ACM-ICPC(2017)北京赛区网络赛-E-Territorial Dispute（计算几何->凸包）

题目5 : Territorial Dispute

时间限制:1000ms

单点时限:1000ms

内存限制: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 of different countries be situated on different sides of the line.

The evil Python programmer, David, notices that there may exist a plan of allocating colonies, which makes the valid border line do not exist. According to human history, this will cause a territorial dispute, and eventually lead to war.

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 ≤ 1000), meaning the location of the i-th colony. There are no two colonies share the same location.

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

220 00 140 00 11 01 1

样例输出

NOYESABBA

比赛已经结束，去题库提交。

题意：给你n个点，让你将这n个点非诚两个集合，分别为A和B，然后问你能否存在一种分配方案，使得，无论如何

画分割线都不能正好将这两个集合完全分开。

题解：随意找四个点，然后分三种情况讨论就好了。

（1）四个点共线，则ABAB就行

（2）4个点都在凸包上，则按对角线分类就好，

（3）三个点在凸包上，则让凸包上的点为A,其他为B就好啦

剩下的就是NO的情况了，很好找，n<=2 || n==3 && 这三个点不在一条直线上  为NO。

#include<set>  #include<map>     #include<stack>            #include<queue>            #include<vector>    #include<string> #include<time.h>#include<math.h>            #include<stdio.h>            #include<iostream>            #include<string.h>            #include<stdlib.h>    #include<algorithm>   #include<functional>    using namespace std;            #define ll long long       #define inf  1000000000       #define mod 1000000007             #define maxn  3600#define lowbit(x) (x&-x)            #define eps 1e-9struct node  {      double x,y;  int id;}a[maxn],st,q[maxn],b[maxn];  double cross(node a,node b,node c)  {      return (a.x-c.x)*(b.y-c.y)-(a.y-c.y)*(b.x-c.x);    }  char s[maxn];int n;bool comp(node a,node b)  {      if(cross(a,b,st)>0)          return 1;      if(cross(a,b,st)==0 && fabs(a.x-st.x)<fabs(b.x-st.x))          return 1;      return 0;  }  int  main(void)  {      double ans=0;      int i,top,T;scanf("%d",&T);while(T--){memset(s,'\0',sizeof(s));scanf("%d",&n);  st.x=st.y=10000000;  for(i=1;i<=n;i++)  {  scanf("%lf%lf",&a[i].x,&a[i].y);  if(a[i].y<st.y || a[i].y==st.y && a[i].x<st.x)  st=a[i];  a[i].id=i;} sort(a+1,a+n+1,comp); q[1]=a[1];q[2]=a[2];top=2;  for(i=3;i<=n;i++)  {  while(top>1 && cross(q[top-1],a[i],q[top])>=0)  top--;  q[++top]=a[i];  }  if(n<3 || n==3 && top==3){printf("NO\n");continue;}printf("YES\n");if(top<n){for(i=1;i<=top;i++)s[q[i].id]='A';for(i=1;i<=n;i++){if(s[a[i].id]!='A' && s[a[i].id]!='B')s[a[i].id]='B';}}else{s[a[1].id]='A';s[a[2].id]='B';s[a[3].id]='A';for(i=4;i<=n;i++)s[a[i].id]='B';}printf("%s\n",s+1);}    return 0;  }