Ural 1028. Stars（树状数组）

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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;// & 0x7FFFFFFFconst 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;}

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