poj2481之排序+树状数组
Cows
| Time Limit: 3000MS | Memory Limit: 65536K | |
| Total 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!
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 <= 10 5), which is the number of cows. Then come N lines, the i-th of which contains two integers: S and E(0 <= S < E <= 10 5) 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.
For each test case, the first line is an integer N (1 <= N <= 10 5), which is the number of cows. Then come N lines, the i-th of which contains two integers: S and E(0 <= S < E <= 10 5) 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
/*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 99999999
typedef 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;
}