Manhattan Positioning System --- Gym

题目

　　https://vjudge.net/problem/Gym-101490G

题意

　　给出 n 个信标的二维坐标 ( X，Y ) 以及唯一接收器距离每个信标的曼哈顿距离 D ( ( x1, y1 ) 和 ( x2, y2 ) 的曼哈顿距离为 | x1 - x2 | + | y1 - y2 | )，问是否能确定接收器的位置。

题解

　　每个信标用曼哈顿距离可以构建出一个旋转 45° 的正方形，取随意一个正方形，用 ( x1, y1 )，( x2, y2 ) 来表示每条边构成边集，将该边集与每个正方形的边集求交集，看剩余的边集是否为唯一一个点。

#include <bits/stdc++.h>
#define ll long long
#define ull unsigned long long
#define met(a, b) memset(a, b, sizeof(a))
#define rep(i, a, b) for(int i = a; i <= b; i++)
#define bep(i, a, b) for(int i = a; i >= b; i--)
#define lowbit(x) (x&(-x))
#define MID (l + r) / 2
#define ls pos*2
#define rs pos*2+1
#define pb push_back
#define ios() ios::sync_with_stdio(0)

using namespace std;

const int maxn = 1e6 + 1010;
const int inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll mod = 123456789;

struct NODE {
    ll x, y, w;
}arr[1010];
struct node {
    ll x1, y1, x2, y2, k, b;
    node () {}
    node (ll t1, ll t2, ll t3, ll t4) {
        x1 = t1, y1 = t2, x2 = t3, y2 = t4;
        if(x1 == x2) k = 1;
        else k = (y2 - y1) / (x2 - x1);
        b = y1 - k*x1;
    }
}t[5];

map<pair<ll, ll>, ll> tran;
queue<node> que[2];
ll tail;

void func(ll op) {
    tran.clear();
    ll now = (op + 1) % 2;
    while(!que[op].empty()) {
        node p = que[op].front();
        que[op].pop();
        ll mp1 = p.x1 + p.y1*10000000;
        ll mp2 = p.x2 + p.y2*10000000;
        rep(k, 1, 4) {
            ll x1 = inf, y1 = inf, x2 = inf, y2 = inf;
            if(p.k == t[k].k) {
                ll t1, tt1, t2, tt2;
                if(p.x1 > t[k].x1) t1 = p.x1, tt1 = p.y1;
                else t1 = t[k].x1, tt1 = t[k].y1;
                if(p.x2 < t[k].x2) t2 = p.x2, tt2 = p.y2;
                else t2 = t[k].x2, tt2 = t[k].y2;
                if(p.b != t[k].b || t1 > t2) continue;
                x1 = t1, y1 = tt1;
                x2 = t2, y2 = tt2;
            }
            else {
                double tt = 1.0 * (p.b - t[k].b) / (t[k].k - p.k);
                if((ll)tt != tt || tt < p.x1 || tt > p.x2 || tt < t[k].x1 || tt > t[k].x2) continue;
                x1 = x2 = tt;
                y1 = y2 = p.k*x1 + p.b;
            }
            mp1 = x1 + y1*10000000;
            mp2 = x2 + y2*10000000;
            pair<ll, ll> pr = make_pair(mp1, mp2);
            if(!tran[pr]) {
                que[now].push((node){x1, y1, x2, y2});
                tran[pr] = 1;
            }
        }
    }
}
int main() {
    ios();
    ll n;
    cin >> n;
    rep(i, 1, n) cin >> arr[i].x >> arr[i].y >> arr[i].w;
    que[1].push((node){arr[1].x - arr[1].w, arr[1].y, arr[1].x, arr[1].y + arr[1].w});
    que[1].push((node){arr[1].x, arr[1].y + arr[1].w, arr[1].x + arr[1].w, arr[1].y});
    que[1].push((node){arr[1].x, arr[1].y - arr[1].w, arr[1].x + arr[1].w, arr[1].y});
    que[1].push((node){arr[1].x - arr[1].w, arr[1].y, arr[1].x, arr[1].y - arr[1].w});
    rep(i, 1, n) {
        t[1] = {arr[i].x - arr[i].w, arr[i].y, arr[i].x, arr[i].y + arr[i].w};
        t[2] = {arr[i].x, arr[i].y + arr[i].w, arr[i].x + arr[i].w, arr[i].y};
        t[3] = {arr[i].x, arr[i].y - arr[i].w, arr[i].x + arr[i].w, arr[i].y};
        t[4] = {arr[i].x - arr[i].w, arr[i].y, arr[i].x, arr[i].y - arr[i].w};
        func(i % 2);
    }
    ll tail = que[(n+1) % 2].size();
    if(tail == 0) cout << "impossible" << endl;
    else {
        node p = que[(n+1) % 2].front();
        if(tail > 1 || p.x1 != p.x2) cout << "uncertain" << endl;
        else cout << p.x1 << ' ' << p.y1 << endl;
    }
    return 0;
}