POJ 2182 Lost Cows [树状数组+二分]

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
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分析

树状数组的 SUM( X ) 用于记录 编号X 后面满足小于等于 X 的已经用掉了的编号的个数；F[ i ] 就是题目给出的 第 i 个牛 前面比 第i 个牛的编号小的编号的个数； 我们需要二分的就是 X，判断 X是不是当前第 i 头牛的编号。

如果 (X-1)-SUM( X - 1) == F[ i ]
(即 编号X 前面剩下的小于 X 的编号的数量恰好等于 第 i 头牛编号的条件， 则 X 就是 第 i 头牛的编号啦）

如果(X-1)-SUM( X - 1) > F[ i ]
（说明编号偏大咯）

如果(X-1)-SUM( X - 1) < F[ i ]
(说明编号偏小咯）
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程序：

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <queue>
using namespace std;
int n,tree[10000],f[10000],num[10000];
int lowbit(int x)
{
    return x & -x;
}
void add(int x,int k)
{
    while(x<=n)
    {
        tree[x]+=k;
        x+=lowbit(x);
    }
}
int sum(int x)
{
    int ans=0;
    while(x>0)
    {
        ans+=tree[x];
        x-=lowbit(x);
    }
    return ans;
}
int main()
{
    scanf("%d",&n);
    num[1]=0;
    for (int i=2;i<=n;i++)
        scanf("%d",&f[i]);
    num[n]=f[n]+1;
    add(num[n],1);
    for (int i=n-1;i>0;i--)
    {
        int l=1,r=n;
        while (r>l)
        {
            int mid=(l+r)>>1;
            if (mid-1-sum(mid)>=f[i]) r=mid; else l=mid+1;
        }
        num[i]=l;
        add(num[i],1);
    }
    for (int i=1;i<=n;i++)
        printf("%d
",num[i]);
    return 0;
}