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Language: Default TOYS
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 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
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 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 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. Source |
题意：给m个点的坐标，落在n+1个区域中，问各个区域有多少个点。
思路：利用叉积去判断点在线段的哪一侧，二分解决。今天开始慢慢接触计算几何了，然而网络流还没玩顺溜=-=
代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define pi acos(-1.0)
#define eps 1e-6
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
#define FRE(i,a,b) for(i = a; i <= b; i++)
#define FREE(i,a,b) for(i = a; i >= b; i--)
#define FRL(i,a,b) for(i = a; i < b; i++)
#define FRLL(i,a,b) for(i = a; i > b; i--)
#define mem(t, v) memset ((t) , v, sizeof(t))
#define sf(n) scanf("%d", &n)
#define sff(a,b) scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf printf
#define DBG pf("Hi\n")
typedef long long ll;
using namespace std;
#define INF 0x3f3f3f3f
#define mod 1000000009
const int maxn = 5500;
const int MAXN = 2005;
const int N = 1005;
int n,m;
int x1,Y1,x2,y2;
struct Point
{
int x,y;
Point(){}
Point(int _x,int _y)
{
x=_x; y=_y;
}
Point operator-(const Point &b)const
{
return Point(x-b.x,y-b.y);
}
int operator*(const Point &b)const
{
return x*b.x+y*b.y;
}
int operator^(const Point &b)const
{
return x*b.y-y*b.x;
}
};
struct Line
{
Point s,e;
Line(){}
Line(Point _s,Point _e)
{
s=_s;e=_e;
}
};
int xmulti(Point p0,Point p1,Point p2) //叉积
{
return (p2-p0)^(p1-p0);
}
Line line[maxn];
int ans[maxn];
void solve()
{
int x,y;
Point p;
mem(ans,0);
while (m--)
{
sff(x,y);
p=Point(x,y);
int l=0,r=n,temp;
while (l<=r)
{
int mid=(l+r)>>1;
if (xmulti(p,line[mid].s,line[mid].e)<0)
l=mid+1;
else
{
temp=mid;
r=mid-1;
}
}
ans[temp]++;
}
for (int i=0;i<=n;i++)
pf("%d: %d\n",i,ans[i]);
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("C:/Users/lyf/Desktop/IN.txt","r",stdin);
#endif
int i,j;
int flag=0;
while (sf(n),n)
{
sf(m);
sff(x1,Y1);
sff(x2,y2);
if (flag)
pf("\n");
flag=1;
int a,b;
for (i=0;i<n;i++)
{
sff(a,b);
line[i]=Line(Point(a,Y1),Point(b,y2));
}
line[n]=Line(Point(x2,Y1),Point(x2,y2));
solve();
}
return 0;
}
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