hdu 1500 Chopsticks 动态规划 比较经典

Chopsticks

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1808    Accepted Submission(s): 836

Problem Description

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.

 

Sample Input

11 401 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

23NoteFor 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

 

先变成非递增序列，因为要保证有一个最大的C在每一组里面。

转移方程： dp[i][j] = Min(dp[i][j-1] , dp[i-1][j-2] + (set[j-1 - set[j] )* (set[j-1] -set[j]))

第i个人 第j个筷子的时候的最小值 

代码：

#include <cstdio>#include <cstring>#include <algorithm>#define INF 1000000000using namespace std ;int dp[1010][5100] ;int min(int a , int b){return a>b?b:a ;}bool cmp(const int a , const int b){return a>b;}int main(){int c[5100];int t;scanf("%d",&t) ;while(t--){int n , k ;scanf("%d%d",&k,&n) ;for(int i = 1 ; i <= n ; ++i){scanf("%d",&c[i]) ;}sort(c+1,c+n+1,cmp);memset(dp,0,sizeof(dp)) ;for(int i = 1 ; i <= k+8 ; ++i){int t = 3*i ;dp[i][t] = dp[i-1][t-2]+(c[t-1]-c[t])*(c[t-1]-c[t])  ; //保证前面i个人至少拿到筷子了 for(int j = t+1 ; j <= n ; ++j){dp[i][j] = min(dp[i][j-1],dp[i-1][j-2]+(c[j-1]-c[j])*(c[j-1]-c[j]))  ;}}printf("%d\n",dp[k+8][n]) ;}return 0 ;}

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