poj 2182 Lost Cows

Lost Cows

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

 1 #include <cstdio>
 2 #include<algorithm>
 3 #include<iostream>
 4 #include<string>
 5 #include<cstring>
 6 #include<stack>
 7 using namespace std;
 8 #define N 8000
 9 struct node{
10     int l,r,num;
11 };
12 node tree[3*N];
13 int ans[N+5];
14 void build(int num,int l,int r){//建线段树
15     tree[num].num=r-l+1;
16     tree[num].l=l;
17     tree[num].r=r;
18     if(l==r){//到根节点
19         //cout<<" "<<l<<endl;
20         return;
21     }
22     int mid=(l+r)>>1;
23     build(num<<1,l,mid);
24     build(num<<1|1,mid+1,r);
25 }
26 int update(int num,int be){//更新
27     tree[num].num--;
28     if(tree[num].l==tree[num].r){
29         return tree[num].l;
30     }
31     if(tree[num<<1].num>=be){
32         return update(num<<1,be);
33     }
34     else{
35         return update(num<<1|1,be-tree[num<<1].num);
36     }
37 }
38 int main(){
39     int n;
40     cin>>n;
41     int i;
42     ans[0]=0;  //第一个前面有0个
43     for(i=1;i<n;i++){
44         scanf("%d",&ans[i]);
45     }
46     build(1,1,n);
47     for(i=n-1;i>=0;i--){
48         ans[i]=update(1,ans[i]+1);
49     }
50     for(i=0;i<n;i++){
51         cout<<ans[i]<<endl;
52     }
53     return 0;
54 }