poj 2182 Lost Cows（线段树）

http://poj.org/problem?id=2182

Lost Cows

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 14055		Accepted: 8944

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

5
1
2
1
0

Sample Output

2
4
5
3
1

Source

USACO 2003 U S Open Orange

---

是个好题

样例 1 2 1 0， 从后往前找，

0代表剩下的数中，没有比他小的，那他就是1， 于是还剩 2 3 4 5

1代表剩下的数中，有一个比他小，就是3， 还剩 2 4 5

。。。。

最终序列就是2 4 5 3 1

利用线段树存储l 到 r中还有多少个数没被去掉，每次面对一个ai 那么就有ai个数比他小，那么就找到ai+1的位置，就是对应的答案

 

#include<iostream>
using namespace std;

const int N = 8005;
struct node
{
	int l, r, sum;
}t[N<<2];
int a[N], ans[N];

void build(int x, int l, int r)
{
	t[x].l = l, t[x].r = r;
	t[x].sum = r - l + 1;
	
	if(l == r) return ;
	
	int mid = (l + r) >> 1;
	build(x<<1, l ,mid);
	build(x<<1|1, mid+1, r);
}

int update(int x, int k)
{
	t[x].sum--;
	
	if(t[x].l == t[x].r) return t[x].l;
	
	if(k <= t[x<<1].sum)//如果左子树剩下的数够 
		return update(x<<1, k);
	else
		return update(x<<1|1, k-t[x<<1].sum);//要减去左子树剩下的数 
}

int main()
{
	int n;
	cin >> n;
	for(int i = 2; i <= n; i++)
	scanf("%d", a+i);
	
	build(1, 1, n);
	
	for(int i = n; i >= 1; i--)
	ans[i] = update(1, a[i]+1);
	
	for(int i = 1; i <= n; i++)
	cout << ans[i] << endl;
	
	return 0;
}