Lost Cows
| Time Limit: 1000MS |
|
Memory Limit: 65536K |
| Total Submissions: 10179 |
|
Accepted: 6525 |
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
每只牛知道自己后边的比自己编号小的牛的个数，求这些牛的编号，之前用树状数组写过一次，不过感觉线段树更简单，还是插空
ac代码
#include<stdio.h>
#include<string.h>
int a[8080],node[8080<<2],ans[8080];
void build(int l,int r,int tr)
{
node[tr]=r-l+1;
if(l==r)
return;
int mid=(l+r)>>1;
build(l,mid,tr<<1);
build(mid+1,r,tr<<1|1);
}
int query(int val,int l,int r,int tr)
{
node[tr]--;
if(l==r)
{
return l;
}
int mid=(l+r)>>1;
if(val<=node[tr<<1])
query(val,l,mid,tr<<1);
else
query(val-node[tr<<1],mid+1,r,tr<<1|1);
}
int main()
{
int n;
while(scanf("%d",&n)!=EOF)
{
int i;
a[1]=0;
memset(node,0,sizeof(node));
build(1,n,1);
for(i=2;i<=n;i++)
{
scanf("%d",&a[i]);
}
for(i=n;i>=1;i--)
{
ans[i]=query(a[i]+1,1,n,1);
}
for(i=1;i<=n;i++)
{
printf("%d\n",ans[i]);
}
}
}
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POJ 题目2182 Lost Cows（线段树）
原文：http://blog.csdn.net/yu_ch_sh/article/details/47378015
