UESTC 电子科大专题训练 数据结构 D

UESTC 1584

题意：平面坐标上有n个怪物，每个怪物有一个rank值，代表x坐标和y坐标都不大于它本身的怪物数（不包括本身）

思路：对x y坐标从小到大排序，x优先排序，用数状数组计算y坐标小于它的数量

AC代码：

#include "iostream"
#include "string.h"
#include "stack"
#include "queue"
#include "string"
#include "vector"
#include "set"
#include "map"
#include "algorithm"
#include "stdio.h"
#include "math.h"
#define ll long long
#define bug(x) cout<<x<<" "<<"UUUUU"<<endl;
#define mem(a) memset(a,0,sizeof(a))
#define mp(x,y) make_pair(x,y)
#define pb(x) push_back(x)
using namespace std;
const long long INF = 1e18+1LL;
const int inf = 1e9+1e8;
const int N=1e5+100;
const ll mod=1e9+7;

struct Point{
    int x, y;
    bool friend operator< (Point a, Point b){
        if(a.x==b.x) return a.y<b.y;
        return a.x<b.x;
    }
};
Point P[N];
int n,ma,Y[N],ans[N];
int lowbit(int x){
    return (-x)&x;
}
void add(int y, int num){
    while(y<=ma){
        Y[y]+=num;
        y+=lowbit(y);
    }
}
int sum(int y){
    int ret=0;
    while(y>0){
        ret+=Y[y];
        y-=lowbit(y);
    }
    return ret;
}
int main(){
    //ios::sync_with_stdio(false),cin.tie(0);cout.tie(0);
    cin>>n;
    for(int i=1; i<=n; ++i){
        cin>>P[i].x>>P[i].y;
        ma=max(ma,P[i].y), ma=max(ma,P[i].y);
    }
    sort(P+1,P+1+n);
    for(int i=1; i<=n; ++i){
        int cnt=sum(P[i].y);
        ans[cnt]++;
        add(P[i].y,1);
    }
    for(int i=0; i<n; ++i){
       cout<<ans[i]<<"
";
    }
    return 0;
}