UVA 10271 Chopsticks

題目鏈接：http://acm.hust.edu.cn/vjudge/problem/19452

題意：將n只筷子分成k份，每份中除去最長的，剩下兩個長度相減的平方作爲一份的值，求k份筷子的值之和最小。

思路：dp[i][j]表示前i只筷子分爲j份和的最小值。我們將筷子從大到小排序。dp[i][j] = min( dp[i-2][j-1] + (a[i-1] - a[i])^2  ) ( 當j*3<= i 時)因爲一定可以分成，那麼第j組的長筷子就可以從1~i-2裏面找沒有用過的。

#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <utility>
using namespace std;

#define rep(i,j,k) for (int i=j;i<=k;i++)
#define Rrep(i,j,k) for (int i=j;i>=k;i--)

#define Clean(x,y) memset(x,y,sizeof(x))
#define LL long long
#define ULL unsigned long long
#define inf 0x7fffffff
#define mod 100000007
const int maxn = 5009;
int T;
int n,k;

int a[maxn];
int dp[maxn][3100];

bool cmp(int x,int y)
{
    return x > y;
}

void init()
{
    cin>>k>>n;
    k = k + 8;
    rep(i,1,n) scanf("%d",&a[i]);
    sort(a+1,a+1+n,cmp);
    Clean(dp,0x3f);
    rep(i,0,n) dp[i][0] = 0;
}

int cal( int x , int y )
{
    return (x-y)*(x-y);
}

void solve()
{
    rep(i,3,n)
    {
        int uplim = min( k , i / 3 );
        rep(j,1,uplim)
        {
            dp[i][j] = min( dp[i-1][j] , dp[i][j] );
            if( j * 3 <= i )
                dp[i][j] = min( dp[i][j] , dp[i-2][j-1] + cal( a[i-1] , a[i] ) );
        }
    }
    cout<<dp[n][k]<<endl;
}

int main()
{
    cin>>T;
    while(T--)
    {
        init();
        solve();
    }
    return 0;
}