POJ 2481 Cows

传送门：POJ 2481

Cows Time Limit: 3000MS Memory Limit: 65536K

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: 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.

Sample Input
3
1 2
0 3
3 4
0

Sample Output
1 0 0

Hint
Huge input and output,scanf and printf is recommended.

题意：
有n头牛，每一头牛都有一个区间[s, e]，对于牛i和牛j来说，满足下面的条件则证明牛i比牛j强壮：
s_i <= s_j 且 e_j <= e_i and e_i - s_i > e_j - s_j。现在已知每一头牛的区间，要求输出每头牛有几头牛比其强壮。

题解：
按照e的降序排序，e相等按s的升序，然后考虑s值，如果当前牛为[s, e]，那么比它强壮的牛的个数，就等于排序在它前面的。下面的就是树状数组部分了。
和poj 3067 Japan 很像。

AC代码：

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<cmath>
#define ll long long
#define inf 0x3f3f3f3f
using namespace std;
const int N=110000;
int n;
ll c[2*N],sum[N];
struct Node
{int s,e;int id;
}cows[2*N];
bool cmp(Node a,Node b)
{if(a.e==b.e){return a.s<b.s;}return a.e>b.e;
}
int lowbit(int x)
{return x&(-x);
}
void update(int x,int k)
{while(x<=n){c[x]+=k;x+=lowbit(x);}
}
ll getsum(int x)
{ll ans=0;for(int i=x;i>0;i-=lowbit(i)){ans+=c[i];}return ans;
}
int main()
{while(scanf("%d",&n)&&n){memset(c,0,sizeof(c));memset(sum,0,sizeof(sum));for(int i=1;i<=n;i++){scanf("%d%d",&cows[i].s,&cows[i].e);cows[i].s++;cows[i].e++;cows[i].id=i;}sort(cows+1,cows+1+n,cmp);sum[cows[1].id]=0;update(cows[1].s,1);for(int i=2;i<=n;i++){if(cows[i].s==cows[i-1].s&&cows[i].e==cows[i-1].e){sum[cows[i].id]=sum[cows[i-1].id];}else{sum[cows[i].id]+=getsum(cows[i].s);}update(cows[i].s,1);}for(int i=1;i<=n;i++){printf("%I64d%c",sum[i],i==n?'\n':' ');}}return 0;
}

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