poj2481之排序+树状数组

Cows

Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 11321 Accepted: 3735

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

31 20 33 40

Sample Output

1 0 0

/*题意大意：FJ有n头牛(编号为1~n)，每一头牛都有一个测验值（S，E），对于牛i和牛j来说，如果它们的测验值满足下面的条件则表示牛i比牛j强壮：Si <= Sj and Ej <= Ei and Ei - Si > Ej - Sj。现在已知每一头牛的测验值，要求输出每头牛有几头牛比其强壮。分析:对n个数根据ei>ej或者ei == ej && si<sj进行排序，这样排完序后只要按顺序扫描数组对于s[j]前面的数组e[k]>=e[j],所以只要求出在s[j]之前有多少s[k]<=s[j]即可，另外如果e[k]=e[j]还需要减去s[k]=s[j]的个数，这个个数可以在扫描数组的过程中进行统计具体看代码 */#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <string>#include <queue>#include <algorithm>#include <map>#include <cmath>#include <iomanip>#define INF 99999999typedef long long LL;using namespace std;const int MAX=100000+10;int n;int num[MAX],c[MAX];struct Node{int s,e,id;bool operator<(const Node &a)const{if(e == a.e)return s<a.s;return e>a.e;}}s[MAX];int lowbit(int x){return x&(-x);}void Update(int x){while(x<MAX){c[x]+=1;x+=lowbit(x);}}int Query(int x){int sum=0;while(x>0){sum+=c[x];x-=lowbit(x);}return sum;}int main(){while(~scanf("%d",&n),n){memset(c,0,sizeof c); for(int i=0;i<n;++i){scanf("%d%d",&s[i].s,&s[i].e);++s[i].s,++s[i].e;s[i].id=i;}sort(s,s+n);num[s[0].id]=0;Update(s[0].s);int ans=0;//ans记录在i之前s[k] == s[i]的个数if(s[0].e == s[1].e && s[0].s == s[1].s)++ans;for(int i=1;i<n;++i){int sum=Query(s[i].s)-ans;if(s[i].e == s[i+1].e && s[i].s == s[i+1].s)++ans;else ans=0;num[s[i].id]=sum;Update(s[i].s);}/*for(int i=1;i<n;++i){if(s[i].e == s[i-1].e && s[i].s == s[i-1].s)num[s[i].id]=num[s[i-1].id];else num[s[i].id]=Query(s[i].s);Update(s[i].s);}*/printf("%d",num[0]);for(int i=1;i<n;++i)printf(" %d",num[i]);printf("\n");}return 0;}