2015
04-14

# Little Sheep

As we all know, sheep are very lazy, they don’t like move and just want to enjoy delicious food. So, Mr. LittleSheep who is in charge of a farm which owns lots of sheep is extremely tired in feeding these sheep. Mr. LittleSheep’s farm is like a circle consists of n corrals, each corral has a certain number of sheep, and every corral has a door for feeding. When it comes to feeding time, Mr. LittleSheep has to decide in which orders should he open those doors, because he can just open one door every time, otherwise these tricky sheep would run away. When the door is opened, the lazy sheep whose distance to the open door less or equal to k will go out, the distance to the open door is defined by the number of corrals between the open door and sheep’s corral. Due to the narrow door, it is allowed just one sheep to leave per minute. In order to let other lazy sheep get fed, Mr. LittleSheep has to move to anther corral to open the door, we assume the distance between the current corral and the destination is D, and the maximum amount of sheep among the rest corrals is M, then it takes D*M minute to move. (Because every step Mr. LittleSheep moves, it takes M minute to calm down the largest group of sheep), Now the question is how Mr. LittleSheep should move to use the least time for feeding all these lazy sheep. Mr. LittleSheep always started his work at 1st corral.

The first line is two integers n, k which indicates the number of corrals and the longest distance which sheep are willing to go.
The second line is n positive integers which describe the amount of sheep in each corral. The ith number a[i] is the number of sheep in ith corral.
(1 <= a[i] <= 10000)(1 <= n <= 2000, 1 <= k <= 500 ，k <= n/2 )

The first line is two integers n, k which indicates the number of corrals and the longest distance which sheep are willing to go.
The second line is n positive integers which describe the amount of sheep in each corral. The ith number a[i] is the number of sheep in ith corral.
(1 <= a[i] <= 10000)(1 <= n <= 2000, 1 <= k <= 500 ，k <= n/2 )

6 1
4 1 2 3 1 3

21

1. 在方法1里面：

//遍历所有的边，计算入度
for(int i=0; i<V; i++)
{
degree = 0;