poj 2318 TOYS (计算几何）

TOYS

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 12453		Accepted: 6013

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.

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.

/*************************************************************************
> File Name: code/poj/2318.cpp
> Author: 111qqz
> Email: rkz2013@126.com
> Created Time: 2015年11月08日 星期日 10时09分41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=5E3+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()
{
//    freopen("in.txt","r",stdin);
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-1,ans[i]);
}
puts("");
//      for ( int i = 1 ; i <= m ;  i++)
//      {
//        if (cnt[i]==0) continue;
//          printf("%d: %d\n",i,cnt[i]);
//      }
}
return 0;
}

View Code

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.

直接上代码