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UVA 10271 Chopsticks（線性DP)

編輯：關於C++

In China, people use a pair of chopsticks to get food on the table, but Mr. L is a bit different. He uses
a set of three chopsticks – one pair, plus an EXTRA long chopstick to get some big food by piercing
it through the food. As you may guess, the length of the two shorter chopsticks should be as close as
possible, but the length of the extra one is not important, as long as it’s the longest. To make things
clearer, for the set of chopsticks with lengths A, B, C (A ≤ B ≤ C), (A ? B)
2
is called the “badness”
of the set.
It’s December 2nd, Mr.L’s birthday! He invited K people to join his birthday party, and would like
to introduce his way of using chopsticks. So, he should prepare K + 8 sets of chopsticks(for himself,
his wife, his little son, little daughter, his mother, father, mother-in-law, father-in-law, and K other
guests). But Mr.L suddenly discovered that his chopsticks are of quite different lengths! He should find
a way of composing the K + 8 sets, so that the total badness of all the sets is minimized.
Input
The first line in the input contains a single integer T, indicating the number of test cases (1 ≤ T ≤ 20).
Each test case begins with two integers K, N (0 ≤ K ≤ 1000, 3K + 24 ≤ N ≤ 5000), the number of
guests and the number of chopsticks.
There are N positive integers Li on the next line in non–decreasing order indicating the lengths of
the chopsticks (1 ≤ Li ≤ 32000).
Output
For each test case in the input, print a line containing the minimal total badness of all the sets.
Note: For the sample input, a possible collection of the 9 sets is:
8,10,16; 19,22,27; 61,63,75; 71,72,88; 81,81,84; 96,98,103; 128,129,148; 134,134,139; 157,157,160
Sample Input
1
1 40
1 8 10 16 19 22 27 33 36 40 47 52 56 61 63 71 72 75 81 81 84 88 96 98
103 110 113 118 124 128 129 134 134 139 148 157 157 160 162 164
Sample Output

要保證有一根長筷子，我們可以從後到前DP：f[i][j]:前i根配了j對的最小代價。

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define CLEAR( a , x ) memset ( a , x , sizeof a )
typedef long long LL;
typedef pairpil;
const int INF = 0x3f3f3f3f;
int t,n,k;
int a[5500];
LL dp[5500][1100];
int main()
{
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&k,&n);
        k+=8;CLEAR(dp,INF);
        for(int i=n;i>=1;i--)
            scanf("%d",&a[i]);
        dp[0][0]=0;
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=k;j++)
            {
                dp[i][j]=dp[i-1][j];
                if(i-j*3>=0)
                    dp[i][j]=min(dp[i][j],dp[i-2][j-1]+(a[i]-a[i-1])*(a[i]-a[i-1]));
            }
        }
        printf("%lld\n",dp[n][k]);
    }
    return 0;
}

NJUPT 1581

題目描述

A 先生有很多雙筷子。確切的說應該是很多根，因為筷子的長度不一，很難判斷出哪兩根是一雙的。這天，A 先生家裡來了K 個客人，A 先生留下他們吃晚飯。加上A 先生，A夫人和他們的孩子小A，共K+3個人。每人需要用一雙筷子。A 先生只好清理了一下筷子，共N 根，長度為T1，T2，T3，……，TN。現在他想用這些筷子組合成K+3 雙，使每雙的筷子長度差的平方和最小。（怎麼不是和最小？？這要去問A 先生了，呵呵）

輸入

共有兩行，第一行為兩個用空格隔開的整數，表示N，K（1≤N≤100，0

輸出

僅一行。如果湊不齊K+3雙，輸出-1，否則輸出長度差平方和的最小值。

樣例輸入

10 1
1 1 2 3 3 3 4 6 10 20

樣例輸出

提示

第一雙 1 1
第二雙 2 3
第三雙 3 3
第四雙 4 6
(1-1)^2+(2-3)^2+(3-3)^2+(4-6)^2=5

弱化版，較簡單

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define CLEAR( a , x ) memset ( a , x , sizeof a )
typedef long long LL;
typedef pairpil;
const int INF = 0x3f3f3f3f;
int dp[110][55];
int L[110];
int n,m;

int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        REPF(i,1,n) scanf("%d",&L[i]);
        sort(L+1,L+n+1);
        m+=3;
        CLEAR(dp,INF);
        dp[0][0]=0;
        for(int i=2;i<=n;i++)
        {
            for(int j=1;j<=m;j++)
                dp[i][j]=min(dp[i-1][j],dp[i-2][j-1]+(L[i-1]-L[i])*(L[i-1]-L[i]));
        }
        if(dp[n][m]!=INF) printf("%d\n",dp[n][m]);
        else puts("-1");
    }
    return 0;
}


/*

*/