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走两次的方格取数问题 - 基础入门+刷题小组 - 美国米群网 - 海外华人留学生最大面试求职社交平台 - 计算机 商科 面试 面经 内推 刷题 算法 职场 绿卡 家庭 休闲 - Powered by MeetQun!
问题 给定n行n列的方阵,每个元素是一个整数,从左上角走到右下角,每次只能向右或者下,走两次,使得得到的总和最大。同一次经过两个元素的话,总和只计算一次。

分析: 如果是只走一次的话,是经典的dp。走两次的话,dp两次是不对的……因为状态之间有影响,第一次和最大的路径可能会影响第二次的路径。这个题还有一个变形,是等价的,就是说从起点到终点只能向右和下,再从终点回起点只能向上或者左,使得总和最大。
           搜索是指数的算法,贪心不靠谱,两次dp也不行。所以还是尝试两个人同时走……,有个好处是,两个人一起走的话,只能在同一步经过相同的格子,这是由距离决定的。因为第step步,所在的格子距离起点的曼哈顿距离|x - x' | + |y - y'| = step……所以走一次时,状态是dp[(x,y)]由(x - 1, y) (x, y - 1)决定,现在状态是dp[(x1,y1)(x2,y2)]由4个状态确定即dp[(x1 - 1, y1)(x2 - 1, y2)] dp[(x1 - 1, x2, y2 - 1)] dp[(x1, y1 - 1)(x2 - 1, y2)] dp[(x1, y1 - 1) (x2 , y2 - 1)]确定。

以上代码的复杂度空间复杂度是O(n^3),稍微想一下可以知道我们在推算dp[step]的时候 只有到了dp[step - 1],所以第一维,我们只开到2就可以了。就是step为奇数时,我们都用dp[1][j]表示状态,step为偶数我们都用dp[0][j]表示状态,这样我们只需要O(n^2)的空间,这就是滚动数组的方法。滚动数组写起来并不复杂,只需要对上面的代码稍作修改即可


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