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：

3 4
1 2
2 1
4 1
5 2

输出样例#1：

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<iostream>
#include<cstdio>
using namespace std;
#define maxn 4010
int map[maxn][maxn],c,n,sum[maxn][maxn],N,M;
int main(){
    scanf("%d%d",&c,&n);
    int x,y;
    for(int i=;i<=n;i++){
        scanf("%d%d",&x,&y);
        map[x][y]++;
        N=max(N,x);
        M=max(M,y);
    }
    int range=max(N,M);
    for(int i=;i<=range;i++)
        for(int j=;j<=range;j++)
            sum[i][j]=sum[i-][j]+sum[i][j-]+map[i][j]-sum[i-][j-];
    for(int i=;i<=range;i++)//正方形的边长
        for(int j=i;j<=range;j++)
            for(int k=i;k<=range;k++)
                if(sum[j][k]-sum[j-i][k]-sum[j][k-i]+sum[j-i][k-i]>=c){
                    printf("%d",i);
                    return ;
                }
}

30分 只用前缀和维护了一下