Separating Pebbles
题目链接：
http://acm.hust.edu.cn/vjudge/contest/127401#problem/H
Description
http://7xjob4.com1.z0.glb.clouddn.com/1e1638de1146450534631815cbf822c6
Input
The first line of the input contains an integer K (K ≤ 20) indicating the number of test cases. The first
line of each test case consists of an integer N (N ≤ 250) indicating the number of pebbles. The next
N lines each contains a triplet (x, y, c), where x and y represent the x and y coordinates (all integers,
0 ≤ x, y ≤ 150) of a pebble point, and c represents the type of pebble: ‘o’ denoted by ‘0’ and ‘+’
denoted by ‘1’.
Output
For each test case, output ‘1’ if Dr. Y can separate the pebbles with a single straight line; if not, output
‘0’.
Sample Input
2
5
1 2 0
2 1 0
4 3 1
5 4 1
6 3 1
4
1 2 0
2 1 0
1 1 1
2 2 1
Sample Output
1
0
##题意：
给出平面上的两类点，判断是否能画一条直线将两类点完全分割开来.
##题解：
枚举任意两点组成的直线, 除了在直线上的点以外，若其余点满足分居两侧的要求，那么我只要稍微旋转一下这条直线就能满足要求.
对直线上点的特殊考虑：当直线上有若干点时，如果出现两个'o'点中间夹一个'x'的情况是无法旋转的.
所以能旋转的条件是直线上不会出现两种类型的点间隔分布.
对重点的特殊考虑：如果有重点的话，特判可能会比较多，所以干脆一开始对所有点去重. 若有两个不同类型的点重叠，那么一定不能划分.
(坑爹的是，题目的数据好像并没有重复点，在uDebug上跑类型不同的重点居然输出1).
不知不觉就打了5000B的代码，稍有点繁琐...
##代码：
``` cpp
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#define LL long long
#define double long long
#define eps 1e-8
#define maxn 300
#define mod 100000007
#define inf 0x3f3f3f3f
#define mid(a,b) ((a+b)>>1)
#define IN freopen("in.txt","r",stdin);
using namespace std;
struct Point{
double x,y;
int tp;
Point(){}
Point(double tx,double ty) {x=tx;y=ty;}
bool operator<(const Point& b)const
{
if(x==b.x) return y<b.y;
return x<b.x;
}
}tmp[maxn];
vector p;
struct Line{
Point a,b;
};
int sign(double x)
{
if(!x) return 0;
return x<0? -1:1;
}
double xmul(Point p0,Point p1,Point p2)
{return (p1.x-p0.x)(p2.y-p0.y)-(p2.x-p0.x)(p1.y-p0.y);}
bool is_inline(Point a, Point b, Point c) {
return xmul(a,b,c) == 0;
}
/判两点在线段异侧,点在线段上返回0/
int opposite_side(Point p1,Point p2,Line l)
{
return sign(xmul(p1,l.a,l.b)*xmul(p2,l.a,l.b))<0;
}
int same_side(Point p1,Point p2,Line l)
{
return sign(xmul(p1,l.a,l.b)*xmul(p2,l.a,l.b))>0;
}
/判断两点是否重合/
int equal_point(Point p1,Point p2)
{
return (sign(p1.x-p2.x)0)&&(sign(p1.y-p2.y)0);
}
int vis[200][200];
int main(int argc, char const *argv[])
{
//IN;
int t; cin >> t;
while(t--)
{
int n; scanf("%d", &n);
memset(vis, -1, sizeof(vis));
p.clear(); p.push_back(Point(0,0));
bool flag = 1;
int m = 0;
for(int i=1; i<=n; i++) {
scanf("%lld %lld %d", &tmp[i].x, &tmp[i].y, &tmp[i].tp);
}
//sort(p+1, p+1+n);
/*去重*/
for(int i=1; i<=n; i++) {
if(vis[tmp[i].x][tmp[i].y] == -1) {
vis[tmp[i].x][tmp[i].y] = tmp[i].tp;
p.push_back(tmp[i]);
m++;
} else {
if(vis[tmp[i].x][tmp[i].y] != tmp[i].tp) {
flag = 0;
break;
}
}
}
if(!flag) { /*如果有两类点重叠，无法划分*/
printf("0\n");
continue;
}
vector<int> in; /*在直线上的点*/
vector<int> a; /*A类点'o'*/
vector<int> b; /*B类点'x'*/
for(int i=1; i<=m; i++) {
for(int j=i+1; j<=m; j++) {
flag = 1;
Line l; l.a = p[i]; l.b = p[j];
a.clear(); b.clear(); in.clear();
for(int k=1; k<=m; k++) {
if(is_inline(p[i], p[j], p[k])) { /*在直线上*/
in.push_back(k);
continue;
}
if(p[k].tp == 0) {
if(a.empty()) {
a.push_back(k);
if(!b.empty()) {
/*因为只能维护是否在同一边而不能具体到是左边还是右边，所以要判断集合中的第一个点是否违背条件，样例2可以说明这一点*/
if(same_side(p[k],p[b[0]],l)) {
flag = 0;
break;
}
}
} else {
if(opposite_side(p[a[0]],p[k],l)) {
flag = 0;
break;
}
else a.push_back(k);
}
continue;
}
if(p[k].tp == 1) {
if(b.empty()) {
b.push_back(k);
if(!a.empty()) {
if(same_side(p[k],p[a[0]],l)) {
flag = 0;
break;
}
}
} else {
if(opposite_side(p[b[0]],p[k],l)) {
flag = 0;
break;
}
else b.push_back(k);
}
continue;
}
}
if(!flag) continue;
/*对直线上的点判断是否有"间隔出现"的情况*/
sort(in.begin(), in.end());
int sz = in.size();
bool change = 0;
int last = p[in[0]].tp;
for(int k=1; k<sz; k++) {
if(p[in[k]].tp != last) {
if(!change) {
last = p[in[k]].tp;
change = 1;
}
else {
flag = 0;
break;
}
}
}
if(flag) {
//printf("%d %d\n", i,j);
goto las;
}
}
}
las:
if(flag) printf("1\n");
else printf("0\n");
}
return 0;
}