Manhattan Positioning System --- Gym - 101490G

Posted 2020-11-24 stay-online

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。

 1 #include <bits/stdc++.h>
 2 #define ll long long
 3 #define ull unsigned long long
 4 #define met(a, b) memset(a, b, sizeof(a))
 5 #define rep(i, a, b) for(int i = a; i <= b; i++)
 6 #define bep(i, a, b) for(int i = a; i >= b; i--)
 7 #define lowbit(x) (x&(-x))
 8 #define MID (l + r) / 2
 9 #define ls pos*2
10 #define rs pos*2+1
11 #define pb push_back
12 #define ios() ios::sync_with_stdio(0)
13 
14 using namespace std;
15 
16 const int maxn = 1e6 + 1010;
17 const int inf = 0x3f3f3f3f;
18 const ll INF = 0x3f3f3f3f3f3f3f3f;
19 const ll mod = 123456789;
20 
21 struct NODE {
22     ll x, y, w;
23 }arr[1010];
24 struct node {
25     ll x1, y1, x2, y2, k, b;
26     node () {}
27     node (ll t1, ll t2, ll t3, ll t4) {
28         x1 = t1, y1 = t2, x2 = t3, y2 = t4;
29         if(x1 == x2) k = 1;
30         else k = (y2 - y1) / (x2 - x1);
31         b = y1 - k*x1;
32     }
33 }t[5];
34 
35 map<pair<ll, ll>, ll> tran;
36 queue<node> que[2];
37 ll tail;
38 
39 void func(ll op) {
40     tran.clear();
41     ll now = (op + 1) % 2;
42     while(!que[op].empty()) {
43         node p = que[op].front();
44         que[op].pop();
45         ll mp1 = p.x1 + p.y1*10000000;
46         ll mp2 = p.x2 + p.y2*10000000;
47         rep(k, 1, 4) {
48             ll x1 = inf, y1 = inf, x2 = inf, y2 = inf;
49             if(p.k == t[k].k) {
50                 ll t1, tt1, t2, tt2;
51                 if(p.x1 > t[k].x1) t1 = p.x1, tt1 = p.y1;
52                 else t1 = t[k].x1, tt1 = t[k].y1;
53                 if(p.x2 < t[k].x2) t2 = p.x2, tt2 = p.y2;
54                 else t2 = t[k].x2, tt2 = t[k].y2;
55                 if(p.b != t[k].b || t1 > t2) continue;
56                 x1 = t1, y1 = tt1;
57                 x2 = t2, y2 = tt2;
58             }
59             else {
60                 double tt = 1.0 * (p.b - t[k].b) / (t[k].k - p.k);
61                 if((ll)tt != tt || tt < p.x1 || tt > p.x2 || tt < t[k].x1 || tt > t[k].x2) continue;
62                 x1 = x2 = tt;
63                 y1 = y2 = p.k*x1 + p.b;
64             }
65             mp1 = x1 + y1*10000000;
66             mp2 = x2 + y2*10000000;
67             pair<ll, ll> pr = make_pair(mp1, mp2);
68             if(!tran[pr]) {
69                 que[now].push((node){x1, y1, x2, y2});
70                 tran[pr] = 1;
71             }
72         }
73     }
74 }
75 int main() {
76     ios();
77     ll n;
78     cin >> n;
79     rep(i, 1, n) cin >> arr[i].x >> arr[i].y >> arr[i].w;
80     que[1].push((node){arr[1].x - arr[1].w, arr[1].y, arr[1].x, arr[1].y + arr[1].w});
81     que[1].push((node){arr[1].x, arr[1].y + arr[1].w, arr[1].x + arr[1].w, arr[1].y});
82     que[1].push((node){arr[1].x, arr[1].y - arr[1].w, arr[1].x + arr[1].w, arr[1].y});
83     que[1].push((node){arr[1].x - arr[1].w, arr[1].y, arr[1].x, arr[1].y - arr[1].w});
84     rep(i, 1, n) {
85         t[1] = {arr[i].x - arr[i].w, arr[i].y, arr[i].x, arr[i].y + arr[i].w};
86         t[2] = {arr[i].x, arr[i].y + arr[i].w, arr[i].x + arr[i].w, arr[i].y};
87         t[3] = {arr[i].x, arr[i].y - arr[i].w, arr[i].x + arr[i].w, arr[i].y};
88         t[4] = {arr[i].x - arr[i].w, arr[i].y, arr[i].x, arr[i].y - arr[i].w};
89         func(i % 2);
90     }
91     ll tail = que[(n+1) % 2].size();
92     if(tail == 0) cout << "impossible" << endl;
93     else {
94         node p = que[(n+1) % 2].front();
95         if(tail > 1 || p.x1 != p.x2) cout << "uncertain" << endl;
96         else cout << p.x1 << ‘ ‘ << p.y1 << endl;
97     }
98     return 0;
99 }