POJ 2182 Lost Cows树状数组

最新推荐文章于 2018-08-06 20:44:30 发布

最新推荐文章于 2018-08-06 20:44:30 发布 · 251 阅读

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本文介绍了一个有趣的排序问题：通过已知每头牛前面有多少头品牌编号较小的牛，逆向推断出牛群原本的品牌编号顺序。利用树状数组进行高效查询与更新，通过二分查找确定每个位置上牛的品牌编号。

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Description

N (2 <= N <= 8,000) cows have unique brands in the range 1..N. In a spectacular display of poor judgment, they visited the neighborhood 'watering hole' and drank a few too many beers before dinner. When it was time to line up for their evening meal, they did not line up in the required ascending numerical order of their brands.

Regrettably, FJ does not have a way to sort them. Furthermore, he's not very good at observing problems. Instead of writing down each cow's brand, he determined a rather silly statistic: For each cow in line, he knows the number of cows that precede that cow in line that do, in fact, have smaller brands than that cow.

Given this data, tell FJ the exact ordering of the cows.

Input

* Line 1: A single integer, N

* Lines 2..N: These N-1 lines describe the number of cows that precede a given cow in line and have brands smaller than that cow. Of course, no cows precede the first cow in line, so she is not listed. Line 2 of the input describes the number of preceding cows whose brands are smaller than the cow in slot #2; line 3 describes the number of preceding cows whose brands are smaller than the cow in slot #3; and so on.

Output

* Lines 1..N: Each of the N lines of output tells the brand of a cow in line. Line #1 of the output tells the brand of the first cow in line; line 2 tells the brand of the second cow; and so on.

Sample Input

Sample Output

本题关键在于知道从后往前找的话，前面有n个点，就是指，该点为第n+1个未使用的点，我们用树状数组的add来确定某点是否被使用，一旦使用就add（x，-1）把他减去，用sum求某点是第几个未被使用的点，之后用二分法确定哪个点是未被使用的第n+1个点，把那点的数值赋给ans，最后按序输出即可。

#include <stdio.h>

int h[10000];
int ans[10000];
int c[10000];
int n;

int lowbit(int k)  
{  
    return (k&(-k));  
}


int sum(int x)  		//求和
{  
    int ret = 0;  
    while(x>0)  
    {  
        ret+=c[x];  
        x-=lowbit(x);  
    }  
    return ret;  
}  

void add(int x,int d)  		//修改节点的值
{  
    while(x<=n)  
    {  
        c[x]+=d;  
        x+=lowbit(x);  
    }  
}

int main()
{
	int i,j;
	scanf("%d",&n);
	for(i=2;i<=n;i++)
	{
		scanf("%d",&h[i]);
	}
	h[1]=0;
	for(i=1;i<=n;i++)
	{
		add(i,1);
	}
	for(i=n;i>=1;i--)
	{
		int l=1,r=n;
		int fou=h[i]+1;
		while(l<=r)
		{
			int mid=(l+r)/2;
			if(sum(mid)<fou)
				l=mid+1;
			else
				r=mid-1;
		}
		ans[i]=l;
		add(l,-1);
	}
	for(i=1;i<=n;i++)
		printf("%d\n",ans[i]);
	return 0;
}