poj 2318 TOYS (二分+叉积)
http://poj.org/problem?id=2318
Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions: 10178 | Accepted: 4880 |
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
Hint
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <iostream>
#include <algorithm>
#include <math.h>
#include <ctype.h>
#define eps 1e-6
#define MAX 5010 using namespace std; typedef struct
{
double x,y;
}point;
typedef struct
{
point a,b;
}line; line li[MAX];
point p[MAX];
int str[MAX]; bool dy(double x,double y){ return x>y+eps; }
bool xy(double x,double y){ return x<y-eps; }
bool dyd(double x,double y){ return x>y-eps; }
bool xyd(double x,double y){ return x<y+eps; }
bool dd(double x,double y){ return fabs(x-y)<eps; } double crossProduct(point a,point b,point c)
{
return (c.x-a.x)*(b.y-a.y)-(c.y-a.y)*(b.x-a.x);
} void BSearch(point a,int n)
{
int l=,r=n-;
while(l<r)
{
int mid=(l+r)/;
if(crossProduct(li[mid].a,a,li[mid].b)>) l=mid+;
else r=mid;
}
if(crossProduct(li[l].a,a,li[l].b)<)str[l]++;
else str[l+]++;
} int main()
{
int n,m,x1,x2,y1,y2;
int i,j;
int part1,part2; point tmp;
while(scanf("%d",&n)!=EOF&&n)
{
scanf("%d%d%d%d%d",&m,&x1,&y1,&x2,&y2);
memset(str,,sizeof(str));
for(i=;i<n;i++)
{
scanf("%d%d",&part1,&part2);
li[i].a.x=part1;
li[i].a.y=y1;
li[i].b.x=part2;
li[i].b.y=y2;
}
for(i=;i<m;i++)
{
scanf("%lf%lf",&p[i].x,&p[i].y);
BSearch(p[i],n);
}
for(i=;i<=n;i++)
{
printf("%d: %d\n",i,str[i]);
}
printf("\n");
}
return ;
}
这个暴力的不超时……
#include <queue>
#include <stack>
#include <math.h>
#include <stdio.h>
#include <stdlib.h>
#include <iostream>
#include <limits.h>
#include <string.h>
#include <algorithm>
using namespace std;
const int MAX = ;
struct SEG{
int x1,y1,x2,y2;
};
SEG s[MAX];
struct point{
int x,y;
};
point toy[MAX];
int sum[MAX];
int crossProduct(point a,point b,point c)//向量 ac 在 ab 的方向
{
return (c.x - a.x)*(b.y - a.y) - (b.x - a.x)*(c.y - a.y);
}
bool inBox(point t,SEG ls,SEG rs)
{
point a,b,c,d;
a.x = ls.x1; a.y = ls.y1;
b.x = ls.x2; b.y = ls.y2;
c.x = rs.x2; c.y = rs.y2;
d.x = rs.x1; d.y = rs.y1;
if( crossProduct(b,t,c) >= && crossProduct(c,t,d) >=
&& crossProduct(d,t,a) >= && crossProduct(a,t,b) >= )
return true;
return false;
}
int main()
{
int n,m,x1,y1,x2,y2,a,b;
while( ~scanf("%d",&n) && n )
{
memset(sum,,sizeof(sum));
scanf("%d %d %d %d %d",&m,&x1,&y1,&x2,&y2);
s[].x1 = x1; s[].y1 = y1;
s[].x2 = x1; s[].y2 = y2;
for(int i=; i<=n; i++)
{
scanf("%d %d",&a,&b);
s[i].x1 = a; s[i].y1 = y1;
s[i].x2 = b; s[i].y2 = y2;
}
n++;
s[n].x1 = x2; s[n].y1 = y1;
s[n].x2 = x2; s[n].y2 = y2;
for(int i=; i<m; i++)
scanf("%d %d",&toy[i].x,&toy[i].y);
for(int i=; i<m; i++)
for(int k=; k<n; k++)
if( inBox(toy[i],s[k],s[k+]) )
{
sum[k]++;
break;
}
for(int i=; i<n; i++)
printf("%d: %d/n",i,sum[i]);
printf("/n");
}
return ;
}
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