首页 > ACM题库 > HDU-杭电 > HDU 1500 Chopsticks-动态规划-[解题报告] C++
2013
12-11

HDU 1500 Chopsticks-动态规划-[解题报告] C++

Chopsticks

问题描述 :

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.

输入:

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).

输出:

For each test case in the input, print a line containing the minimal total badness of all the sets.

样例输入:

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

样例输出:

23

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

一道不算是很难的DP,一些人的爱好还真是奇怪的说!

嘿嘿

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 

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#define NN 5010
#define Min(a,b) (a>b?b:a)
int set[NN];
int dp[1010][NN];
int main()
{
	int t,k,n,i,j;
	scanf("%d",&t);
	while(t--){
		scanf("%d%d",&k,&n);
		k += 8;
		for(i=n;i>=1;i--)
			scanf("%d",&set[i]);
		for(i=1;i<=k;i++){
			dp[i][i*3] = dp[i-1][i*3-2] + (set[i*3-1] - set[i*3])*(set[i*3-1] - set[i*3]);
			for(j=i*3+1;j<=n;j++)
				dp[i][j] = Min(dp[i][j-1],dp[i-1][j-2] +(set[j-1] -set[j])*(set[j-1] - set[j]));
		}
		printf("%d\n",dp[k][n]);
	}
	return 0;
}

解题报告转自:http://blog.csdn.net/surfacedust/article/details/6685681


  1. 算法是程序的灵魂,算法分简单和复杂,如果不搞大数据类,程序员了解一下简单点的算法也是可以的,但是会算法的一定要会编程才行,程序员不一定要会算法,利于自己项目需要的可以简单了解。