Ural 1028. Stars（树状数组）

题目链接：点击打开链接

思路：

为了满足第一个条件， 我们可以先按照x坐标排序，  然后我们用树状数组来维护y坐标大小关系， 就可以在O(nlogn)的时间内求出答案了。

细节参见代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <string>
#include <vector>
#include <stack>
#include <bitset>
#include <cstdlib>
#include <cmath>
#include <set>
#include <list>
#include <deque>
#include <map>
#include <queue>
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
using namespace std;
typedef long long ll;
typedef long double ld;
const double eps = 1e-6;
const double PI = acos(-1);
const int mod = 1000000000 + 7;
const int INF = 0x3f3f3f3f;
// & 0x7FFFFFFF
const int seed = 131;
const ll INF64 = ll(1e18);
const int maxn = 15111;
int T,n,m,bit[32111],len,ans[maxn];
struct node {
int x, y;
node(int x=0, int y=0):x(x), y(y) {}
bool operator < (const node& rhs) const {
if(x != rhs.x) return x < rhs.x;
else return y < rhs.y;
}
}a[maxn];
void add(int x, int d) {
while(x <= len) {
bit[x] += d;
x += x & -x;
}
}
int sum(int x) {
int ans = 0;
while(x > 0) {
ans += bit[x];
x -= x & -x;
}
return ans;
}
int main() {
while(~scanf("%d",&n)) {
memset(bit, 0, sizeof(bit));
len = 0;
for(int i = 1; i <= n; i++) {
scanf("%d%d", &a[i].x, &a[i].y);
a[i].x++; a[i].y++;
ans[i-1] = 0;
len = max(len, a[i].y);
}
sort(a+1, a+n+1);
for(int i = 1; i <= n; i++) {
int cur = sum(a[i].y);
ans[cur]++;
add(a[i].y, 1);
}
for(int i = 0; i < n; i++) printf("%d\n", ans[i]);
}
return 0;
}