dp...
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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 ) ) << "\n";
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
[Submit][Status][Discuss]
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.