poj1696/hlg1318 蛋疼的蚂蚁

Space Ant

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 2934		Accepted: 1874

Description

The most exciting space discovery occurred at the end of the 20th century. In 1999, scientists traced down an ant-like creature in the planet Y1999 and called it M11. It has only one eye on the left side of its head and just three feet all on the right side of its body and suffers from three walking limitations: 

1. It can not turn right due to its special body structure. 
2. It leaves a red path while walking. 
3. It hates to pass over a previously red colored path, and never does that.

The pictures transmitted by the Discovery space ship depicts that plants in the Y1999 grow in special points on the planet. Analysis of several thousands of the pictures have resulted in discovering a magic coordinate system governing the grow points of the plants. In this coordinate system with x and y axes, no two plants share the same x or y. 
An M11 needs to eat exactly one plant in each day to stay alive. When it eats one plant, it remains there for the rest of the day with no move. Next day, it looks for another plant to go there and eat it. If it can not reach any other plant it dies by the end of the day. Notice that it can reach a plant in any distance. 
The problem is to find a path for an M11 to let it live longest. 
Input is a set of (x, y) coordinates of plants. Suppose A with the coordinates (xA, yA) is the plant with the least y-coordinate. M11 starts from point (0,yA) heading towards plant A. Notice that the solution path should not cross itself and all of the turns should be counter-clockwise. Also note that the solution may visit more than two plants located on a same straight line. 

Input

The first line of the input is M, the number of test cases to be solved (1 <= M <= 10). For each test case, the first line is N, the number of plants in that test case (1 <= N <= 50), followed by N lines for each plant data. Each plant data consists of three integers: the first number is the unique plant index (1..N), followed by two positive integers x and y representing the coordinates of the plant. Plants are sorted by the increasing order on their indices in the input file. Suppose that the values of coordinates are at most 100.

Output

Output should have one separate line for the solution of each test case. A solution is the number of plants on the solution path, followed by the indices of visiting plants in the path in the order of their visits.

Sample Input

2
10
1 4 5
2 9 8
3 5 9
4 1 7
5 3 2
6 6 3
7 10 10
8 8 1
9 2 4
10 7 6
14
1 6 11
2 11 9
3 8 7
4 12 8
5 9 20
6 3 2
7 1 6
8 2 13
9 15 1
10 14 17
11 13 19
12 5 18
13 7 3
14 10 16

Sample Output

10 8 7 3 4 9 5 6 2 1 10
14 9 10 11 5 12 8 7 6 13 4 14 1 3 2

Source

Tehran 1999

 

 

大意：

给你n个点 每次从最坐下点往前走  只能想左转弯   问最后最多能走多少个点 并输出顺序

 

分析：此题考查对于极角排序的理解

极角排序的cmp函数能使得对于 某一个点  把其他的点 按照顺时针或者逆时针的方向进行排序

拿原点为例

则极角排序之后就是顺或逆的点的顺序

拿tan 的a角度比较好理解

 

手写一下 cmp函数

//首先定义一下cross叉积函数

int cross(point p0, point p1, point p2)//如果是正值 p1在p2的顺时针方向

{

　　　return (p1.x - p0.x)*(p2.y - p0.y)　－　(p2.x - p0.x)*(p1.y - p0.y);

}

int pos;

bool cmp(point p1, point p2)

{

　　return cross(point[pos], p1, p2) > 0;//若为>0 则p1在p2的顺时针方向//即按照逆时针排序

}

 

 

代码：

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;

const int maxn = 55;
struct point
{
    int id;
    int x;
    int y;
}poin[maxn];

bool cmp(point p1, point p2){
    if(p1.y != p2.y)
        return p1.y < p2.y;
    return p1.x < p2.x;
}

int cross(point p0, point p1, point p2)
{
    return (p1.x - p0.x)*(p2.y - p0.y) - (p2.x - p0.x)*(p1.y - p0.y);
}

int Distance(point p1, point p2)
{
    return (p2.x - p1.x)*(p2.x - p1.x) + (p2.y - p1.y)*(p2.y - p1.y);
}
int pos;

bool Polar_cmp(point p1, point p2)
{
    int tmp = cross(poin[pos], p1, p2);
    if(tmp)
        return tmp > 0;
    return Distance(poin[pos], p1) <= Distance(poin[pos], p2);
}

int main()
{
    int t, n;
    scanf("%d",&t);
    while(t--){
        scanf("%d", &n);
        for(int i = 0; i < n; i++)
            scanf("%d %d %d",&poin[i].id, &poin[i].x, &poin[i].y);
        int ans[maxn];
        int k = 0;
        sort(poin, poin + n, cmp);
        pos = 0;
        int cnt = n - 1;
        for(int i = 1; i <= cnt; i++)
        {
            sort(poin + i, poin + n, Polar_cmp);
            //printf("*%d ",poin[i].id);
            pos++;
        }
        printf("%d",n);
        for(int i = 0; i < n; i++)
            printf(" %d",poin[i].id);
        puts("");
    }
    return 0;
}