POJ 2182 Lost Cows（牛排序，线段树）

最新推荐文章于 2018-05-25 08:40:52 发布

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最新推荐文章于 2018-05-25 08:40:52 发布

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分类专栏：线段树

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本文详细介绍了USACO 2003 US Open Orange赛题中关于牛排序的问题，通过提供输入示例、输出解答和核心算法实现（使用C++），深入探讨了如何根据给定的牛的品牌和它们之间的相对顺序，推断出牛的正确排列顺序。重点介绍了二叉搜索树（BST）的构建和查询过程，以及如何优化排序过程。

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Lost Cows

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9207		Accepted: 5922

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

Source

USACO 2003 U S Open Orange

给牛排序，n头牛，然后从第二头牛开始给出他在比他序号小这么多牛的地位，最后确定牛的排名

第一次看别人题解惊呆了，f[pos].va--,居然就可以把这个位置腾出来，以后都不会访问，还有一点，从后往前给牛排名

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>

#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)
using namespace std;
#define N  8005

struct stud{
int le,ri;
int len;
}f[N*4];

int a[N];
int ans[N];

void build(int pos,int le,int ri)
{
    f[pos].le=le;
    f[pos].ri=ri;
    f[pos].len=ri-le+1;
    if(ri==le)  return ;

    int mid=MID(le,ri);

    build(L(pos),le,mid);
    build(R(pos),mid+1,ri);

}

int query(int pos,int le)
{
   f[pos].len--;

   if(f[pos].le==f[pos].ri)
      return f[pos].le;

   if(f[L(pos)].len>=le)
      return query(L(pos),le);
   return query(R(pos),le-f[L(pos)].len);
}

int main()
{
    int i,j,n;
    while(~scanf("%d",&n))
    {
        build(1,1,n);

        for(i=2;i<=n;i++)
            scanf("%d",&a[i]);

        a[1]=0;

        for(i=n;i>=1;i--)
            ans[i]=query(1,a[i]+1);

        for(i=1;i<=n;i++)
            printf("%d\n",ans[i]);

        return 0;
    }
    return 0;
}