2018HDU多校训练-3-Problem G. Interstellar Travel

链接：http://acm.hdu.edu.cn/showproblem.php?pid=6325                                 

　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　 Interstellar Travel 

　　

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

 

 

Recommend

chendu

题解：WA了无数发，忘了一句话，“有的星球由于太近可以认为坐标相同，但输出的时候应该输出编号小的”，后来重读了题目，家里个条件就过了； 

这个题目可以转化为凸包问题，只求上凸包，求上凸包时注意排除横坐标相同的点，只取纵坐标大的点，重要的一点：对于纵坐标相同而横坐标不同的点，我们需要判断中间点的编号是否比下一个点大，因为按字典序输出，故如果大于则去掉该点，小于等于则保留，注意细节；

参考代码：

 

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn=2e5+10;
LL vis[maxn],T,n;
LL num[maxn];
struct Point{
    LL x,y;
    LL id;
    Point(double xx=0,double yy=0) : x(xx),y(yy) {}
} p[maxn],ch[maxn];
typedef Point Vector; 
Vector operator + (Vector a,Vector b) { return Vector(a.x+b.x,a.y+b.y); }
Vector operator - (Vector a,Vector b) { return Vector(a.x-b.x,a.y-b.y); }
Vector operator * (Vector a,Vector b) { return Vector(a.x*b.x,a.y*b.y); }
Vector operator / (Vector a,Vector b) { return Vector(a.x/b.x,a.y/b.y); }
bool operator < (const Point &a,const Point &b){ return a.x==b.x? (a.y==b.y? a.id<b.id : a.y>b.y) : a.x<b.x ; }
LL Cross(Vector a,Vector b) { return a.x*b.y-a.y*b.x; }

void ConvexHull() 
{
    LL m=0; memset(vis,0,sizeof vis);
    for(int i=1;i<=n;i++)
    {
        if(i>1 && p[i].x == p[i-1].x) continue;
        while(m>1 && Cross(ch[m]-ch[m-1],p[i]-ch[m])>0) m--;
        ch[++m]=p[i];
    }
    
    vis[1]=vis[m]=1;
    for(int i=2;i<m;i++) 
        if(Cross(ch[i+1]-ch[i],ch[i]-ch[i-1])!=0) vis[i]=1;
    for(int i=m;i>0;i--)
    {
        if(vis[i]) num[i]=ch[i].id;
        else num[i]=min(num[i+1],ch[i].id);
    }
    for(int i=1;i<m;i++) 
        if(num[i]==ch[i].id) printf("%lld ",num[i]);
    printf("%lld
",num[m]);
}

int main()
{
    scanf("%lld",&T);
    while(T--)
    {
        scanf("%lld",&n);
        for(int i=1;i<=n;i++) scanf("%lld%lld",&p[i].x,&p[i].y),p[i].id=i;
        sort(p+1,p+n+1);
        ConvexHull();
    }
    return 0;
}