HDU 4946 area of mushroom

The question is on a two-dimensional plane.

Given n people

Coordinates and moving speed of each person v

For a certain point, only X can arrive first (that is, no one can arrive at this point or at the same time than X)

This point is called X.

If someone can occupy an infinite area, 1 is output; otherwise, 0 is output.

Ideas:

1. Sort all vertices by speed, and remove vertices that are not the maximum speed.

The remaining points may occupy an infinite area.

2. Create a convex hull for the vertex with the highest speed. Only the vertices on the convex hull can occupy an infinite number.

If there are multiple people in a single position, these people cannot possess an infinite number.

The vertices on the top of the convex hull must be retained ,,

A wave of data is provided

#include <cstdio>#include <vector>#include <algorithm>#include <set>using namespace std;#define INF 999999999.9#define PI acos(-1.0)#define ll intconst int MAX_N = 507;const double eps = 1e-6;struct Point {    int x, y, v;    int id;    Point () {}    Point (int _x, int _y, int _v, int _id) {        x = x, y = _y, v = _v, id = _id;    }    bool operator < (const Point &rhs) const {        if (x != rhs.x) return x < rhs.x;        return y < rhs.y;    }};int v[MAX_N];bool vis[MAX_N];int n, top;int ans[MAX_N];Point P[MAX_N], p1[MAX_N];double cross(Point a, Point b, Point c) {    return (a.x - c.x) * (b.y - c.y) - (b.x - c.x) * (a.y - c.y);}bool cmp(Point a, Point b) {   if (a.y == b.y) return a.x < b.x;   return a.y < b.y;}void graham() {    sort(p1, p1 + n, cmp);    top = 1;    for (int i = 0; i < 2; i++) v[i] = i;    for (int i = 2; i < n; i++) {       while (top > 0 && cross(p1[i], p1[v[top]], p1[v[top - 1]]) > 0) top--;       v[++top] = i;    }    int len = top;    v[++top] = n - 2;    for (int i = n - 3; i >= 0; i--) {        while (top > len && cross(p1[i], p1[v[top]], p1[v[top - 1]]) > 0) top--;        v[++top] = i;    }}void Clear() {    memset(ans, 0, sizeof ans);    memset(p1, 0, sizeof p1);    memset(P, 0, sizeof P);    memset(v, 0, sizeof v);    memset(vis, 0, sizeof vis);}const int N = 505;struct E {    int x, y, v, id, ok;}s[N];vector<E> G;bool cmp1(const E a, const E b) {    if(a.v != b.v) return a.v > b.v;    if(a.x != b.x) return a.x < b.x;    if(a.y != b.y) return a.y < b.y;    return a.id < b.id;}int nn;void put(int ttop){    for(int i = 1; i <= nn; i ++) printf("%d", ans[i]);    puts("");}int main() {    int cas = 0;    while(~scanf("%d", &nn), nn) {Clear();        printf("Case #%d: ", ++cas);        memset(ans, 0, sizeof ans);        for(int i = 0; i < nn; i ++) {            scanf("%d%d%d", &s[i].x, &s[i].y, &s[i].v);            s[i].id = i+1;            s[i].ok = 1;            for(int j = 0; j < i; j++)                if(s[i].x==s[j].x && s[i].y==s[j].y && s[i].v==s[j].v)                {                    s[i].ok = 0;                    break;                }        }        sort(s, s + nn, cmp1);if(s[0].v==0){put(12);continue;}        int ttop = 0;        while(ttop < nn && s[ttop].v == s[0].v) ttop ++;        //----------------------------        G.clear();        for(int i = 0; i < ttop; i++)if( s[i].ok ) G.push_back(s[i]);        bool gongxian = true;        int x = 0, y = 0;        for(int i = 1; i < ttop; i++)        {            if(s[i].x == s[i-1].x && s[i].y == s[i-1].y)continue;            if(x==0&&y==0) {                x = s[i].x-s[i-1].x;                y = s[i].y - s[i-1].y;            }            else if(s[i].x - s[i-1].x != x || s[i].y - s[i-1].y != y)            {                gongxian =false;                break;            }        }        if(G.size()<=2 || gongxian) {            for(int i = 0; i < G.size(); i++)            {                bool ok = true;                for(int j = 0; ok && j < ttop; j++)                    if(G[i].id != s[j].id && G[i].x==s[j].x && G[i].y==s[j].y)                        ok = false;                ans[G[i].id] = ok;            }            put(ttop); continue;        }        for(int i = 0; i < G.size(); i++)        {            p1[i].x = G[i].x;            p1[i].y = G[i].y;            p1[i].id = G[i].id;        }n = G.size();graham();        for(int i = 0; i <= top; i++)        {            bool ok = true;            for(int j = 0; ok && j < ttop; j++)            {                if(p1[v[i]].id != s[j].id && p1[v[i]].x==s[j].x && p1[v[i]].y==s[j].y)                    ok = false;            }            ans[p1[v[i]].id] = ok;        }        put(ttop);    }    return 0;}/*90 0 10 0 10 10 10 10 110 0 110 0 110 10 110 10 15 5 1100 0 10 0 10 10 10 10 110 0 110 0 110 10 110 10 15 5 15 5 140 0 11 0 12 0 11 1 110 0 060 0 1-1 0 11 0 10 1 10 -1 10 -1 160 0 1-1 0 11 0 10 1 10 -1 10 -1 1*/