POJ 2481 Cows - 树状数组/线段树

Cows

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: cowi and cowj, their favourite clover range is [Si, Ei] and [Sj, Ej]. If Si <= Sj and Ej <= Ei and Ei - Si > Ej - Sj, we say that cowi is stronger than cowj.  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 cowi. 

大体思路同POJ 2352 Stars，只是需要自己写排序，注意在排序前记录排序前的顺序cnt，最后用num数组统计顺序输出即可。

N.B.

1.两头牛完全相等的情况下，若是第二头牛进行求和则必然包含了第一头牛，需要进行判断，跳过相等的情况；

2.每一次case循环后需要memset每个数组，切记！！！

#include
#include
#include
#include
#include
using namespace std;

const int maxn=32100;

int c[maxn],num[maxn];
int n,m;

inline int lowbit(int x)
{
	return x&(-x);
}
void update(int x)
{
	for(int i=x;i<=maxn;i+=lowbit(i))
			c[i]++;
}
int query(int x)
{
	int ans=0;
	for(int i=x;i>0;i-=lowbit(i))
			ans+=c[i];
	return ans;
}
int main()
{
	scanf("%d",&n);
	int x,y;
	for(int i=1;i<=n;i++)
	{
		scanf("%d%d",&x,&y);	
		num[query(x+1)]++;
		update(x+1);
	}
	for(int i=0;i