2724: Cows

描述

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!

输入

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.

输出

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.

样例输入

 3

1 2
0 3
3 4
0

样例输出

 1 0 0

题意

每头奶牛都有一个吃草区域[S,E];如果满足Si <= Sj and Ej <= Ei and Ei - Si > Ej - Sj，说明第i头牛比第j头牛强壮。对每头奶牛求比其强壮的奶牛数量。

代码

#include<bits/stdc++.h>
using namespace std;
const int N=1e5+5;
struct node
{
    int S,E,id;
}e[N];
int n,maxn;
int cmp(node x,node y)
{
    return x.S<y.S||x.S==y.S&&x.E>y.E;
}
int f[N],num[N];
void update(int x,int data)
{
      for(int i=x;i<N;i+=i&-i)
      {
          f[i]+=data;
      }
}
int Getsum(int x)
{
    int result=0;
    for(int i=x;i>=1;i-=i&-i)
    {
       result+=f[i];
    }
    return result;
}
int main()
{
     while(~scanf("%d",&n),n)
     {
          for(int i=1;i<=n;i++)
          {
              scanf("%d%d",&e[i].S,&e[i].E);
              e[i].id=i;
          }
          sort(e+1,e+n+1,cmp);
          memset(f,0,sizeof(f));
          num[e[1].id]=0;
          update(e[1].E,1);
          maxn=e[1].E;
          for(int i=2;i<=n;i++)
          {
              if(e[i].S==e[i-1].S&&e[i].E==e[i-1].E)
              num[i]=num[i-1];
              num[e[i].id]=Getsum(maxn)-Getsum(e[i].E-1);
              update(e[i].E,1);
              maxn=max(maxn,e[i].E);
          }
          for(int i=1;i<=n;i++)
          {
              if(i!=1)printf(" ");
              printf("%d",num[i]);
          }
          printf("\n");
     }
}

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