poj 2398 Toy Storage (计算几何,判断点和线段关系)

http://poj.org/problem?id=2398

题意大概是说将一个盒子用n个board分成n+1 部分

然后往里面放toy,给定盒子,board,和toy的坐标

问所有的toy放完后,有多少部分中有t个toy;

简单计算几何

需要判断的是点和直线的关系.

/*************************************************************************
> File Name: code/2015summer/0718/B.cpp
> Author: 111qqz
> Email: rkz2013@126.com
> Created Time: 2015年07月18日 星期六 11时58分14秒
************************************************************************/
#include<iostream>
#include<iomanip>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<map>
#include<set>
#include<queue>
#include<vector>
#include<stack>
using namespace std;
#define REP(i, n) for (int i=0;i<int(n);++i)
typedef long long LL;
typedef unsigned long long ULL;
const int N=2E3+5;
struct node
{
int x,y;
};
struct node rec,rec2;
struct node par
,par2
;
struct node toy
;
int ans
,cnt
;
int n,m;
bool judge(node p1,node p2,node p3) //判断点是否在直线的[右侧!!]
{
int s = (p1.x-p3.x)*(p2.y-p3.y)-(p1.y-p3.y)*(p2.x-p3.x);
if (s>0) return false;
if (s<0) return true;
}
bool cmp(node a,node b)
{
if (a.x<b.x) return true;
if (a.x==b.x&&a.y<b.y) return true;
return false;
}
int main()
{
while (scanf("%d",&n)!=EOF&&n)
{
memset(ans,0,sizeof(ans));
memset(par,0,sizeof(par));
memset(par2,0,sizeof(par2));
memset(toy,0,sizeof(toy));
cin>>m>>rec.x>>rec.y>>rec2.x>>rec2.y;
for ( int i = 1 ;  i <= n ; i++)
{
cin>>par[i].x>>par2[i].x;
par[i].y=rec.y;
par2[i].y=rec2.y;
}
for ( int i = 1 ; i <= n-1 ; i++)
{
for ( int j = i+1 ; j <= n ; j++)
{
if (par[i].x>par[j].x)
{
swap(par[i].x,par[j].x);
//  swap(par[i].y,par[j].y);
swap(par2[i].x,par2[j].x);
//   swap(par2[i].y,par2[j].y);
}
}
}
//      for ( int i = 1 ;  i <= n ; i++)
//        cout<<par[i].x<<endl;
for ( int i = 1 ;  i <= m ; i++ )
{
cin>>toy[i].x>>toy[i].y;
}
int p;
sort(toy+1,toy+m+1,cmp);  //如果第i个娃娃在第k个分划中,那么排序后第i+1个娃娃至少在第k个分划中....(某大神说过,顺手就能写的优化顺手
//      for ( int i = 1 ; i <= m ; i++)  cout<<"x[i]:"<<toy[i].x<<" y[i]:"<<toy[i].y<<endl;
for ( int i = 1 ; i  <= m ;  i++)
{
p = n + 1;  //如果在所有board的右侧,那么一定是在最后一个分划中(n个板子形成n+1个分划)
bool ok=false;
for ( int j = 1 ; j  <= n ; j++)
{
ok=judge(par2[j],par[j],toy[i]);
if (!ok)
{
//      cout<<"i:"<<i<<" j:"<<j<<" "<<par2[j].x<<" "<<par2[j].y<<" "<<par[j].x<<" "<<par[j].y<<endl;
p = j;
break;
//    cout<< "hhhhhh"<<"I:"<<i<<" j:"<<j<<endl;
}
}
ans[p]++;
}
cout<<"Box"<<endl;
memset(cnt,0,sizeof(cnt));
for ( int i = 1 ; i <= n+1 ; i++)
{
if (ans[i]==0) continue;
cnt[ans[i]]++;
//    printf("%d: %d\n",i,ans[i]);
}
for ( int i = 1 ; i <= m ;  i++)
{
if (cnt[i]==0) continue;
printf("%d: %d\n",i,cnt[i]);
}
}
return 0;
}