poj2481之线段树单点更新
Cows
Time Limit: 3000MS Memory Limit: 65536K
Total Submissions: 9813 Accepted: 3214
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
Time Limit: 3000MS Memory Limit: 65536K
Total Submissions: 9813 Accepted: 3214
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
/*
思路:对所有牛按先左端点s从小到大排序,如果s相等则按e从大到小排序
每次更新e点,更新的目的是使得包含e点的区间含有点的个数+1
对于比牛[s,e]强壮的牛的个数则只要查询区间[e,MAX]内的点的个数
(由于前面更新的牛s<=当前的牛的s,所以只要前面的牛的e>=当前的牛的e即可,即求[e,MAX]含有的点的个数)
*/
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<queue>
#include<algorithm>
#include<map>
#include<iomanip>
#define INF 99999999
using namespace std;
const int MAX=100000+10;
int sum[MAX<<2];//区间内含有点的个数
int num[MAX];//记录比第i只牛壮的数量
struct node{
int s,e,ID;
node(){}
node(int S,int E,int pos):s(S),e(E),ID(pos){}
bool operator <(const node &a)const {//比较函数
if(s == a.s)return e>a.e;
return s<a.s;
}
}s[MAX];
void Update(int pos,int n,int left,int right){
++sum[n];
if(left == right)return;
int mid=left+right>>1;
if(pos<=mid)Update(pos,n<<1,left,mid);
else Update(pos,n<<1|1,mid+1,right);
}
int Query(int L,int R,int n,int left,int right){
if(L<=left && right<=R)return sum[n];
int mid=left+right>>1,ans=0;
if(L<=mid)ans+=Query(L,R,n<<1,left,mid);
if(R>mid)ans+=Query(L,R,n<<1|1,mid+1,right);
return ans;
}
int main(){
int n,a,b;
while(cin>>n,n){
for(int i=1;i<=n;++i){
scanf("%d%d",&a,&b);
s[i]=node(a,b,i);
}
sort(s+1,s+n+1);
memset(sum,0,sizeof sum);
for(int i=1;i<=n;++i){
if(s[i].s == s[i-1].s && s[i].e == s[i-1].e){
num[s[i].ID]=num[s[i-1].ID];
}else num[s[i].ID]=Query(s[i].e,MAX,1,1,MAX);
Update(s[i].e,1,1,MAX);
}
for(int i=1;i<=n;++i)printf("%d%c",num[i],i == n?'\n':' ');
}
return 0;
}