[hdu][3480][Division][四边形不等式]

题目：http://acm.hdu.edu.cn/showproblem.php?pid=3480

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <algorithm>

#define ll long long
#define clr(a,b) memset(a, b, sizeof(a))
using namespace std;

const int N = 10000+10;
const int M = 10000+10;
const int inf = 0x3f3f3f3f;

int m, n;
int a[N], dp[N/2][N], s[N/2][N];

// dp[i][j] = min{dp[i-1][k] + sqr(a[j]-a[k+1])}, 0<k<j

void input()
{
    scanf("%d%d", &n, &m);
    for (int i=1; i<=n; i++)
        scanf("%d", &a[i]);
    sort(a+1, a+1+n);
}

int main()
{
    //freopen("D:/a.txt", "r", stdin);
    int T, cas=1;
    scanf("%d", &T);
    while (T--)
    {
        input();
        for (int j=1; j<=n; j++)
            s[1][j]=0, dp[1][j]=(a[j]-a[1])*(a[j]-a[1]);
        for (int i=2; i<=m; i++)
        {
            for (int j=n; j>=i; j--)
            {
                dp[i][j] = inf;
                int st = s[i-1][j], ed = s[i][j+1];
                if (j==n) st=i-1, ed=n;
                for (int k=st; k<=ed; k++)
                {
                    int tmp = (a[j]-a[k+1])*(a[j]-a[k+1]);
                    if (dp[i][j] > dp[i-1][k] + tmp)
                        s[i][j] = k, dp[i][j] = dp[i-1][k] + tmp;
                }
            }
        }
        printf("Case %d: %d\n", cas++, dp[m][n]);
    }
    return 0;
}