HDU-6325 Interstellar Travel(几何...上凸包)

                                                                                Interstellar Travel
                                                                Time Limit: 4000/2000 MS (Java/Others) 

Problem Description
After trying hard for many years, Little Q has finally received an astronaut license. To celebrate the fact, he intends to buy himself a spaceship and make an interstellar travel.
Little Q knows the position of n planets in space, labeled by 1 to n. To his surprise, these planets are all coplanar. So to simplify, Little Q put these n planets on a plane coordinate system, and calculated the coordinate of each planet (xi,yi).
Little Q plans to start his journey at the 1-th planet, and end at the n-th planet. When he is at the i-th planet, he can next fly to the j-th planet only if xi Please write a program to help Little Q find the best route with minimum total cost.
 

Input
The first line of the input contains an integer T(1≤T≤10), denoting the number of test cases.
In each test case, there is an integer n(2≤n≤200000) in the first line, denoting the number of planets.
For the next n lines, each line contains 2 integers xi,yi(0≤xi,yi≤109), denoting the coordinate of the i-th planet. Note that different planets may have the same coordinate because they are too close to each other. It is guaranteed that y1=yn=0,0=x1  

Output
For each test case, print a single line containing several distinct integers p1,p2,...,pm(1≤pi≤n), denoting the route you chosen is p1→p2→...→pm−1→pm. Obviously p1 should be 1 and pm should be n. You should choose the route with minimum total cost. If there are multiple best routes, please choose the one with the smallest lexicographically.
A sequence of integers a is lexicographically smaller than a sequence of b if there exists such index j that ai=bi for all i  

Sample Input
1
3
0 0
3 0
4 0
 

Sample Output
1 2 3
题意：从X最小都到X最大，每次只能往x增大的方向走，花费为xi*yj-xj*yi,为负数则赚了.问最小花费选的点且字典序最小.

思路：首先我们想每个x只能选一次,重合点也只能选一个.其实就是i向量与j向量叉乘，我们想尽量为负数,j在i的顺时针方向则为负，反之则为正,所以我们想要得到一个上凸包~~几个点x相同当然要选y比较大的那个，这也就决定了我们的排序顺序.

最后还要保证字典序最小，这里还要说一点，求上凸包的过程下面我写的那个函数，< 则会留下在同一条直线的几个点，<= 则同一条直线只留两个端点,假如我们同一条直线的几个点都留下了,则最后我们还要看一看那些留下，哪些删去.端点必须留下，也就是说直线上可使字典序变小的点我们也留下~

代码：

#include
#define mem(a,b) memset(a,b,sizeof(a))
#define mod 1000000007
using namespace std;
typedef long long ll;
const int maxn = 2e5+5;
const double eps = 1e-12;
const int inf = 0x3f3f3f3f;
map::iterator it;

struct node
{
	int x,y;
	int pos;
	node(){}
	node(int x,int y):x(x),y(y){pos = 0;}
	friend node operator - (node A,node B)
	{
		return node(A.x-B.x,A.y-B.y);
	}
} p[maxn],q[maxn];

int n,cnt;
int ans[maxn];
bool vis[maxn];

bool cmp(node x,node y)//注意排序规则
{
	if(x.x != y.x) return x.x< y.x;
	if(x.y != y.y)return x.y> y.y;
	return x.pos< y.pos;
}

ll cross(node A,node B)//叉乘
{
	return 1ll*A.x*B.y-1ll*A.y*B.x;
}

bool judge(node A,node B,node C)//判断是否共线
{
	return cross(B-A,C-A)!= 0;
}

void ConvexHull(node p[],node q[])//求上凸包
{
	cnt = 0;
    for(int i=1;i<=n;i++)  
    {
        if(i> 1&&p[i].x==p[i-1].x)continue;
        while(cnt>=2&&1ll*(q[cnt].y-q[cnt-1].y)*(p[i].x-q[cnt].x)< 1ll*(q[cnt].x-q[cnt-1].x)*(p[i].y-q[cnt].y))cnt--;
        q[++cnt]=p[i];
    }
}

int main()
{
	int t;
	cin>>t;
	
	while(t--)
	{
		mem(vis,0);
		scanf("%d",&n);
		for(int i = 1;i<= n;i++)
			scanf("%d %d",&p[i].x,&p[i].y),p[i].pos = i;
		
		sort(p+1,p+n+1,cmp);
		
                ConvexHull(p,q);
		
		vis[1] = vis[cnt] = 1;
		for(int i = 2;i< cnt;i++)
			if(judge(q[i-1],q[i],q[i+1])) vis[i] = 1;//先只留端点
		
		for(int i = cnt;i>= 1;i--)//看看非端点是否可是字典序变小
			if(vis[i]) ans[i] = q[i].pos;
			else ans[i] = min(q[i].pos,ans[i+1]);
		
		for(int i = 1;i< cnt;i++)
			if(ans[i] == q[i].pos) printf("%d ",ans[i]);
		printf("%d\n",ans[cnt]);
	}
	
	return 0;
}