%d%d%d",&n,&m,&k);
   	for(int i=0;i<m;i++)
   	{
   		scanf("%d",&a[i].v);
   		a[i].id=i+1;
   		que.push(a[i]);
   	}
   	node SS,S2;
   	int tt=n;
   	int last=0;
   	int lastnum=0;
   	while(tt)
   	{
   		int t;
   		SS=que.top();
   		que.pop();
   		if(SS.id==last)//最后一个数和当前最大数量的数字相同时 
   		{
   			if(lastnum==k)//已经达到最大连续个数  换下一个队列值 
   			{			
   			node T=que.top();
   			que.pop();
   			que.push(SS);
   			SS=T;
   			t=min(k,SS.v);
			}
			else//未达最大连续个数 贪心继续加 
			{
				t=min(k-lastnum,SS.v);
			}
		}
		else t=min(k,SS.v);//和最后一个数字不同 
   		S2=que.top();
   		que.pop();
   		SS.v-=t;
   		if(t==0) //tt!=0 当前第二大的已经为0  则不合法 
   		{
   			printf("-1\n");
   			return 0;
		}
   		for(int i=0;i<t;i++)
   		{
   			ans.push_back(SS.id);
   			last=SS.id;
   			lastnum=t;
   		}
   		tt-=t;
   		if(S2.v==0&&tt)
   		{
   			printf("-1\n");
   			return 0;
   		}
   		else if(tt)
   		{
   			ans.push_back(S2.id);
   			last=S2.id;
   			lastnum=1;
   			tt-=1;
   			S2.v-=1;
   		}
   		que.push(SS);
   		que.push(S2);
   	}
   	for(int i=0;i<n;i++)
   	{
   		printf("%d ",ans[i]);
   	}
   	
}

//12 3 2
//5 5 2 
//1 1 2 2 1 2 2 3 1 1 2 3