http://poj.org/problem?id=2318
TOYS
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 10178 | Accepted: 4880 |
Description
Calculate the number of toys that land in each bin of a partitioned toy box.
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.
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
The input file contains one or more problems. The first line of a problem consists of six integers, n m x1 y1 x2 y2. The number of cardboard partitions is n (0 < n <= 5000) and the number of toys is m (0 < m <= 5000). The coordinates of the upper-left corner and the lower-right corner of the box are (x1,y1) and (x2,y2), respectively. The following n lines contain two integers per line, Ui Li, indicating that the ends of the i-th cardboard partition is at the coordinates (Ui,y1) and (Li,y2). You may assume that the cardboard partitions do not intersect each other and that they are specified in sorted order from left to right. The next m lines contain two integers per line, Xj Yj specifying where the j-th toy has landed in the box. The order of the toy locations is random. You may assume that no toy will land exactly on a cardboard partition or outside the boundary of the box. The input is terminated by a line consisting of a single 0.
Output
The output for each problem will be one line for each separate bin in the toy box. For each bin, print its bin number, followed by a colon and one space, followed by the number of toys thrown into that bin. Bins are numbered from 0 (the leftmost bin) to n (the rightmost bin). Separate the output of different problems by a single blank line.
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
As the example illustrates, toys that fall on the boundary of the box are "in" the box.
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题意不难理解，有那么多的格子，问当有m个点放上去时，每个格子有多少点
自己写了个暴力的超时了，但是有人暴力过了，真是好无语啊～～
用二分求出精确区间，再判断就可以了
#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 ;
}