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
3 4
1 2
2 1
4 1
5 2
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
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