[hdu][3480][Division][四边形不等式]

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

View Code

#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;

}