poj 2398Toy Storage

最新推荐文章于 2025-03-18 20:46:33 发布

Modiz

最新推荐文章于 2025-03-18 20:46:33 发布

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Toy Storage

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 4063		Accepted: 2410

Description

Mom and dad have a problem: their child, Reza, never puts his toys away when he is finished playing with them. They gave Reza a rectangular box to put his toys in. Unfortunately, Reza 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 Reza to find his favorite toys anymore.
Reza's parents came up with the following idea. They put cardboard partitions into the box. Even if Reza keeps throwing his toys into the box, at least toys that get thrown into different partitions stay separate. The box looks like this from the top:

We want for each positive integer t, such that there exists a partition with t toys, determine how many partitions have t, toys.

Input

The input consists of a number of cases. The first line consists of six integers n, m, x1, y1, x2, y2. The number of cardboards to form the partitions is n (0 < n <= 1000) and the number of toys is given in m (0 < m <= 1000). 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 each consists of two integers Ui Li, indicating that the ends of the ith cardboard is at the coordinates (Ui, y1) and (Li, y2). You may assume that the cardboards do not intersect with each other. The next m lines each consists of two integers Xi Yi specifying where the ith toy has landed in the box. You may assume that no toy will land on a cardboard.

A line consisting of a single 0 terminates the input.

Output

For each box, first provide a header stating "Box" on a line of its own. After that, there will be one line of output per count (t > 0) of toys in a partition. The value t will be followed by a colon and a space, followed the number of partitions containing t toys. Output will be sorted in ascending order of t for each box.

Sample Input

4 10 0 10 100 0
20 20
80 80
60 60
40 40
5 10
15 10
95 10
25 10
65 10
75 10
35 10
45 10
55 10
85 10
5 6 0 10 60 0
4 3
15 30
3 1
6 8
10 10
2 1
2 8
1 5
5 5
40 10
7 9
0

Sample Output

Box
2: 5
Box
1: 4
2: 1

题意：跟上一题类似不过这题线段(也就是题目中所说的纸板)不是按照顺序给出的，需要进行排次序，然后题目要求的是有多少个空间里放t个玩具。比如第一组样例中每个空间里都是2个玩具，所以输出2:5.第二组样例有四个空间里有1个玩具，有1个空间里有2个玩具，所以如上输出。

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <cmath>
#include <queue>
#include <map>
#include <stack>
#include <list>
#include <vector>
#include <ctime>
#define LL __int64
#define eps 1e-8
using namespace std;
int gg[1010],ans[1010],a[1010];
struct node
{
	int x1,x2;
}f[1010];
int across(int x1,int y1,int x2,int y2)
{
	return (x1*y2)-(x2*y1);
}
int cmp1(node a,node b)
{
	return a.x1<b.x1;
}
int cmp2(int a,int b)
{
	return a<b;
}
int main()
{
	int n,x1,m,x,y,x2,y1,y2,i,j;
	while (~scanf("%d",&n) && n)
	{
		scanf("%d%d%d%d%d",&m,&x1,&y1,&x2,&y2);
		for (i=0;i<n;i++)
			scanf("%d%d",&f[i].x1,&f[i].x2);
		f[n].x1=x2;f[n].x2=x2;
		sort(f,f+n+1,cmp1);
		memset(ans,0,sizeof(ans));
		memset(gg,0,sizeof(gg));
		memset(a,0,sizeof(a));
		for (i=1;i<=m;i++)
		{
			scanf("%d%d",&x,&y);
			for (j=0;j<=n;j++){
				int k=across(f[j].x1-x,y1-y,f[j].x2-x,y2-y);
				if (k<0){
					ans[j]++;
					break;
				}
			}
		}
		map<int,int>mp;
		int t=0;
		for (i=0;i<=n;i++)
		{
			if (ans[i]>0)
			{
				gg[ans[i]]++;
				if (mp[ans[i]]==0)
				{
					mp[ans[i]]=1;
					a[t++]=ans[i];
				}
			}
		}
		sort(a,a+t,cmp2);
		cout<<"Box"<<endl;
		for (i=0;i<t;i++)
		{
			printf("%d: %d\n",a[i],gg[a[i]]);
		}
	}
	return 0;
}