洛谷 P2862 [USACO06JAN]把牛Corral the Cows

P2862 [USACO06JAN]把牛Corral the Cows

题目描述

Farmer John wishes to build a corral for his cows. Being finicky beasts, they demand that the corral be square and that the corral contain at least C (1 <= C <= 500) clover fields for afternoon treats. The corral's edges must be parallel to the X,Y axes.

FJ's land contains a total of N (C <= N <= 500) clover fields, each a block of size 1 x 1 and located at with its lower left corner at integer X and Y coordinates each in the range 1..10,000. Sometimes more than one clover field grows at the same location; such a field would have its location appear twice (or more) in the input. A corral surrounds a clover field if the field is entirely located inside the corral's borders.

Help FJ by telling him the side length of the smallest square containing C clover fields.

约翰打算建一个围栏来圈养他的奶牛.作为最挑剔的兽类，奶牛们要求这个围栏必须是正方 形的，而且围栏里至少要有C< 500)个草场，来供应她们的午餐.

约翰的土地上共有C<=N<=500)个草场，每个草场在一块1x1的方格内，而且这个方格的 坐标不会超过10000.有时候，会有多个草场在同一个方格内，那他们的坐标就会相同.

告诉约翰，最小的围栏的边长是多少？

输入输出格式

输入格式：

 

Line 1: Two space-separated integers: C and N

Lines 2..N+1: Each line contains two space-separated integers that are the X,Y coordinates of a clover field.

 

输出格式：

 

Line 1: A single line with a single integer that is length of one edge of the minimum size square that contains at least C clover fields.

 

输入输出样例

输入样例#1： 复制

3 4
1 2
2 1
4 1
5 2

输出样例#1： 复制

4

说明

Explanation of the sample:

|* *

| * *

+------Below is one 4x4 solution (C's show most of the corral's area); many others exist.

|CCCC

|CCCC

|*CCC*

|C*C*

+------

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
int c,n,l,r,mid;
struct nond{
    int x,y;
}v[501];
bool judge(){
    int bns=0,cns=0,dns=0,ens=0;
    for(int i=1;i<=n;i++){
        int tx1=v[i].x,dx1=v[i].x+mid-1;
        int ty1=v[i].y,dy1=v[i].y+mid-1;
        int tx2=v[i].x-mid+1,dx2=v[i].x;
        int ty2=v[i].y-mid+1,dy2=v[i].y;
        int tx3=v[i].x,dx3=v[i].x+mid-1;
        int ty3=v[i].y-mid+1,dy3=v[i].y;
        int tx4=v[i].x-mid+1,dx4=v[i].x;
        int ty4=v[i].y,dy4=v[i].y+mid-1;    
        for(int j=1;j<=n;j++){
            if(v[j].x>=tx1&&v[j].x<=dx1&&v[j].y>=ty1&&v[j].y<=dy1)
                bns++;
            if(v[j].x>=tx2&&v[j].x<=dx2&&v[j].y>=ty2&&v[j].y<=dy2)
                cns++;
            if(v[j].x>=tx3&&v[j].x<=dx3&&v[j].y>=ty3&&v[j].y<=dy3)
                dns++;
            if(v[j].x>=tx4&&v[j].x<=dx4&&v[j].y>=ty4&&v[j].y<=dy4)
                ens++;
        }
        if(bns>=c||cns>=c||dns>=c||ens>=c)    return true;
        bns=0;cns=0;dns=0;ens=0;
    }
    return false;
}
int main(){
    //freopen("testdata.in","r",stdin);
    scanf("%d%d",&c,&n);
    for(int i=1;i<=n;i++)
        scanf("%d%d",&v[i].x,&v[i].y);
    l=0;r=10010;
    while(l<=r){
        mid=(l+r)/2;
        if(judge())    r=mid-1;
        else l=mid+1;
    }
    cout<<l;
}

https://www.luogu.org/problemnew/solution/P2862