本文介绍了一个有趣的问题，即如何为一种特殊生物M11（仅能在平面特定条件下左转并移动）找到一条路径，使其能够访问最多的植物点以维持生命。文章提供了两种解题思路：一是通过递归搜索的方法来确定每一步的最佳选择；二是利用计算几何中的凸包算法解决该问题，并给出了相应的代码实现。
Space Ant
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 4353 | Accepted: 2740 |
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:
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.
- It can not turn right due to its special body structure.
- It leaves a red path while walking.
- 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
题意：一只蚂蚁,只会向左转,现在给出平面上很多个点,求解一种走法,能使得蚂蚁能经过的点最多,每个顶点该蚂蚁只能经过一次,且所行走的路线不能发生交叉
解题思路：蚂蚁的起点必然是所有点中纵坐标最小的点，然后暴力搜其他没走过的点，使得向左转的角度尽量小，就一定能走完所有点，或者用凸包
#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
using namespace std;
#define LL long long
const int INF=0x3f3f3f3f;
const double pi=acos(-1.0);
int id[60];
struct node
{
double x,y;
} a[60];
int visit[60];
int n;
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
int di,kk;
double mi=1.0*INF;
for(int i=1; i<=n; i++)
{
scanf("%d%lf%lf",&di,&a[i].x,&a[i].y);
if(a[i].y<mi)
{
mi=a[i].y;
kk=i;
}
}
memset(visit,0,sizeof visit);
visit[kk]=1;
int cnt=1;
double z=0;
id[cnt++]=kk;
for(int i=1; i<n; i++)
{
int k=-1;
double mii=1.0*INF;
for(int j=1; j<=n; j++)
{
if(!visit[j])
{
double xx=a[j].x-a[id[cnt-1]].x;
double yy=a[j].y-a[id[cnt-1]].y;
double ll=sqrt(xx*xx+yy*yy);
if(a[j].y>a[id[cnt-1]].y)
{
if(z>=0&&z<=pi)
{
if(acos(xx/ll)<z&&acos(xx/ll)+2*pi-z<mii)
{
k=j;
mii=acos(xx/ll)+2*pi-z;
}
else if(acos(xx/ll)>z&&acos(xx/ll)-z<mii)
{
k=j;
mii=acos(xx/ll)-z;
}
}
else if(acos(xx/ll)+2*pi-z<mii)
{
k=j;
mii=acos(xx/ll)+2*pi-z;
}
}
else if(a[j].y<a[id[cnt-1]].y)
{
if(z>=0&&z<=pi&&4*pi-acos(xx/ll)-z<mii)
{
mii=4*pi-acos(xx/ll)-z;
k=j;
}
else if(z>pi&&z<2*pi)
{
if(2*pi-acos(xx/ll)>z&&2*pi-acos(xx/ll)-z<mii)
{
k=j;
mii=2*pi-acos(xx/ll)-z;
}
else if(4*pi-acos(xx/ll)-z<mii)
{
k=j;
mii=4*pi-acos(xx/ll)-z;
}
}
}
else if(a[j].y==a[id[cnt-1]].y&&a[j].x>a[id[cnt-1]].x&&2*pi-z<mii)
{
k=j;
mii=2*pi-z;
}
else if(a[j].y==a[id[cnt-1]].y&&a[j].x<a[id[cnt-1]].x)
{
if(pi>z&&pi-z<mii)
{
k=j;
mii=pi-z;
}
else if(pi<z&&3*pi<mii)
{
k=j;
mii=3*pi-z;
}
}
}
}
if(k!=-1)
{
id[cnt++]=k;
visit[k]=1;
z=mii+z;
while(z>2*pi) z-=2*pi;
}
}
printf("%d",cnt-1);
for(int i=1; i<cnt; i++) printf(" %d",id[i]);
printf("\n");
}
return 0;
}
凸包：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
using namespace std;
#define LL long long
const int INF = 0x3f3f3f3f;
const double eps = 1e-8;
int sgn(double x)
{
if(fabs(x) < eps)return 0;
if(x < 0)return -1;
else return 1;
}
struct Point
{
int id;
double x,y;
};
const int MAXN = 1010;
Point list[MAXN];
int Stack[MAXN], top, pos;
double dist(Point a, Point b)
{
return sqrt((a.x - b.x)*(a.x - b.x)+(a.y-b.y)*(a.y-b.y));
}
bool _cmp(Point p1, Point p2)
{
double tmp = (p1.x - list[pos].x)*(p2.y - list[pos].y) - (p2.x - list[pos].x)*(p1.y - list[pos].y);
if (sgn(tmp) > 0) return true;
else if (sgn(tmp) == 0 && sgn(dist(p1, list[pos]) - dist(p2, list[pos])) <= 0) return true;
else return false;
}
/*
bool _cmp(Point p1, Point p2)
{
double tmp = (p1.x - list[0].x)*(p2.y-list[0].y) - (p2.x - list[0].x)*(p1.y-list[0].y);
if (sgn(tmp) > 0) return true;
else if (sgn(tmp) == 0 && sgn(dist(p1, list[0]) - dist(p2, list[0])) <= 0) return true;
else return false;
}
void Graham(int n)
{
Point p0=list[0];
int k = 0;
for(int i = 1;i < n;i++)
{
if( (p0.y > list[i].y) || (p0.y == list[i].y && p0.x > list[i].x) )
{
p0 = list[i]; k = i;
}
}
swap(list[k],list[0]);
sort(list+1,list+n,_cmp);
if(n == 1) { top = 1; Stack[0] = 0; return; }
if(n == 2) { top = 2; Stack[0] = 0; Stack[1] = 1; return ; }
Stack[0] = 0; Stack[1] = 1; top = 2;
for(int i = 2;i < n;i++)
{
while(top > 1 && sgn((list[Stack[top-1]].x-list[Stack[top-2]].x)*(list[i].y-list[Stack[top-2]].y)-(list[i].x- list[Stack[top - 2]].x)*(list[Stack[top - 1]].y - list[Stack[top - 2]].y)) <= 0) top--;
Stack[top++] = i;
}
for(int i=0;i<top;i++)
{
printf("%d ", Stack[i]);
}
printf("\n");
}
*/
int main()
{
int n,t;
scanf("%d", &t);
while (t--)
{
scanf("%d", &n);
for (int i = 0; i < n; i++) scanf("%d%lf%lf",&list[i].id, &list[i].x, &list[i].y);
Point p0 = list[0];
int k = 0;
for (int i = 1; i < n; i++)
{
if ((p0.y > list[i].y) || (p0.y == list[i].y && p0.x > list[i].x))
{
p0 = list[i]; k = i;
}
}
swap(list[k], list[0]);
pos = 0;
for (int i = 1; i < n; i++)
{
sort(list + i, list + n, _cmp);
pos++;
}
printf("%d", n);
for (int i = 0; i < n; i++) printf(" %d", list[i].id);
printf("\n");
}
return 0;
}
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