HDU 1500 Chopsticks(动态规划)
来源:互联网 发布:头戴式蓝牙推荐 知乎 编辑:程序博客网 时间:2024/05/19 14:15
Chopsticks
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 1885 Accepted Submission(s): 882
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.
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
终于是自己独自想出的一道DP题了,心里激动万分!
分析:一看给出的数据那么大,1000 和5000,用搜索很显然不实际,所以想到了DP。人数从i =1~ k+8,先找到1个人3只筷子时,
最小困难度,再逐渐找到1个人n只筷子的最小困难度,然后再找2个人6只筷子的最小困难度...
状态转移方程为:
dp[ i ][ j ]=min{ dp[ i ][ j-1 ],dp[ i-1 ][ j-2 ] + (a[ j ]-a[ j-1 ])*(a[ j ]-a[ j-1 ]) }
代码如下:
#include <stdio.h>int dp[1010][5005];int main(){int i,j;int a[5005];int T,n,k;scanf("%d",&T);while(T--){scanf("%d %d",&k,&n);for(i=n;i>=1;i--)scanf("%d",a+i);for(i=0;i<=k+8;i++){for(j=0;j<=n;j++)dp[i][j]=32005;}for(j=0;j<=n;j++)dp[0][j]=0;for(i=1;i<=k+8;i++){for(j=3*i;j<=n;j++){if(dp[i][j-1]>dp[i-1][j-2]+(a[j]-a[j-1])*(a[j]-a[j-1]))dp[i][j]=dp[i-1][j-2]+(a[j]-a[j-1])*(a[j]-a[j-1]);elsedp[i][j]=dp[i][j-1];}}printf("%d\n",dp[k+8][n]);}return 0;}
0 0
- HDU 1500 Chopsticks(动态规划)
- 动态规划 hdu 1500 Chopsticks
- hdu 1500 Chopsticks(动态规划)
- hdu 1500 Chopsticks 动态规划 比较经典
- ZJU1234 Chopsticks - 动态规划
- [动态规划]Zju1234--Chopsticks
- !HDU 1500 Chopsticks-dp-(分组问题)
- HDU 1500 Chopsticks
- hdu 1500 chopsticks
- HDU-1500 Chopsticks DP
- HDU-OJ-1500 Chopsticks
- HDU 1500 Chopsticks(DP)
- hdu 1500 Chopsticks (dp)
- hdu Chopsticks
- HDU 1421 (动态规划)
- hdu 2084(动态规划)
- hdu 4001 (动态规划)
- HDU 1506(动态规划)
- 学会使用WorkSpace
- 文件操作
- 项目从mysql移植到SQL sqlserver问题汇总
- jquery图片轮播,三种方式
- Xcode 报错: value of type InerfaceController does not conform to HKWorkoutSessionDelegate in assignmen
- HDU 1500 Chopsticks(动态规划)
- 基础篇——数论基础
- POJ 2251Dungeon Master
- Web Uploader图片上传 java实现
- VC6.0工程设置
- android监听应用缓存和任务栈顶的方法
- java设计模式:观察者模式实例详解(基于java内置观察者)
- 对单一职责原则的理解
- Method Swizzling(iOS的hook机制)