LintCode 476 - Stone Game

https://zhengyang2015.gitbooks.io/lintcode/stone_game_476.html

There is a stone game.At the beginning of the game the player picks n piles of stones in a line.

The goal is to merge the stones in one pile observing the following rules:

At each step of the game,the player can merge two adjacent piles to a new pile.

The score is the number of stones in the new pile.

You are to determine the minimum of the total score.

Example

For [4, 1, 1, 4], in the best solution, the total score is 18:

1. Merge second and third piles => [4, 2, 4], score +2
2. Merge the first two piles => [6, 4]，score +6
3. Merge the last two piles => [10], score +10

Other two examples:

[1, 1, 1, 1] return 8

[4, 4, 5, 9] return 43

这道题既能用DP解，也能用记忆化搜索的方法解。

+

dp[i][j]表示合并i到j的石头需要的最小代价。

转移函数：

dp[i][j]=dp[i][k]+dp[k+1][j]+sum[i][j] （i<=k<j）。即合并i－j的代价为合并左边部分的代价＋合并右边部分的代价＋合并左右部分的代价（即i－j所有元素的总和）。找到使dp[i][j]最小的k。

需要初始化sum。DP以长度和不同起点为循环条件，而记忆化搜索需要start和end来确定搜索范围，然后找分割点k，再递归搜索左右部分，有点D&C的味道。

    public int stoneGame(int[] A) {
        // DP
        if(A == null || A.length == 0){
            return 0;
        }

        int n = A.length;
        int[][] sum = new int[n][n];
        for(int i = 0; i < n; i++){
            sum[i][i] = A[i];
            for(int j = i + 1; j < n; j++){
                sum[i][j] = sum[i][j - 1] + A[j];
            }
        }

        int[][] dp = new int[n][n];
        for(int i = 0; i < n; i++){
            dp[i][i] = 0;
        }

        for(int len = 2; len <= n; len++){
            for(int i = 0; i + len - 1 < n; i++){
                int j = i + len - 1;
                int min = Integer.MAX_VALUE;
                for(int k = i; k < j; k++){
                    min = Math.min(min, dp[i][k] + dp[k + 1][j]);
                }
                dp[i][j] = min + sum[i][j];
            }
        }

        return dp[0][n - 1];
    }

Memorized Search
这题可以用prefixSum数组来优化sum[i][j]数组，把空间复杂度从O(n ^ 2)降到O(n)

    public int stoneGame(int[] A) {
        // Memorized search
        if(A == null || A.length == 0){
            return 0;
        }

        int n = A.length;
        int[][] dp = new int[n][n];
        int[] sum = new int[n + 1];

        for(int i = 0; i < n - 1; i++){
            for(int j = i + 1; j < n; j++){
                dp[i][j] = -1;
            }
        }

        sum[0] = 0;
        for(int i = 0; i < n; i++){
            dp[i][i] = 0;
            sum[i + 1] = sum[i] + A[i];
        }

        return search(0, n - 1, sum, dp);
    }

    private int search(int start, int end, int[] sum, int[][] dp){
        if(dp[start][end] >= 0){
            return dp[start][end];
        }

        int min = Integer.MAX_VALUE;
        for(int k = start; k < end; k++){
            int left = search(start, k, sum, dp);
            int right = search(k + 1, end, sum, dp);
            int now = sum[end + 1] - sum[start];
            min = Math.min(min, left + right + now);
        }
        dp[start][end] = min;
        return dp[start][end];
    }

https://www.jiuzhang.com/qa/806/

由于这两部的分析，那么我们的其他状态f[i][j]就清晰了是dp[i][j] = min(dp[i][k]+dp[k+1][j]+sum[i,j]。

有了上面的状态分析，下面我们直接给出这题动态规划的四要素：

State: dp[i][j] 表示把第i到第j个石子合并到一起的最小花费

Function: 

         预处理sum[i,j]

         dp[i][j] = min(dp[i][k]+dp[k+1][j]+sum[i,j]) 对于所有k属于{i,j}

Intialize: 

         dp[i][i] = 0 for each i.  

Answer:

dp[0][n]

http://janexiehappy.blogspot.com/2016/04/stone-game-lintcode-dp.html

    public int stoneGame(int[] A) {

        // Write your code here

        int m=A.length;

        if(m==0)return 0;

        if(m==1)return 0;

        int[][] sum = new int[m][m];

        int[][] dp = new int[m][m];

        

        for(int i=0;i<m;++i){

            sum[i][i]=A[i];

            for(int j=i+1;j<m;++j){

                sum[i][j]=sum[i][j-1]+A[j];

            }

        }

        

        for(int j=1;j<m;++j){

            //dp edge case

            dp[j][j]=0;

            

            for(int i=j-1;i>=0;--i){

                

                int min = Integer.MAX_VALUE;

                //dp common case func

                for(int k=i;k<=j-1;++k){

                    int tmp = dp[i][k] + dp[k+1][j];

                    if(tmp < min){

                        min = tmp;

                    }

                }

                dp[i][j] = min + sum[i][j];

            }

        }

        return dp[0][m-1];

    }

http://www.cnblogs.com/lz87/p/6951217.html

here is a stone game.At the beginning of the game the player picks n piles of stones in a circle.

The goal is to merge the stones in one pile observing the following rules:

At each step of the game,the player can merge two adjacent piles to a new pile.
The score is the number of stones in the new pile.
You are to determine the minimum of the total score.

Example

For [1, 4, 4, 1], in the best solution, the total score is 18:

1. Merge second and third piles => [2, 4, 4], score +2
2. Merge the first two piles => [6, 4]，score +6
3. Merge the last two piles => [10], score +10

Other two examples:
[1, 1, 1, 1] return 8
[4, 4, 5, 9] return 43

https://zyfu0408.gitbooks.io/interview-preparation/content/%E5%8C%BA%E9%97%B4%E7%B1%BBdp.html
求一段区间的解 min / max / count
相比划分类DP，区间类DP为连续相连的subproblem，中间不留空，更有divide & conquer的味道。
转移方程通过区间更新
从大到小的更新
http://lixinzhang.github.io/qu-jian-xing-dp.html
区间形DP特征：

F[i][j]=F[i][k]+F[k+1][j]+CostFunction(i,j)

设状态为F[i][j]$F[i][j]$，表示第i堆到第j堆石子合并之后的最小分值，那么其上一状态一定是由两个子堆合并而来，那么枚举中间分割位置k$k$为决策状态,因此状态转移方程：

F[i][j]=F[i][k]+F[k+1][j]+sum(i,j)$$F[i][j]=F[i][k]+F[k+1][j]+sum(i,j)$$

其中sum(i,j)$sum(i,j)$为两个自堆合并时，所产生的分值。

此类DP，先计算小区间，然后再通过小区间迭代得到大区间的值。

说有n个节点n条边组成一个圈，每个节点上面有一个数，边上有一个+或*，如果消掉某条边，其相邻两个节点就用这个运算符合并。这样一路消边到底，问用什么过程能让最后得到的数最大。