UVaLive 7375 Hilbert Sort (递归，四分图，模拟)

题意：告诉你一条希尔伯特曲线的大小，然后给你n 个人，及n 个人的坐标，你的起点是左下角，终点是右下角，按照希尔伯特的曲线去走，按照这个顺序给n个人排序，

按顺序输出每个人的名字！

析：这就是一个四分图，每次都把当前的图分成四份，左下角的是顺时针旋转，左上角和右上角不变，右下角逆时针旋转90，那么我们就递归这个过程，给所给的人编号，

最后再按编号输出即可。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <stack>
using namespace std ;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-10;
const int maxn = 2e5 + 5;
const int mod = 1e9 + 7;
const char *mark = "+-*";
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
int n, m;
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
struct node{
    string id;
    string name;
    bool operator < (const node &p) const{
        return id < p.id;
    }
};
node a[maxn];

string dfs(double x, double y, double s, int d){
    double mid = s / 2;
    string ans;
    if(d >= 1){
        if(x <= mid && y <= mid)  ans = "1" + dfs(y, x, mid, d-1);
        else if(x <= mid && y > mid)  ans = "2" + dfs(x, y-mid, mid, d-1);
        else if(x > mid && y > mid)  ans = "3" + dfs(x-mid, y-mid, mid, d-1);
        else if(x > mid && y <= mid) ans = "4" + dfs(mid-y, s-x, mid, d-1);
    }
    return ans;
}

int main(){
    while(scanf("%d %d", &n, &m) == 2){
        int x, y;
        for(int i = 0; i < n; ++i){
            string s;
            cin >> x >> y >> s;
            a[i].name = s;
            a[i].id = dfs(x, y, m, 30);
        }
        sort(a, a+n);
        for(int i = 0; i < n; ++i)
            cout << a[i].name << endl;
    }
    return 0;
}