丢失的牛？_Lost Cows_POJ2182_线段树

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

题目大意：

有n头牛，编号为1~n，它们并没有按照编号的顺序排好队列。现在只知道每一个牛前面有多少只牛的编号比它小。求所有牛的编号。

 

太水，懒得讲

源代码/pas:

 

#include <stdio.h>#include <stack>using namespace std;struct tree{int l,r,c;}t[24001];stack<int>ans;int s[8001];void build(int f,int x,int y){t[f]=(tree){x,y,y-x+1};int mid=(x+y)>>1;if (x==y)return;build(f+f,x,mid);build(f+f+1,mid+1,y);}int count(int f,int x){--t[f].c;if (t[f].l==t[f].r)return t[f].r;if (t[f+f].c<x)return count(f+f+1,x-t[f+f].c);elsereturn count(f+f,x);}int main(){int n;scanf("%d",&n);for (int i=2;i<=n;i++)scanf("%d",&s[i]);build(1,1,n);for (int i=n;i>=1;i--)ans.push(count(1,s[i]+1));while (!ans.empty()){printf("%d\n",ans.top());ans.pop();}return 0;}

typetree=record  l,r,c:longint;end;var  t:array[0..24000]of tree;  a:array[0..8001]of longint;  n:longint;procedure build(f,x,y:longint);var  mid:longint;begin  t[f].l:=x;  t[f].r:=y;  t[f].c:=y-x+1;  mid:=(x+y)div 2;  if x=y then exit;  build(f*2,x,mid);  build(f*2+1,mid+1,y);end;function count(f,x:longint):longint;begin  dec(t[f].c);  if t[f].l=t[f].r then  exit(t[f].l)  else  if t[f*2].c<x then  exit(count(f*2+1,x-t[f*2].c))  else  exit(count(f*2,x));end;procedure main;var  i:longint;  s:array[0..8001]of longint;begin  for i:=n downto 1 do  s[i]:=count(1,a[i]+1);  for i:=1 to n do  writeln(s[i]);end;procedure init;var  i:longint;begin  readln(n);  for i:=2 to n do read(a[i]);  build(1,1,n);end;begin  init;  main;end.