POJ 2182 解题报告

Lost Cows

Time Limit: 1000 MS Memory Limit: 65536 KB

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

大致题意：要求出给给定n头牛的位置排列，由Sample Output 2 4 5 3 1 可以看出 4 前面比其小的牛的 头数 为 1，5 前面比其小的牛的头数为 2 ，

以此类推，可以得到input数据 1 2 1 0 ，所以我们倒序求解，用树状数组+二分
代码：

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<math.h>
using namespace std;

const int maxn=8005;

int n;
int num[maxn],C[maxn];

///树状数组
int lowbit(int x)
{
return x&(-x);
}

int sum(int x)
{
int ret=0;
while(x>0)
{
ret+=C[x];
x-=lowbit(x);
}
return ret;
}

void add(int x)
{
while(x<n)
{
C[x]++;
x+=lowbit(x);
}
}

///开始二分查找

int Search(int x)
{
int L=1,R=n,mid;
while(L<R)
{
mid=(L+R)/2;
int a=sum(mid);///统计num[mid]前比其小的牛的头数
if(mid-1-a>=x)
R=mid;
else
L=mid+1;
}
return L;
}

int main()
{
while(~scanf("%d",&n))
{
for(int i=1;i<n;i++)///注意从num[1]输到num[4],因为第一头牛前比其
/// 小的牛的头数永远为0
{
scanf("%d",&num[i]);///输入每个位置上的牛其前面比它小的牛的头数
}

for(int i=n-1;i>=0;i--)
{
int x=Search(num[i]);///对num[4]=0查找
num[i]=x;
add(x);
}
for(int i=0;i<n;i++)///输出num[0]至num[4]共5头牛的序号
{
printf("%d\n",num[i]);
}
}
return 0;
}