Poj 2182 Lost Cows 线段树

最新推荐文章于 2019-09-25 15:05:59 发布

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本文介绍了一种解决LostCows问题的方法，通过记录每头牛前面有多少头编号较小的牛，利用线段树数据结构逆向确定每头牛的具体编号。文章详细解释了算法思路，并提供了完整的代码实现。

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Lost Cows

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions:12167		Accepted: 7819

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.

弱校ACMer一直偷懒没怎么搞懂线段树。。。

从头来一遍。。

有n头奶牛都有自己的编号。

现在他们站在一排。。

每个奶牛都知道前边的 i-1 头奶牛中有几个比他编号大的。。

我们如果从前面走第二个如果是1 的话那么它的答案可以使（2-n）中任意一个数字

所以我们从后开始遍历。

最后一个牛前面有s【i】头比它大的因为她是最后一头了。。所以它的编号就是s【i】+1；

然后往前推。。

暴力的话枚举每个点数据小也可以把。。

这里不行。。

上代码。

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
#define N 80005
struct Node
{
    int l,r;
    int val;
} tree[N];
int s[N],res[N];
void build(int v,int l,int r)
{
    tree[v].l=l;
    tree[v].r=r;
    tree[v].val=r-l+1;
    if(l == r)return;
    int mid = (l + r)/2;
    build(v*2, l, mid);
    build(v*2+1, mid+1, r);
    return;
}
int query(int root,int Target)
{
    tree[root].val--;
    if(tree[root].l == tree[root].r)
        return tree[root].l;
    if(Target<=tree[2*root].val)
    {
        return query(root*2,Target);
    }
    else return query(root*2+1,Target-tree[2*root].val);
}
int main()
{
    int n;
    scanf("%d",&n);
    for(int i=2; i<=n; i++)
    {
        scanf("%d",&s[i]);
    }
    s[1]=0;
    build(1,1,n);
    for(int i=n; i>=1; i--)
    {
        res[i]=query(1,s[i]+1);
    }
    for(int i=1; i<=n; i++)
    {
        printf("%d\n",res[i]);
    }
}