Cows - POJ 2481 树状数组_each of farmer john's n (1 <= n <= 1000) cows has -优快云博客

本文链接：https://blog.youkuaiyun.com/u014733623/article/details/25799637

本文解析了一道经典的算法题目——牛群吃草问题。该问题要求确定每头牛相对于其他牛的优势地位，通过巧妙地利用二元素排序及树状数组来解决。文章详细介绍了问题背景、解题思路及AC代码实现。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Cows

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 11846		Accepted: 3915

Description

Farmer John's cows have discovered that the clover growing along the ridge of the hill (which we can think of as a one-dimensional number line) in his field is particularly good.

Farmer John has N cows (we number the cows from 1 to N). Each of Farmer John's N cows has a range of clover that she particularly likes (these ranges might overlap). The ranges are defined by a closed interval [S,E].

But some cows are strong and some are weak. Given two cows: cow_i and cow_j, their favourite clover range is [Si, Ei] and [Sj, Ej]. If Si <= Sj and Ej <= Ei and Ei - Si > Ej - Sj, we say that cow_i is stronger than cow_j.

For each cow, how many cows are stronger than her? Farmer John needs your help!

Input

The input contains multiple test cases.
For each test case, the first line is an integer N (1 <= N <= 10⁵), which is the number of cows. Then come N lines, the i-th of which contains two integers: S and E(0 <= S < E <= 10⁵) specifying the start end location respectively of a range preferred by some cow. Locations are given as distance from the start of the ridge.

The end of the input contains a single 0.

Output

For each test case, output one line containing n space-separated integers, the i-th of which specifying the number of cows that are stronger than cow_i.

Sample Input

Sample Output

1 0 0

题意：给你每头牛的吃草范围，如果一头牛A的吃草范围包括另一头牛B的范围，且他们的范围不相同，那么A牛比B牛更牛（更壮）。问你对于每头牛，有多少头牛比它牛（壮）。

思路：二元素排序，先以左范围从小到大排，再以右范围从大到小排，这样每次就可以得到有多少头牛比它壮了。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
int a[100010];
int ans[100010];
int lowbit(int x)
{ return x&(-x);
}
void update(int x)
{ while(x<=100000)
  { a[x]++;
    x+=lowbit(x);
  }
}
int sum(int x)
{ int ret=0;
  while(x)
  { ret+=a[x];
    x-=lowbit(x);
  }
  return ret;
}
struct node
{ int l,r,pos;
}cow[100010];
bool cmp(node a,node b)
{ if(a.l<b.l)
   return true;
  else if(a.l==b.l)
   return a.r>b.r;
  else
   return false;
}
int main()
{ int n,i,j;
  while(~scanf("%d",&n) && n>0)
  { for(i=1;i<=n;i++)
    { scanf("%d%d",&cow[i].l,&cow[i].r);
      cow[i].pos=i;
    }
    memset(a,0,sizeof(a));
    sort(cow+1,cow+1+n,cmp);
    for(i=1;i<=n;i++)
    { if(cow[i].l==cow[i-1].l && cow[i].r==cow[i-1].r)
       ans[cow[i].pos]=ans[cow[i-1].pos];
      else
      ans[cow[i].pos]=i-1-sum(cow[i].r-1);
      update(cow[i].r);
    }
    printf("%d",ans[1]);
    for(i=2;i<=n;i++)
     printf(" %d",ans[i]);
    printf("\n");
  }
}