POJ 2318 几何初步 + 二分 及其 姊妹题 POJ 2398
叉积+二分
如果ｔｏｙｓ在当前板的左边 cross(L,toys,U) < 0
反之在右边会 > 0
根据这个,我们进行二分；
为什么最后得到的不是ｍｉｄ而是ｌ呢?
我们来看这张图.
加入现在是 l = 2 , r = 5；ｍｉｄ = 3； toy在4纸板右边,5纸板左边,
执行ｌ = ｌ +1 = 3, ｒ = 5, ｍｉｄ = 4；依然>0 执行,
ｌ = ｌ + 1 = 4,ｒ = 5, ｍｉｄ = 4； 依然大于0 ,执行
ｌ = ｌ + 1 = 5, ｒ= 5, 退出； 此时l = 5, 之前ｔｏｙ确实在ｌ右边,可是现在在它左边了.(有的情况还是会在右边)
所以我们后面要加一个判断:
if(cross(CardB[r].L, Toys, CardB[r].U) < 0)
N_Toys[l]++;
else
N_Toys[l+1]++;
显然,我们同样可以:
if(cross(CardB[r].L, Toys, CardB[r].U) > 0)
N_Toys[r+1]++;
else
N_Toys[r]++;
那么,如果利用ｍｉｄ 改怎么改写代码呢?
for(int i = 0; i < m; i++)
{
l = 0, r = n-1;
scanf("%d%d",&Toys.x,&Toys.y);
while(l <= r)
{
mid = (l + r) >> 1;
if(cross(CardB[mid].L, Toys, CardB[mid].U) > 0)
l = mid + 1;
else
r = mid - 1;
}
if(cross(CardB[mid].L, Toys, CardB[mid].U) > 0)
N_Toys[mid+1]++;
else
N_Toys[mid]++;
}
/*
*POJ 2318
*fuqiang
*几何初步
*2013/8/2
*/
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 5000+3;
const double PI = 4.0*atan(1.0);
const double eps = 1e-8;
struct point
{
int x;
int y;
} Toys;
struct Line
{
point U;
point L;
} CardB[maxn];
int N_Toys[maxn];
int cross(const point &p0, const point &p1, const point &p2) //计算叉积
{
return (p1.x-p0.x)*(p2.y-p0.y) - (p2.x-p0.x)*(p1.y-p0.y);
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("in","r",stdin);
#endif
int n,m,x1,y1,x2,y2;
while(scanf("%d",&n) && n)
{
scanf("%d%d%d%d%d",&m,&x1,&y1,&x2,&y2);
for(int i = 0; i < n; i++)
{
scanf("%d%d", &CardB[i].U.x, &CardB[i].L.x);
CardB[i].U.y = y1;
CardB[i].L.y = y2;
}
memset(N_Toys,0,sizeof(N_Toys));
int l, r, mid, ans, p;
for(int i = 0; i < m; i++)
{
l = 0, r = n-1;
scanf("%d%d",&Toys.x,&Toys.y);
while(l < r)
{
mid = (l + r) >> 1;
if(cross(CardB[mid].L, Toys, CardB[mid].U) > 0)
l = mid + 1;
else
r = mid;
}
if(cross(CardB[r].L, Toys, CardB[r].U) < 0)
N_Toys[l]++;
else
N_Toys[l+1]++;
}
for(int i = 0; i <= n; i++)
{
printf("%d: %d\n",i,N_Toys[i]);
}
printf("\n");
}
}
还有个更好的点子,别人的代码:
将最后一个点也作为纸板,此时 我们只需要返回ｌ:
while (l <= r)
{
mid = (l + r) >> 1;
if ( cal (low[mid], high[mid], cc) > 0 )
l = mid + 1;
else r = mid - 1;
}
return l;
}
POJ 2398
加个排序,统计后输出,注意ｔ>0
/*
*POJ 2318
*fuqiang
*几何初步
*2013/8/2
*/
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 5000+3;
const double PI = 4.0*atan(1.0);
const double eps = 1e-8;
struct point
{
int x;
int y;
} Toys;
struct Line
{
point U;
point L;
} CardB[maxn];
int N_Toys[maxn];
int cross(const point &p0, const point &p1, const point &p2) //计算叉积
{
return (p1.x-p0.x)*(p2.y-p0.y) - (p2.x-p0.x)*(p1.y-p0.y);
}
int cmp(Line a, Line b)//sort...
{
return a.L.x < b.L.x;
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("in","r",stdin);
#endif
int n,m,x1,y1,x2,y2;
while(scanf("%d",&n) && n)
{
scanf("%d%d%d%d%d",&m,&x1,&y1,&x2,&y2);
for(int i = 0; i < n; i++)
{
scanf("%d%d", &CardB[i].U.x, &CardB[i].L.x);
CardB[i].U.y = y1;
CardB[i].L.y = y2;
}
sort(CardB,CardB+n,cmp); //sort
memset(N_Toys,0,sizeof(N_Toys));
int l, r, mid, ans, p;
for(int i = 0; i < m; i++)
{
l = 0, r = n-1;
scanf("%d%d",&Toys.x,&Toys.y);
while(l <= r)
{
mid = (l + r) >> 1;
if(cross(CardB[mid].L, Toys, CardB[mid].U) > 0)
l = mid + 1;
else
r = mid - 1;
}
if(cross(CardB[mid].L, Toys, CardB[mid].U) > 0)
N_Toys[mid+1]++;
else
N_Toys[mid]++;
}
sort(N_Toys,N_Toys+n+1);
printf("Box\n");
int st = 0;
while(N_Toys[st] == 0)
st++;
for(int i = st; i <= n; )
{
int num = 1;
printf("%d: ",N_Toys[i]);
for(int j = i++; j <= n; j++,i++)
{
if(N_Toys[i]==N_Toys[j])
num++;
else
break;
}
printf("%d\n",num);
}
// printf("\n");
}
}