BZOJ 1742: [Usaco2005 nov]Grazing on the Run 边跑边吃草( dp )

dp...

dp( l , r , k )  , 表示 吃了[ l , r ] 的草 , k = 1 表示最后在 r 处 , k = 0 表示最后在 l 处 .

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#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 ) )

 

using namespace std;

 

const int maxn = 1000 + 5;

const int inf = 0x3f3f3f3f;

 

int d[ maxn * maxn >> 1 ][ 2 ];

int h[ maxn ];

int n;

 

#define f( l , r ) ( ( ( ( ( n << 1 ) - l ) * l ) >> 1 ) + r )

 

// k 0 L , k 1 R

int dp( int l , int r , int k ) {

int &ans = d[ f( l , r ) ][ k ];

if( ans != inf ) return ans;

if( k ) {

ans = min( dp( l , r - 1 , 1 ) + ( h[ r ] - h[ r - 1 ] ) * ( n - r + l ) , dp( l , r - 1 , 0 ) + ( h[ r ] - h[ l ] ) * ( n - r + l ) );

} else {

ans = min( dp( l + 1 , r , 1 ) + ( h[ r ] - h[ l ] ) * ( n - r + l ) , dp( l + 1 , r , 0 ) + ( h[ l + 1 ] - h[ l ] ) * ( n - r + l ) );

}

return ans;

}

 

int main() {

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

clr( d , inf );

int p;

cin >> n >> p;

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

sort( h , h + n );

rep( i , n ) {

int t = f( i , i );

d[ t ][ 0 ] = d[ t ][ 1 ] = abs( p - h[ i ] ) * n;

}

cout << min( dp( 0 , n - 1 , 0 ) , dp( 0 , n - 1 , 1 ) ) << " ";

return 0;

}

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1742: [Usaco2005 nov]Grazing on the Run 边跑边吃草

Time Limit: 5 Sec  Memory Limit: 64 MB
Submit: 191  Solved: 102
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Description

A long, linear field has N (1 <= N <= 1,000) clumps of grass at unique integer locations on what will be treated as a number line. Think of the clumps as points on the number line. Bessie starts at some specified integer location L on the number line (1 <= L <= 1,000,000) and traverses the number line in the two possible directions (sometimes reversing her direction) in order to reach and eat all the clumps. She moves at a constant speed (one unit of distance in one unit of time), and eats a clump instantly when she encounters it. Clumps that aren't eaten for a while get stale. We say the ``staleness'' of a clump is the amount of time that elapses from when Bessie starts moving until she eats a clump. Bessie wants to minimize the total staleness of all the clumps she eats. Find the minimum total staleness that Bessie can achieve while eating all the clumps.

John养了一只叫Joseph的奶牛。一次她去放牛，来到一个非常长的一片地，上面有N块地方长了茂盛的草。我们可以认为草地是一个数轴上的一些点。Joseph看到这些草非常兴奋，它想把它们全部吃光。于是它开始左右行走，吃草。John和Joseph开始的时候站在p位置。Joseph的移动速度是一个单位时间一个单位距离。不幸的是，草如果长时间不吃，就会腐败。我们定义一堆草的腐败值是从Joseph开始吃草到吃到这堆草的总时间。Joseph可不想吃太腐败的草，它请John帮它安排一个路线，使得它吃完所有的草后，总腐败值最小。John的数学很烂，她不知道该怎样做，你能帮她么？ 

Input

输入(ontherun.in) 
输入文件第一行有两个整数，N(N<=3000)和p，分别代表草的堆数和起始位置。下面N行，每行一个整数ai，代表N堆草的位置。(1 <= ai , p <= 1000000) 

Output

输出一个整数，最小总腐败值。结果保证在2^31-1内。 

Sample Input

Hint

提示 
0时刻，在位置10。 
移动到9，1时刻到达，吃草。 
移动到11，3时刻到达，吃草。 
移动到19，11时刻到达，吃草。 
移动到1，29时刻到达，吃草。 
总腐败值1 + 3 + 11 + 29 = 44最优。 
数据规模 
对于30%的数据，N <= 10。 
对于60%的数据，N <= 300。 
对于100%的数据，N <= 3000。 

Input

* Line 1 : Two space-separated integers: N and L. * Lines 2..N+1: Each line contains a single integer giving the position P of a clump (1 <= P <= 1,000,000).

Output

* Line 1: A single integer: the minimum total staleness Bessie can achieve while eating all the clumps.

Sample Input

4 10
1
9
11
19

INPUT DETAILS:

Four clumps: at 1, 9, 11, and 19. Bessie starts at location 10.

Sample Output

44

OUTPUT DETAILS:

Bessie can follow this route:

* start at position 10 at time 0
* move to position 9, arriving at time 1
* move to position 11, arriving at time 3
* move to position 19, arriving at time 11
* move to position 1, arriving at time 29

giving her a total staleness of 1+3+11+29 = 44. There are other routes
with the same total staleness, but no route with a smaller one.

HINT

Source

Gold