题意是在二维平面上
给定n个人
每个人的坐标和移动速度v
若对于某个点，只有 x 能最先到达（即没有人能比x先到这个点或者同时到这个点）
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、把所有点按速度排个序，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，则只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
*/
HDU 4946 Area of Mushroom 共线凸包,布布扣,bubuko.com
HDU 4946 Area of Mushroom 共线凸包
原文：http://blog.csdn.net/qq574857122/article/details/38561783