BZOJ 1734: [Usaco2005 feb]Aggressive cows 愤怒的牛( 二分答案 )

最新推荐文章于 2020-01-13 11:42:49 发布

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最新推荐文章于 2020-01-13 11:42:49 发布

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原文链接：http://www.cnblogs.com/JSZX11556/p/4592459.html

本文探讨了在农业布局中如何通过最大最小距离算法优化牛的分配，确保每头牛之间的最小距离最大化，从而避免冲突。通过案例分析和代码实现，详细解释了算法原理和具体步骤。

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最小最大...又是经典的二分答案做法..

--------------------------------------------------------------------------

#include<cstdio>

#include<cstring>

#include<algorithm>

#include<iostream>

#define rep( i , n ) for( int i = 0 ; i < n ; ++i )

#define clr( x , c ) memset( x , c , sizeof( x ) )

#define Rep( i , n ) for( int i = 1 ; i <= n ; ++i )

using namespace std;

const int maxn = 100000 + 5;

int pos[ maxn ];

int n , c;

bool jud( int x ) {

int cnt = c - 1 , p = pos[ 0 ];

Rep( i , n - 1 ) if( pos[ i ] - p >= x )

--cnt , p = pos[ i ];

return cnt > 0 ? false : true;

}

int main() {

freopen( "test.in" , "r" , stdin );

cin >> n >> c;

rep( i , n ) scanf( "%d" , pos + i );

sort( pos , pos + n );

int L = 1 , R = pos[ n - 1 ] , ans;

while( L <= R ) {

int m = ( L + R ) >> 1;

if( jud( m ) )

ans = m , L = m + 1;

else

R = m - 1;

}

cout << ans << "\n";

return 0;

}

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1734: [Usaco2005 feb]Aggressive cows 愤怒的牛

Time Limit: 5 Sec Memory Limit: 64 MB
Submit: 221 Solved: 179
[ Submit][ Status][ Discuss]

Description

Farmer John has built a new long barn, with N (2 <= N <= 100,000) stalls. The stalls are located along a straight line at positions x1,...,xN (0 <= xi <= 1,000,000,000). His C (2 <= C <= N) cows don't like this barn layout and become aggressive towards each other once put into a stall. To prevent the cows from hurting each other, FJ want to assign the cows to the stalls, such that the minimum distance between any two of them is as large as possible. What is the largest minimum distance?

农夫 John 建造了一座很长的畜栏，它包括NN (2 <= N <= 100,000)个隔间，这些小隔间依次编号为x1,...,xN (0 <= xi <= 1,000,000,000). 但是，John的C (2 <= C <= N)头牛们并不喜欢这种布局，而且几头牛放在一个隔间里，他们就要发生争斗。为了不让牛互相伤害。John决定自己给牛分配隔间，使任意两头牛之间的最小距离尽可能的大，那么，这个最大的最小距离是什么呢

Input

* Line 1: Two space-separated integers: N and C * Lines 2..N+1: Line i+1 contains an integer stall location, xi

第一行：空格分隔的两个整数N和C

第二行---第N+1行：i+1行指出了xi的位置

Output

* Line 1: One integer: the largest minimum distance

第一行：一个整数，最大的最小值

Sample Input

5 3
1
2
8
4
9

Sample Output

3

把牛放在1，4，8这样最小距离是3

HINT

Source

Gold

转载于:https://www.cnblogs.com/JSZX11556/p/4592459.html