hdu6325 Interstellar Travel 凸包变形

题目传送门

题目大意：

给出n个平面坐标，保证第一个点和第n个点y值为0，其余点的x坐标都在中间，要从 i 点走到 j 点的要求是 i 点的横坐标严格小于 j 的横坐标，并且消耗的能量是(x_i * y_j - x_j * y_i),要求消耗的能量最小（能量可以为负），并且字典序要求最小。

思路：

消耗能量的式子就是两个坐标的叉积，叉积的几何意义就是两个向量对应的平行四边形的面积，但是这个面积会有正负，如果向量 j 在向量 i 的顺时针方向，则这个面积是负的，如果我希望能量最小，那么就尽可能使向量是顺时针方向的，由此发现其实就得到了一个凸包，而且是一个上凸包。经过凸包上的点都符合能量的要求，但是由于要求字典序最小，所以如果凸包上的某一条边上有很多点，那么就需要判断一下这些点的id，如果线段中间的点的id比较小，那么这些点选上了之后，字典序肯定会变小，所以在做凸包的时候就要对凸包算法加一点点变形。

#include<cstdio>
#include<algorithm>
#include<iostream>
#include<vector>
#include<map>
#include<set>
#include<cstring>
#include<queue>
#include<stack>
#define CLR(a,b) memset(a,b,sizeof(a))
#define mkp(a,b) make_pair(a,b)
typedef long long ll;
using namespace std;
const int inf = 0x3f3f3f3f;
const int maxn = ;
struct dian {
    double x, y;
    int id;
    dian (){}
    dian(double x, double y, int id) :x(x), y(y), id(id){}
}a[maxn],ch[maxn];
bool cmp(dian &a, dian &b)
{
    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;
}
typedef dian Vector;
Vector operator -(Vector a, Vector b) {
    return Vector ( a.x - b.x, a.y - b.y, );
}
bool operator == (Vector a,Vector b){
    return ((a.x==b.x) && (a.y==b.y));
}
double cross(Vector a, Vector b)
{
    return a.x*b.y - a.y*b.x;
}
int andrew(dian *p, int n, dian *ch)
{
    int m = ;
    sort(p, p + n, cmp);
    for (int i = ; i < n; i++)
    {
        if(i>&&p[i]==ch[m-])continue;
        while (m >  && cross(ch[m - ] - ch[m - ], p[i] - ch[m - ] )<= )
        {
            if(cross(ch[m - ] - ch[m - ], p[i] - ch[m - ] )< )
            m--;
            else if(ch[m-].id>p[i].id){
                m--;
            }else break;
        }
        ch[m++] = p[i];
    }
    return m;
}
bool vis[maxn];
int main() {
    int n,T;
    cin >> T;
    while (T--)
    {
        CLR(vis, inf);
        scanf("%d", &n);
        for (int i = ; i < n; i++)
        {
            scanf("%lf%lf", &a[i].x, &a[i].y);
            a[i].id = i+;
        }
        int m = andrew(a, n, ch);
        for (int i = ; i<=m-; i++)
        {
            printf("%d", ch[i].id);
            if (i < m-)printf(" ");
            else printf("\n");
        }
    }

}

/*

1
7
0 0
9 0
3 6
1 2
3 6
2 4
10 0

*/

Problem G. Interstellar Travel

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 2098    Accepted Submission(s): 544

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<xj, which will cost his spaceship xi×yj−xj×yi units of energy. Note that this cost can be negative, it means the flight will supply his spaceship.
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<x2,x3,...,xn−1<xn.

 

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<j, but aj<bj.

 

Sample Input

1
3
0 0
3 0
4 0

 

Sample Output

1 2 3

 

Source

2018 Multi-University Training Contest 3