Copy Books @Lintcode

Given an array A of integer with size of n( means n books and number of pages of each book) and k people to copy the book. You must distribute the continuous id books to one people to copy. (You can give book A[1],A[2] to one people, but you cannot give book A[1], A[3] to one people, because book A[1] and A[3] is not continuous.) Each person have can copy one page per minute. Return the number of smallest minutes need to copy all the books.

Example

Given array A = [3,2,4], k = 2.

Return 5( First person spends 5 minutes to copy book 1 and book 2 and second person spends 4 minutes to copy book 3. )

很明显地用dp来做。dp[i][j] 代表前i本书由j个工作者完成需要的最小时间。状态转移方程如下：

dp[i+1][j+1] = Min(Max(dp[i][k], A[k]+...+A[j+1])),  where k from i to j+1

意思就是每加入一个新worker的时候，从后往前累加他需要copy的book，每次更新总体所需要的时间；而总体需要的时间是之前所有worker分配方案中的最优时间和新worker所需时间两者中的较大值。

时间复杂度 O(N^2*k), 暂时没想到 O(N*k) 的方法 :( 麻烦各路大神指点。。。

	public int copyBooks(int[] pages, int k) {
	// write your code here
	int[][] dp = new int[k+1][pages.length+1];
	int sum = 0;
	int max = 0;
	for(int i=1; i<=pages.length; i++){
	sum += pages[i-1];
	dp[1][i] = sum;
	max = Math.max(max, pages[i-1]);
	}
	if(k>=pages.length){
	return max;
	}
	for(int i=2; i<=k; i++){
	for(int j=pages.length-1; j>=i-1; j--){
	int current = pages[j];
	int min = Math.max(pages[j], dp[i-1][j]);
	dp[i][j+1] = min;
	for(int l=j-1; l>=i-1; l--){
	current += pages[l];
	int curMin = Math.max(current, dp[i-1][l]);
	if(curMin<min){
	dp[i][j+1] = curMin;
	min = curMin;
	}
	}
	}
	}
	return dp[k][pages.length];
	}

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