[leetcode-573-Squirrel Simulation]

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There‘s a tree, a squirrel, and several nuts. Positions are represented by the cells in a 2D grid. Your goal is to find the minimal distance for the squirrel to collect all the nuts and put them under the tree one by one. The squirrel can only take at most one nut at one time and can move in four directions - up, down, left and right, to the adjacent cell. The distance is represented by the number of moves.

Example 1:

Input: 
Height : 5
Width : 7
Tree position : [2,2]
Squirrel : [4,4]
Nuts : [[3,0], [2,5]]
Output: 12
Explanation:

Note:

1. All given positions won‘t overlap.
2. The squirrel can take at most one nut at one time.
3. The given positions of nuts have no order.
4. Height and width are positive integers. 3 <= height * width <= 10,000.
5. The given positions contain at least one nut, only one tree and one squirrel.

思路：

由于松鼠每次只能走一个单位，而且每次只能带一个坚果到树下，那么除了第一个要捡的坚果外，从树出发，捡其他每一个

坚果的距离都要乘以2（从树到坚果i的距离 +从坚果i到树的距离）。

不妨认为捡每一个坚果都是从树出发，那么最终的距离：

dis = 2*(所有getDis(tree,nut[i])的和) - getDis(tree[i],nut[i])) + getDis(squireel,nut[i])

 int getDis(vector<int>& a,vector<int>& b)
 {
     return abs(a[0] - b[0]) + abs(a[1] - b[1]);
 }
int minDistance(int height, int width, vector<int>& tree, vector<int>& squirrel, vector<vector<int> >& nuts)
{
    //dis = 2*(所有getDis(tree,nut[i])) - getDis(tree,nut[i])) + getDis(squireel,nut[i])
    int tempdis = 0;
     int distance = 1000000;
    for(int i=0;i<nuts.size();i++)
    {
         tempdis += 2*getDis(tree,nuts[i]);
    }
    for(int i=0;i<nuts.size();i++)
    {
         int distanceTemp = tempdis - getDis(tree,nuts[i]) + getDis(squirrel,nuts[i]);
         if(distance >distanceTemp) distance = distanceTemp;
    }
    return distance;
}

[leetcode-573-Squirrel Simulation]

标签：output   tor   out   span   imu   其他   ima   res   stat   

原文地址：http://www.cnblogs.com/hellowooorld/p/6820396.html