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hdu1421 搬寝室

2013-01-14 18:47 246 查看
给n个数。选出k个数。使疲劳度最小

疲劳度是两手物品重量差的平方

dp[i][j]表示从i个物品选j个的最小疲劳度



所以dp[i][j] = min ( dp[i - 1][j] , dp[i - 2][j] + ( w[i] - w[i - 1] ) * ( w[i] - w[i - 1] ) )



#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;

int ct, k, i, j, w[2005], dp[2005][2005];

int main ( ) {
while ( scanf ( "%d %d", &ct, &k ) != EOF ) {
memset ( dp, 127, sizeof ( dp ) );
for ( i = 1; i <= ct; ++i )
scanf ( "%d", &w[i] );
sort ( w + 1, w + ct + 1 );
for ( i = 0; i <= ct; ++i ) dp[i][0] = 0;
for ( i = 2; i <= ct; ++i )
for ( j = 1; 2 * j <= i; ++j )
dp[i][j] = min ( dp[i - 2][j - 1] + ( w[i] - w[i - 1] ) * ( w[i] - w[i - 1] ), dp[i - 1][j] );
printf ( "%d\n", dp[ct][k] );
}
}
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