Uvalive-5908 (uva1517) Tracking RFIDs (violent)

Uvalive-5908 (uva1517) Tracking RFIDs (violent)

Question Link

I will give you a second sensor, and the sensing radius of each sensor is R, W wall, and the sensing range of the sensor will be reduced by 1 because of a wall, if there are n walls between the sensor and an item, the sensing range is r-n. I will give you p items and ask you to find out which sensors can sense each item.

Solution: Because the s of this question is very large, the P is relatively small, and the sensor positions are all Integer Points, then enumerate the Integer Points not greater than the R around each product (a maximum of 2? R? 2? R). Then, determine whether there is a difference product of the wall vector between two points. Complexity: 1000040 40*10. It is acceptable because extreme data may not exist. Set is much faster than map.

Code:

#include <cstdio>#include <cstring>#include <vector>#include <set>#include <cmath>using namespace std;const int N = 1e4 + 5;typedef long long ll;int S, R, W, P;struct Point {    int x, y;    Point (int x = 0, int y = 0) : x(x) , y(y) {}    bool operator < (const Point &a) const {        return a.x == x ? y < a.y : x < a.x;     }    Point operator - (const Point &a) const {        Point ans;        ans.x = x - a.x;        ans.y = y - a.y;        return ans;    }    Point operator + (const Point &a) const {        Point ans;        ans.x = x + a.x;        ans.y = y + a.y;        return ans;    }    int operator ^ (const Point &a) const {        return x * a.y - y * a.x;    }};struct Line {    Point b, e;    void init (int bx = 0, int by = 0, int ex = 0, int ey = 0) {        b.x = bx;        b.y = by;        e.x = ex;        e.y = ey;    }}l[15];vector<Point> p;set<Point> vis;ll dis (Point a, Point b) {    ll dx = a.x - b.x;    ll dy = a.y - b.y;    return dx * dx + dy * dy;}bool check (Point a, Point b, Point c, Point d) {    if (min (a.x, b.x) > max (c.x, d.x) ||        min (a.y, b.y) > max (c.y, d.y) ||        min (c.x, d.x) > max (a.x, b.x) ||        min (c.y, d.y) > max (a.y, b.y))        return false;    ll i = (a - b) ^ (a - c);    ll j = (a - b) ^ (a - d);    ll k = (c - d) ^ (c - a);    ll l = (c - d) ^ (c - b);    return i * j <= 0 && k * l <= 0;}bool judge (Point a, Point b) {    if (vis.find(a) == vis.end())        return false;    ll d = dis (a, b);    if (d > R * R)        return false;    int cnt = 0;    for (int i = 0; i < W; i++)         if (check(a, b, l[i].b, l[i].e))             cnt++;    if (cnt > R)        return false;    return d <= (R - cnt) * (R - cnt);}int main () {    int T;    int x, y;    Point tmp;    scanf ("%d", &T);    while (T--) {        scanf ("%d%d%d%d", &S, &R, &W, &P);        vis.clear();        for (int i = 0; i < S; i++) {            scanf ("%d%d", &tmp.x, &tmp.y);            vis.insert (tmp);        }        int bx, by, ex, ey;        for (int i = 0; i < W; i++) {            scanf ("%d%d%d%d", &bx, &by, &ex, &ey);            l[i].init(bx, by, ex, ey);        }        Point a;        int d;        for (int i = 0; i < P; i++) {            p.clear();            scanf ("%d%d", &tmp.x, &tmp.y);            for (int j = -R; j <= R; j++) {                d = sqrt (R * R - j * j);                for (int k = max (-R, -d) ; k <= min (R, d); k++) {                    a.x = tmp.x + j;                    a.y = tmp.y + k;                    if (judge(a, tmp))                        p.push_back (a);                }            }            printf ("%d", p.size());            for (int j = 0; j < p.size(); j++)                 printf (" (%d,%d)", p[j].x, p[j].y);            printf ("\n");        }        }    return 0;}

Uvalive-5908 (uva1517) Tracking RFIDs (violent)