POJ 2182 Lost Cows（牛排序，线段树）

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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

5
1
2
1
0

Sample Output

2
4
5
3
1

Source

USACO 2003 U S Open Orange

给牛排序，n头牛，然后从第二头牛開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--,竟然就能够把这个位置腾出来，以后都不会訪问，另一点，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;
}