poj 2481 cows

Cows

Time Limit : 6000/3000ms (Java/Other) Memory Limit : 131072/65536K (Java/Other)

Total Submission(s) : 59 Accepted Submission(s) : 14

Problem 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

3
1 2
0 3
3 4
0

Sample Output

1 0 0

Source

PKU

AC代码：

#include<iostream>   
#include<cstdio>   
#include<cstring>   
#include<algorithm>   
#define low_bit(k) k&-k;   
  
using namespace std;  
  
int n,mx,my;  
int tol[100100];//用于树状数组    
int val[100100];//每个点的结果(左上方的点的个数)    
struct Node{  
    int x,y;  
    int d;  
}dq[100100];  
  
int cmp(Node a,Node b){//排序很重要   
    if(a.y==b.y) return a.x<b.x;  //按纵坐标由小到大排序 
    return a.y>b.y;  //当纵坐标相同时横坐标按由小到大排序 
}  
  
int _cmp(Node a,Node b){  
    return a.d<b.d;  
}  
  
void add(int x){  
    while(x<=mx){  
        tol[x]+=1;  
        x+=low_bit(x);  
    }  
}  
  
int sum(int x){  
    int cnt=0;  
    while(x) cnt+=tol[x],x-=low_bit(x);  
    return cnt;  
}  
  
int main(){  
    int i,tot;  
    while(scanf("%d",&n),n){  
        memset(tol,0,sizeof(tol));  
        memset(val,0,sizeof(val));  
        for(i=1,mx=0,my=0;i<=n;i++){  
            scanf("%d%d",&dq[i].x,&dq[i].y);  
            dq[i].x++,dq[i].y++;  
            dq[i].d=i;  
            mx<dq[i].x?mx=dq[i].x:0;  
        }  
        sort(dq+1,dq+n+1,cmp);   
        val[dq[1].d]=sum(dq[1].x);  
        add(dq[1].x);  
        for(i=2;i<=n;i++){  
            if(dq[i].y==dq[i-1].y&&dq[i].x==dq[i-1].x){//如果有相同点的时候   
            //排序后相邻了，所以直接等于前一个，但要add();    
                val[dq[i].d]=val[dq[i-1].d];  
                add(dq[i].x);  
                continue;  
            }  
            val[dq[i].d]=sum(dq[i].x);  
            add(dq[i].x);  
        }   
        for(i=1;i<n;i++) printf("%d ",val[i]);  
        printf("%d\n",val[n]);  
    }  
    return 0;  
}