poj2481 Cows

Description

Input

Output

Sample Input

3
1 2
0 3
3 4
0

Sample Output

1 0 0

这题的做法和star那道题差点儿相同，先按x坐标进行升序排列，然后x同样的取对y进行降序排列，然后每次循环推断当前线段和上一条线段是不是x。y都一样，假设一样就直接等于上一条算出的值。不等于就计算。

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define maxn 100006
struct node{
	int x,y,id,num;
}a[maxn];
int b[maxn];
bool cmp(node a,node b){
	if(a.x==b.x)return a.y>b.y;
	return a.x<b.x;
}
bool cmp1(node a,node b){
	return a.id<b.id;
}

int lowbit(int x){
	return x&(-x);
}
void update(int pos,int num){
	while(pos<=maxn){
		b[pos]+=num;pos+=lowbit(pos);
	}
}
int getsum(int pos)
{
	int num=0;
	while(pos>0){
		num+=b[pos];pos-=lowbit(pos);
	}
	return num;
}

int main()
{
	int n,m,i,j,t;
	while(scanf("%d",&n)!=EOF && n!=0)
	{
		memset(a,0,sizeof(a));
		for(i=1;i<=n;i++){
			scanf("%d%d",&a[i].x,&a[i].y);
			a[i].x++;a[i].y++;
			a[i].id=i;
		}
		memset(b,0,sizeof(b));
		sort(a+1,a+1+n,cmp);
		for(i=1;i<=n;i++){
			if(a[i].x==a[i-1].x && a[i].y==a[i-1].y){
				a[i].num=a[i-1].num;
			}
			else{
				a[i].num=getsum(maxn)-getsum(a[i].y-1);
			}
			update(a[i].y,1);
		}
		sort(a+1,a+1+n,cmp1);
		for(i=1;i<=n;i++){
			if(i==n)printf("%d\n",a[i].num);
			else printf("%d ",a[i].num);
		}
	}
	return 0;
}