POJ 2318 TOYS(叉积+二分)
题目传送门:POJ 2318 TOYS
Description
Mom and dad have a problem - their child John never puts his toys away when he is finished playing with them. They gave John a rectangular box to put his toys in, but John is rebellious and obeys his parents by simply throwing his toys into the box. All the toys get mixed up, and it is impossible for John to find his favorite toys.
John's parents came up with the following idea. They put cardboard partitions into the box. Even if John keeps throwing his toys into the box, at least toys that get thrown into different bins stay separated. The following diagram shows a top view of an example toy box.
For this problem, you are asked to determine how many toys fall into each partition as John throws them into the toy box.
Input
Output
Sample Input
5 6 0 10 60 0
3 1
4 3
6 8
10 10
15 30
1 5
2 1
2 8
5 5
40 10
7 9
4 10 0 10 100 0
20 20
40 40
60 60
80 80
5 10
15 10
25 10
35 10
45 10
55 10
65 10
75 10
85 10
95 10
0
Sample Output
0: 2
1: 1
2: 1
3: 1
4: 0
5: 1 0: 2
1: 2
2: 2
3: 2
4: 2 题目大意: 给你一个盒子,里面有n个隔板,n+1个区间:0~n 里面放m个物品,问每个区间有多少个物品。 解题思路:
对于每个物品都进行二分来查找区间,通过叉积来判断点与直线的位置关系,进而确定这个物品在哪个区间。
(我之前是想分别将线段和点进行排序然后依次比较就行了 太天真了...
#include<cmath>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N = +;
int num[N];
struct point
{
double x,y;
point(double a = , double b = ) { x = a, y = b; }
double operator ^(const point& b) const { return x * b.y - y * b.x; }/// 叉积
point operator - (const point& b) const { return point(x - b.x, y - b.y); }
}p;
struct line
{
int id;
point s,e;
}l[N]; bool _is(line c,point p)///判断点是否在直线右边
{
point p1=c.s-p,p2=c.e-p;
double ans=p1^p2;
if (ans>) return true;
return false;
}
int main()
{
int n,m;
point b,d;
double u,v;
while(~scanf("%d",&n)&&n)
{
memset(num,,sizeof(num));
scanf("%d%lf%lf%lf%lf",&m,&b.x,&b.y,&d.x,&d.y);
l[].s=b;l[].e.x=b.x;l[].e.y=d.y;
l[n+].s.x=d.x;l[n+].s.y=b.y;l[n+].e=d;
l[n+].id=n+;
for (int i=;i<=n;i++)
{
scanf("%lf%lf",&u,&v);
l[i].s.x=u,l[i].s.y=b.y;
l[i].e.x=v,l[i].e.y=d.y;
l[i].id=i;
}
for (int i=;i<m;i++)
{
scanf("%lf%lf",&p.x,&p.y);
int L=,R=n+;
while(L<R)
{
int mid=(L+R)/;
if (_is(l[mid],p)) L=mid+;
else R=mid-;
}
while (!_is(l[L],p)) --L;
num[L]++;
}
for (int i=;i<=n;i++) printf("%d: %d\n",i,num[i]);
printf("\n");
}
return ;
}
同类型题:Toy Storage POJ - 2398
#include<cmath>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N = +;
int num1[N],num2[N];
struct point
{
double x,y;
point(double a = , double b = ) { x = a, y = b; }
double operator ^(const point& b) const { return x * b.y - y * b.x; }/// 叉积
point operator - (const point& b) const { return point(x - b.x, y - b.y); }
}p;
struct line
{
int id;
point s,e;
}l[N];
bool cmp(line a,line b)
{
return a.s.x<b.s.x;
}
bool _is(line c,point p)///判断点是否在直线右边
{
point p1=c.s-p,p2=c.e-p;
double ans=p1^p2;
if (ans>) return true;
return false;
}
int main()
{
int n,m;
point b,d;
double u,v;
while(~scanf("%d",&n)&&n)
{
memset(num1,,sizeof(num1));
memset(num2,,sizeof(num2));
scanf("%d%lf%lf%lf%lf",&m,&b.x,&b.y,&d.x,&d.y);
l[].s=b;l[].e.x=b.x;l[].e.y=d.y;
l[n+].s.x=d.x;l[n+].s.y=b.y;l[n+].e=d;
l[n+].id=n+;
for (int i=;i<=n;i++)
{
scanf("%lf%lf",&u,&v);
l[i].s.x=u,l[i].s.y=b.y;
l[i].e.x=v,l[i].e.y=d.y;
l[i].id=i;
}
sort(l+,l+n+,cmp);//与上题比多了个排序,因为它的输入不是按顺序的
for (int i=;i<m;i++)
{
scanf("%lf%lf",&p.x,&p.y);
int L=,R=n+;
while(L<R)
{
int mid=(L+R)/;
if (_is(l[mid],p)) L=mid+;
else R=mid-;
}
while (!_is(l[L],p)) --L;
num1[L]++;
}
printf("Box\n");
for (int i=;i<=n;i++) num2[num1[i]]++;
//这题所求有点不同
for (int i=;i<=n;i++)
if (num2[i]) printf("%d: %d\n",i,num2[i]);
}
return ;
}
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