Lost Cows(线段树 POJ2182)

Lost Cows
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 10354 Accepted: 6631

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

类似排队买票,从后先前查找

#include <map>
#include <set>
#include <queue>
#include <cstring>
#include <string>
#include <cstdio>
#include <iostream>
#include <algorithm>

using namespace std;

typedef long long LL;

int Tree[32000];

int a[8500];

void Build(int L,int R,int site)
{
    Tree[site]=(R-L+1);//每个区间中牛的个数

    if(L==R)
    {
        return ;
    }
    int mid=(L+R)>>1;

    Build(L,mid,site<<1);

    Build(mid+1,R,site<<1|1);
}

int Query(int L,int R,int site,int num)//查询牛的编号
{
    Tree[site]--;

    if(L==R)
    {

        return L;
    }

    int mid=(L+R)>>1;

    if(Tree[site<<1]>=num)
    {
        return Query(L,mid,site<<1,num);
    }
    else
    {
        return Query(mid+1,R,site<<1|1,num-Tree[site<<1]);
    }
}

int main()
{
    int n;

    while(~scanf("%d",&n))
    {
        a[1]=0;//第一个元素前面比他小的牛自然是0个人

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

        Build(1,n,1);

        for(int i=n;i>=1;i--)
        {
            a[i]=Query(1,n,1,a[i]+1);//a[i]+1为这个牛的剩余牛中的位置
        }

        for(int i=1;i<=n;i++)
        {
            printf("%d\n",a[i]);
        }
    }
    return 0;
}