主题链接：http://poj.org/problem?
id=2182
|
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 Sample Output 2 Source |
[Submit] [Go Back] [Status]
[Discuss]
线段树的一道经典题目。这个题目能够转化成从后向前依次查询，比方当前奶牛的前面有x个号码比它小的奶牛，那么它就应该在剩余的数的序列中排第x+1。当某个号码确定时，就在线段树中去除这个号码。我们遍历一下线段树若左子树区间内未删除元素个数满足当前要找的数成为第a+1个。能则递归左子树。否则递归右子树，直至到叶子节点。那么叶子节点的值就是其初始编号。
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<cstdio>
#include<vector>
#include<bitset>
#include<string>
#include<queue>
#include<stack>
#include<set>
#include<map>
#include<cstdlib>
using namespace std;
#define CLR(A) memset(A,0,sizeof(A))
const int MAX=8010;
struct Node{
int l,r,len,v;
}tree[MAX*4];
int ans[MAX];
int f[MAX];
inline void pushup(int id){
tree[id].len=tree[id<<1].len+tree[id<<1|1].len;
}
void build(int id,int l,int r){
tree[id].l=l;tree[id].r=r;
if(l==r){
tree[id].v=l;
tree[id].len=1;
return;
}
int m=(l+r)>>1;
build(id<<1,l,m);
build(id<<1|1,m+1,r);
pushup(id);
}
void query(int id,int cur,int pos){
if(tree[id].l==tree[id].r){
tree[id].len=0;
ans[pos]=tree[id].v;
return;
}
if(tree[id<<1].len>=cur) query(id<<1,cur,pos);
else query(id<<1|1,cur-tree[id<<1].len,pos);
pushup(id);
}
int main(){
int n;
while(~scanf("%d",&n)&&n){
for(int i=2;i<=n;i++){
scanf("%d",&f[i]);
f[i]++;
}
build(1,1,n);
f[1]=1;
for(int i=n;i>=1;i--){
query(1,f[i],i);
}
for(int i=1;i<=n;i++){
printf("%d\n",ans[i]);
}
}
return 0;
}
版权声明：本文博主原创文章，博客，未经同意不得转载。