根据题意我们要维护前n-1个点中按照x值排序的前缀和,那么我们就可以用树状数组来维护一个大小为32000+的数组的前缀和,然后logn的更新,logn的查询就能过了。
#pragma warning(disable:4996)
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int M = 32005;
const int N = 15005;
int C[M], len = 32002;
int ans[N];
int lowbit(int x) {
return x&(-x);
}
void update(int i, int v) {
while (i <= len) {
C[i] += v;
i += lowbit(i);
}
}
int sum(int i) {
int ret = 0;
while (i > 0) {
ret += C[i];
i -= lowbit(i);
}
return ret;
}
int main() {
int n;
while (~scanf("%d", &n)) {
memset(C, 0, sizeof C);
memset(ans, 0, sizeof ans);
for (int i = 1; i <= n; i++) {
int x, y;
scanf("%d%d", &x, &y);
x++; y++;
ans[sum(x)]++;
update(x, 1);
}
for (int i = 0; i < n; i++) {
printf("%d\n", ans[i]);
}
}
return 0;
}