#1582 : Territorial Dispute
1000ms
1000ms
256MB
描述
In 2333, the C++ Empire and the Java Republic become the most powerful country in the world. They compete with each other in the colonizing the Mars.
There are n colonies on the Mars, numbered from 1 to n. The i-th colony's location is given by a pair of integers (xi, yi). Notice that latest technology in 2333 finds out that the surface of Mars is a two-dimensional plane, and each colony can be regarded as a point on this plane. Each colony will be allocated to one of the two countries during the Mars Development Summit which will be held in the next month.
After all colonies are allocated, two countries must decide a border line. The Mars Development Convention of 2048 had declared that: A valid border line of two countries should be a straight line, which makes colonies ofdifferent countries be situated on different sides of the line.
The evil Python programmer, David, notices that there may exist a plan
David wants to change the colony allocation plan secretly during the Mars Development Summit. Now he needs you to give him a specific plan of allocation which will cause a territorial dispute. He promises that he will give you 1000000007 bitcoins for the plan.
输入
The first line of the input is an integer T, the number of the test cases (T ≤ 50).
For each test case, the first line contains one integer n (1 ≤ n ≤ 100), the number of colonies.
Then n lines follow. Each line contains two integers xi, yi (0 ≤ xi, yi
There are no more than 10 test cases with n > 10.
输出
For each test case, if there exists a plan of allocation meet David's demand, print "YES" (without quotation) in the first line, and in the next line, print a string consisting of English letters "A" and "B". The i-th character is "A" indicates that the i-th colony was allocated to C++ Empire, and "B" indicates the Java Republic.
If there are several possible solutions, you could print just one of them.
If there is no solution, print "NO".
注意
This problem is special judged.
样例输入
2 2 0 0 0 1 4 0 0 0 1 1 0 1 1
样例输出
NO YES ABBA
【分析】：
题意，给出一些坐标点，任务是把这些点分成A类和B类，使得无法用一条直线分隔开这些点，输出YES即每个点放在A类还是B类（任意分）。如果做不到，就printf（“NO\n”）；
对于大于三个点时，跑一遍凸包。
如果凸包点个数==n;说明所有点都在凸包边上，选择两个不相邻的点归为一类即可。
如果凸包点个数<n；说明是凸包多边形里面包着一些点。直接把边归为一类，里面的另一类。
n==3是特判是不是直线。
n<3时全NO
【代码】：
#include<stdio.h>
#include<math.h>
#include<stdlib.h>
#include<algorithm>
using namespace std;
struct point{
double x,y;
int i;
}p[1100],stk[1100],base;
int n,top;
double lenth(point a,point b)
{
return sqrt(pow(a.x-b.x,2)+pow(a.y-b.y,2));
}
double across(point c,point m,point n)//求叉乘
{
return (m.x-c.x)*(n.y-c.y)-(m.y-c.y)*(n.x-c.x);
}
bool cmp(point p1,point p2)
{
if(across(base,p1,p2)==0)//三点一线
return lenth(base,p1)<lenth(base,p2);//按距离从近到远
return across(base,p1,p2)>0;
}
void getside()//求凸包点，存入数组stk[]
{
stk[0]=p[0];
stk[1]=p[1];
top=1;//top始终表示上一个确定下来的点
for(int i=2;i<n;i++)
{
while(top>0&&across(stk[top-1],stk[top],p[i])<=0)//上一个确定的点是凹的
top--;
stk[++top]=p[i];
}
}
char s[1010];
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
scanf("%d",&n);
base={1000,1000};
int mini=0;
for(int i=0;i<n;i++)
{
scanf("%lf%lf",&p[i].x,&p[i].y);
p[i].i=i;
if(p[i].y<base.y||(p[i].y==base.y&&p[i].x<base.x))
{
base.x=p[i].x;
base.y=p[i].y;
mini=i;
}
}
swap(p[mini],p[0]);
sort(p+1,p+n,cmp);//排序
getside();//得出凸包点
if(n<=2 || n==3&&top+1==n)
puts("NO");
else
{
for(int i=0;i<120;i++)s[i]='B';
s[n]=0;
puts("YES");
if(top+1<n)
{
for(int i=0;i<=top;i++)
s[stk[i].i]='A';
}
else if(top+1==n)
{
s[stk[0].i]='A';
s[stk[2].i]='A';
}
puts(s);
}
}
return 0;
}
