POJ 2182 Lost Cows(点的查找)

Lost Cows

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 9350 Accepted: 6022

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

51210

Sample Output

24531

本题的大致意思为：有N头牛，编号为1~N ，乱序排成一列，现在已知每头牛前面有多少头牛比他的编号小，求排队后从前往后数，每头牛的编号

• Source Code

  #include<cstdio>#include<iostream>#include<algorithm>using namespace std;int small[10000],ans[10000];struct Se{    int rc,lc,Len; }s[30000];void build(int root, int lc, int rc){s[root].lc = lc;s[root].rc = rc;s[root].Len = rc - lc + 1;if(lc==rc)         return;int mid = (lc +rc)/2;build(2*root,lc, mid);build(2*root+1,mid+1,rc);}int query(int root ,int k){s[root].Len--;if(s[root].lc == s[root].rc)return s[root].rc;else if(k <= s[2*root].Len)return query(2*root,k);else      return query(root*2+1 ,k- s[2*root].Len);}int main(){int n,i;while(scanf("%d",&n)!=EOF){for(int i=2;i <= n; i++)scanf("%d",&small[i]);small[1] = 0;build(1,1,n);for(int i=n; i >=1; i--){ans[i] = query(1, small[i]+1);}for(int i=1; i<= n;i++)      printf("%d\n",ans[i]);}return 0;}