hdu 1500 Chopsticks DP
题目链接:HDU - 1500
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.
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).
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#define inf 0x7fffffff
using namespace std;
const int maxn=+; int k,n,num[maxn];
int dp[+][maxn]; int main()
{
int t;
scanf("%d",&t);
while (t--)
{
scanf("%d%d",&k,&n);
for (int i=n ;i>= ;i--) scanf("%d",&num[i]);
k += ;
memset(dp,,sizeof(dp));
for (int i= ;i<=k ;i++)
{
dp[i][*i]=dp[i-][*i-]+(num[*i-]-num[*i])*(num[*i-]-num[*i]);
for (int j=*i+ ;j<=n ;j++)
dp[i][j]=min(dp[i][j-],dp[i-][j-]+(num[j-]-num[j])*(num[j-]-num[j]));
}
printf("%d\n",dp[k][n]);
}
return ;
}
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