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题意  中文

那么就有转移方程d[i][j]=min(d[i-1][j],d[i-2][j-1]+(w[i]-w[i-1])^2)

d是初始化为无穷大的 所以i=j*2时 d[i-1][j]是等于无穷大的 所以状态转移方程可以统一为

d[i][j]=min(d[i-1][j],d[i-2][j-1]+(w[i]-w[i-1])^2)

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N = 2014;
int w[N], d[N][N / 2], n, k;

int main()
{
    while (~scanf ("%d%d", &n, &k))
    {
        for (int i = 1; i <= n; ++i)
            scanf ("%d", &w[i]);
        sort (w + 1, w + n + 1);
        memset (d, 0x7f, sizeof (d));
        for (int i = 0; i <= n; ++i)  d[i][0] = 0;
        
        for (int i = 2; i <= n; ++i)
            for (int j = 1; j * 2 <= i; ++j)
                d[i][j] = min (d[i - 1][j], d[i - 2][j - 1] + (w[i] - w[i - 1]) * (w[i] - w[i - 1]));
                
        printf ("%d\n", d[n][k]);
    }
    return 0;
}

搬寝室

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标签：动态规划   dp   hdu   acm   

原文地址：http://blog.csdn.net/iooden/article/details/38487887