UVALive 4639 - Separate Points
2012-08-16 21:04
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Time Limit:3000MS Memory Limit:0KB 64bit IO Format:%lld
& %llu
Submit Status Practice UVALive
4639
Description
Numbers of black and white points are placed on a plane. Let's imagine that a straight line of infinite length is drawn on the plane. When the line does not meet any of the points, the line divides these points
into two groups. If the division by such a line results in one group consisting only of black points and the other consisting only of white points, we say that the line ``separates black and white points".
Let's see examples in Figure 3. In the leftmost example, you can easily find that the black and white points can be perfectly separated by the dashed line according to their colors. In the remaining three examples,
there exists no such straight line that gives such a separation.
[align=center]Figure 3: Example planes[/align]
In this problem, given a set of points with their colors and positions, you are requested to decide whether there exists a straight line that separates black and white points.
Input
The input is a sequence of datasets, each of which is formatted as follows.
n m
x1 y1
...
xn yn
xn+1 yn+1
...
xn+m yn+m
The first line contains two positive integers separated by a single space; n is the number of black points, and m is the number of white points. They are less than or equal to 100. Then n + m lines
representing the coordinates of points follow. Each line contains two integersxi and yi separated by a space, where (xi, yi) represents
the x-coordinate and the y-coordinate of the i-th point. The color of the i-th point is black for 1
i
n,
and is white for n + 1
i
n + m.
All the points have integral x- and y-coordinate values between 0 and 10000 inclusive. You can also assume that no two points have the same
position.
The end of the input is indicated by a line containing two zeros separated by a space.
Output
For each dataset, output ``YES" if there exists a line satisfying the condition. If not, output ``NO". In either case, print it in one line for each input dataset.
Sample Input
Sample Output
比赛时，WA了三遍的题。大体思路比较简单,但细节比较纠结，就是在黑点和白点中任意取两点组成一条直线，只要可以将白点和黑点分成两部分则具备一条直线可以将其分成黑白两部分，但当所有点都在一条直线上时需要特殊处理（按x值排序比较黑白临界点大小）。
比赛时的代码（跑了1s多）：
比赛完又大体剪了一下枝，时间减到280ms，但代码却已出奇的长了，抽空再建子函数缩一下代码吧，今儿个实在累了。
粗减之后的代码：
& %llu
Submit Status Practice UVALive
4639
Description
Numbers of black and white points are placed on a plane. Let's imagine that a straight line of infinite length is drawn on the plane. When the line does not meet any of the points, the line divides these points
into two groups. If the division by such a line results in one group consisting only of black points and the other consisting only of white points, we say that the line ``separates black and white points".
Let's see examples in Figure 3. In the leftmost example, you can easily find that the black and white points can be perfectly separated by the dashed line according to their colors. In the remaining three examples,
there exists no such straight line that gives such a separation.
[align=center]Figure 3: Example planes[/align]
In this problem, given a set of points with their colors and positions, you are requested to decide whether there exists a straight line that separates black and white points.
Input
The input is a sequence of datasets, each of which is formatted as follows.
n m
x1 y1
...
xn yn
xn+1 yn+1
...
xn+m yn+m
The first line contains two positive integers separated by a single space; n is the number of black points, and m is the number of white points. They are less than or equal to 100. Then n + m lines
representing the coordinates of points follow. Each line contains two integersxi and yi separated by a space, where (xi, yi) represents
the x-coordinate and the y-coordinate of the i-th point. The color of the i-th point is black for 1
i
n,
and is white for n + 1
i
n + m.
All the points have integral x- and y-coordinate values between 0 and 10000 inclusive. You can also assume that no two points have the same
position.
The end of the input is indicated by a line containing two zeros separated by a space.
Output
For each dataset, output ``YES" if there exists a line satisfying the condition. If not, output ``NO". In either case, print it in one line for each input dataset.
Sample Input
3 3 100 700 200 200 600 600 500 100 500 300 800 500 3 3 100 300 400 600 400 100 600 400 500 900 300 300 3 4 300 300 500 300 400 600 100 100 200 900 500 900 800 100 1 2 300 300 100 100 500 500 1 1 100 100 200 100 2 2 0 0 500 700 1000 1400 1500 2100 2 2 0 0 1000 1000 1000 0 0 1000 3 3 0 100 4999 102 10000 103 5001 102 10000 102 0 101 3 3 100 100 200 100 100 200 0 0 400 0 0 400 3 3 2813 1640 2583 2892 2967 1916 541 3562 9298 3686 7443 7921 0 0
Sample Output
YES NO NO NO YES YES NO NO NO YES
比赛时，WA了三遍的题。大体思路比较简单,但细节比较纠结，就是在黑点和白点中任意取两点组成一条直线，只要可以将白点和黑点分成两部分则具备一条直线可以将其分成黑白两部分，但当所有点都在一条直线上时需要特殊处理（按x值排序比较黑白临界点大小）。
比赛时的代码（跑了1s多）：
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<cstdlib>
using namespace std;
struct point
{
double x, y;
} b[105], w[105];
int cmp(const void *a, const void *b)
{
point *aa = (point *)a;
point *bb = (point *)b;
if(aa->x != bb->x)
return aa->x > bb->x ? 1 : -1;
return aa->y > bb->y ? 1 : -1;
}
const double eps = 1e-12;
int main()
{
int m,n;
while(cin>>m>>n)
{
if(!m&&!n)break;
for(int i=0; i<m; i++)
cin>>b[i].x>>b[i].y;
for(int i=0; i<n; i++)
cin>>w[i].x>>w[i].y;
qsort(b, m, sizeof(b[0]), cmp);
qsort(w, n, sizeof(w[0]), cmp);
int flag=0, cct, fflag = 0;
for(int i = 0; i < m; i++) // 先检验黑点
{
for(int j = i + 1; j < m; j++)
{
cct = 0;
int ll = 0, rr = 0, ll2 = 0, rr2 = 0;
if(!(fabs(b[i].x - b[j].x) < eps))
{
double k = (b[i].y - b[j].y) / (b[i].x - b[j].x);
double bb = b[i].y - k * b[i].x;
for(int kk = 0; kk < n; kk++)
if(w[kk].y - k * w[kk].x - bb < -eps)
++rr;
else if(w[kk].y - k * w[kk].x - bb > eps)
++ll;
else if(fabs(w[kk].y - k * w[kk].x - bb) < eps)
++cct;
for(int kk = 0; kk < m; kk++)
if(kk != i && kk != j)
{
if(b[kk].y - k * b[kk].x - bb < -eps)
++rr2;
else if(b[kk].y - k * b[kk].x - bb > eps)
++ll2;
else if(fabs(b[kk].y - k * b[kk].x - bb) < eps)
++cct;
}
}
else // 注意垂直直线的处理
{
for(int kk = 0; kk < n; kk++)
if(w[kk].x < b[j].x)
++rr;
else if(w[kk].x > b[j].x)
++ll;
for(int kk = 0; kk < m; kk++)
if(kk != i && kk != j)
{
if(b[kk].x < b[j].x)
++rr2;
else if(b[kk].x > b[j].x)
++ll2;
}
}
if(cct == m + n - 2)
{
fflag = 1;
break;
}
if((ll == n && !ll2) || (rr == n && !rr2))
{
flag = 1;
break;
}
}
if(flag)
break;
}
if(!flag && ! fflag) // 检验白点
{
for(int i = 0; i < n; i++)
{
for(int j = i + 1; j < n; j++)
{
int ll = 0, rr = 0, ll2 = 0, rr2 = 0;
if(!(fabs(w[i].x - w[j].x) < eps))
{
double k = (w[i].y - w[j].y) / (w[i].x - w[j].x);
double bb = w[i].y - k * w[i].x;
for(int kk = 0; kk < m; kk++)
if(b[kk].y - k * b[kk].x - bb < -eps)
++rr;
else if(b[kk].y - k * b[kk].x - bb > eps)
++ll;
for(int kk = 0; kk < n; kk++)
if(kk != i && kk != j)
{
if(w[kk].y - k * w[kk].x - bb < -eps)
++rr2;
else if(w[kk].y - k * w[kk].x - bb > eps)
++ll2;
}
}
else
{
for(int kk = 0; kk < m; kk++)
if(b[kk].x < w[j].x)
++rr;
else if(b[kk].x > w[j].x)
++ll;
for(int kk = 0; kk < n; kk++)
if(kk != i && kk != j)
{
if(w[kk].x < w[j].x)
++rr2;
else if(w[kk].x > w[j].x)
++ll2;
}
}
if((ll == m && !ll2) || (rr == m && !rr2))
{
flag = 1;
break;
}
}
if(flag)
break;
}
}
if(fabs(b[0].x - b[m - 1].x) < eps && fabs(w[0].x - w[n - 1].x) < eps && fabs(b[0].x - w[0].x) < eps) // 所有点在一条垂直直线时，特殊处理
{
if(b[m - 1].y < w[0].y || w[n - 1].y < b[0].y)
flag = 1;
}
if(fflag) // 所有点在一条斜线时，特殊处理
{
if(b[m - 1].x < w[0].x || w[n - 1].x < b[0].x)
flag = 1;
}
if(flag == 1||(m==1&&n==1))cout<<"YES"<<endl;
else cout<<"NO"<<endl;
}
return 0;
}比赛完又大体剪了一下枝，时间减到280ms，但代码却已出奇的长了，抽空再建子函数缩一下代码吧，今儿个实在累了。
粗减之后的代码：
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<cstdlib>
using namespace std;
struct point
{
double x, y;
} b[105], w[105];
int cmp(const void *a, const void *b)
{
point *aa = (point *)a;
point *bb = (point *)b;
if(aa->x != bb->x)
return aa->x > bb->x ? 1 : -1;
return aa->y > bb->y ? 1 : -1;
}
const double eps = 1e-12;
int main()
{
freopen("in.txt", "r", stdin);
int m,n;
while(cin>>m>>n)
{
if(!m&&!n)break;
for(int i=0; i<m; i++)
cin>>b[i].x>>b[i].y;
for(int i=0; i<n; i++)
cin>>w[i].x>>w[i].y;
qsort(b, m, sizeof(b[0]), cmp);
qsort(w, n, sizeof(w[0]), cmp);
int flag=0, cct, fflag = 0;
for(int i = 0; i < m; i++)
{
for(int j = i + 1; j < m; j++)
{
cct = 0;
int ll = 0, rr = 0, ll2 = 0, rr2 = 0;
if(!(fabs(b[i].x - b[j].x) < eps))
{
int flag_l = 0, flag_r = 0;
double k = (b[i].y - b[j].y) / (b[i].x - b[j].x);
double bb = b[i].y - k * b[i].x;
for(int kk = 0; kk < n; kk++)
{
if(w[kk].y - k * w[kk].x - bb < -eps)
{
++rr;
flag_l = 1;
}
else if(w[kk].y - k * w[kk].x - bb > eps)
{
++ll;
flag_r = 1;
}
else if(fabs(w[kk].y - k * w[kk].x - bb) < eps)
++cct;
if(flag_l && flag_r)
break;
}
if(flag_l && flag_r) // 加了许多诸如此类的粗剪枝
continue;
flag_l = 0, flag_r = 0;
for(int kk = 0; kk < m; kk++)
if(kk != i && kk != j)
{
if(b[kk].y - k * b[kk].x - bb < -eps)
{
++rr2;
flag_r = 1;
}
else if(b[kk].y - k * b[kk].x - bb > eps)
{
++ll2;
flag_l = 1;
}
else if(fabs(b[kk].y - k * b[kk].x - bb) < eps)
++cct;
if(flag_l && flag_r)
break;
}
if(flag_l && flag_r)
continue;
}
else
{
int flag_l = 0, flag_r = 0;
for(int kk = 0; kk < n; kk++)
{
if(w[kk].x < b[j].x)
{
flag_r = 1;
++rr;
}
else if(w[kk].x > b[j].x)
{
flag_l = 1;
++ll;
}
if(flag_l && flag_r)
break;
}
if(flag_l && flag_r)
continue;
flag_l = 0, flag_r = 0;
for(int kk = 0; kk < m; kk++)
if(kk != i && kk != j)
{
if(b[kk].x < b[j].x)
{
++rr2;
flag_r = 1;
}
else if(b[kk].x > b[j].x)
{
++ll2;
flag_l = 1;
}
if(flag_l && flag_r)
break;
}
if(flag_l && flag_r)
continue;
}
if(cct == m + n - 2)
{
fflag = 1;
break;
}
if((ll == n && !ll2) || (rr == n && !rr2))
{
flag = 1;
break;
}
}
if(flag)
break;
}
if(!flag && ! fflag)
{
for(int i = 0; i < n; i++)
{
for(int j = i + 1; j < n; j++)
{
int ll = 0, rr = 0, ll2 = 0, rr2 = 0;
if(!(fabs(w[i].x - w[j].x) < eps))
{
int flag_l = 0, flag_r = 0;
double k = (w[i].y - w[j].y) / (w[i].x - w[j].x);
double bb = w[i].y - k * w[i].x;
for(int kk = 0; kk < m; kk++)
{
if(b[kk].y - k * b[kk].x - bb < -eps)
{
++rr;
flag_r = 1;
}
else if(b[kk].y - k * b[kk].x - bb > eps)
{
++ll;
flag_l = 1;
}
if(flag_l && flag_r)
break;
}
if(flag_l && flag_r)
continue;
flag_l = 0, flag_r = 0;
for(int kk = 0; kk < n; kk++)
if(kk != i && kk != j)
{
if(w[kk].y - k * w[kk].x - bb < -eps)
{
++rr2;
flag_r = 1;
}
else if(w[kk].y - k * w[kk].x - bb > eps)
{
++ll2;
flag_l = 1;
}
if(flag_l && flag_r)
break;
}
if(flag_l && flag_r)
continue;
}
else
{
int flag_l = 0, flag_r = 0;
for(int kk = 0; kk < m; kk++)
{
if(b[kk].x < w[j].x)
{
++rr;
flag_r = 1;
}
else if(b[kk].x > w[j].x)
{
++ll;
flag_l = 1;
}
if(flag_l && flag_r)
break;
}
if(flag_l && flag_r)
continue;
flag_l = 0, flag_r = 0;
for(int kk = 0; kk < n; kk++)
if(kk != i && kk != j)
{
if(w[kk].x < w[j].x)
{
++rr2;
flag_r = 1;
}
else if(w[kk].x > w[j].x)
{
++ll2;
flag_l = 1;
}
if(flag_l && flag_r)
break;
}
if(flag_l && flag_r)
continue;
}
if((ll == m && !ll2) || (rr == m && !rr2))
{
flag = 1;
break;
}
}
if(flag)
break;
}
}
if(fabs(b[0].x - b[m - 1].x) < eps && fabs(w[0].x - w[n - 1].x) < eps && fabs(b[0].x - w[0].x) < eps)
{
if(b[m - 1].y < w[0].y || w[n - 1].y < b[0].y)
flag = 1;
}
if(fflag)
{
if(b[m - 1].x < w[0].x || w[n - 1].x < b[0].x)
flag = 1;
}
if(flag == 1||(m==1&&n==1))cout<<"YES"<<endl;
else cout<<"NO"<<endl;
}
return 0;
}
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