HDU 4946 Area of Mushroom 共线凸包

题意是在二维平面上

给定n个人

每一个人的坐标和移动速度v

若对于某个点，仅仅有 x 能最先到达（即没有人能比x先到这个点或者同一时候到这个点）

则这个点称作被x占有

若有人能占有无穷大的面积 则输出1 。否则输出0

思路：

1、把全部点按速度排个序。然后把不是最大速度的点去掉

剩下的点才有可能是占有无穷大的面积

2、给速度最大的点做个凸包，则仅仅有在凸包上的点才有可能占有无穷大

若一个位置有多个人，则这几个人都是不能占有无穷大的。

凸包上边里共线的点是要保留的，，

附赠一波数据

#include <cstdio>
#include <vector>
#include <algorithm>
#include <set>
using namespace std;
#define INF 999999999.9
#define PI acos(-1.0)
#define ll int
const 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;
}
/*
9
0 0 1
0 0 1
0 10 1
0 10 1
10 0 1
10 0 1
10 10 1
10 10 1
5 5 1

10
0 0 1
0 0 1
0 10 1
0 10 1
10 0 1
10 0 1
10 10 1
10 10 1
5 5 1
5 5 1

4
0 0 1
1 0 1
2 0 1
1 1 1

1
0 0 0

6
0 0 1
-1 0 1
1 0 1
0 1 1
0 -1 1
0 -1 1

6
0 0 1
-1 0 1
1 0 1
0 1 1
0 -1 1
0 -1 1

*/