375. Guess Number Higher or Lower II

We are playing the Guess Game. The game is as follows:

I pick a number from 1 to n. You have to guess which number I picked.

Every time you guess wrong, I'll tell you whether the number I picked is higher or lower.

However, when you guess a particular number x, and you guess wrong, you pay $x. You win the game when you guess the number I picked.

Example:

n = 10, I pick 8.

First round:  You guess 5, I tell you that it's higher. You pay $5.
Second round: You guess 7, I tell you that it's higher. You pay $7.
Third round:  You guess 9, I tell you that it's lower. You pay $9.

Game over. 8 is the number I picked.

You end up paying $5 + $7 + $9 = $21.

Given a particular n ≥ 1, find out how much money you need to have to guarantee a win.

题意分析：

首先定义dp[i][j]，代表如果我们在区间[i,j]内进行查找，所需要最少cost来保证找到结果。我们要求的就是dp[1][n];

给定范围1--n,你想要获取win，最少要花多少money才能保证win.

对于给定区间[i,j],因为目标数字是i~j中的任何一个都有可能。对于目标数字每取一个数，我们猜测k(i<=k<=j)我们可能出现以下三种结果：

1. k 就是答案，此时子问题的额外 cost = 0 ，当前位置总 cost  = k + 0;
2. k 过大，此时我们的有效区间缩小为 [i , k - 1] 当前操作总 cost  = k + dp[start][k - 1];
3. k 过小，此时我们的有效区间缩小为 [k + 1 , j] 当前操作总 cost  = k + dp[k + 1][j];

由于我们需要 “保证得到结果”，也就是说对于指定 k 的选择，我们需要准备最坏情况 cost 是上述三种结果生成的 subproblem 中cost 最大的那个；

我们只需要求得所有的结果下的最小的一个满足题意的条件就可以保证能够使用这个money赢下一种可能性的目标。

class Solution {
public:
   
    int getMinCost(int s,int e,vector>& data)
    {
         if(s>=e)return 0;
         if(data[s][e]!=0)return data[s][e];
         int minCost = INT_MAX;
         for(int i=s;i<=e;i++)
         {
            minCost = min(minCost,i+max(getMinCost(s,i-1,data),getMinCost(i+1,e,data)));
         }
         data[s][e]=minCost;
         return data[s][e];

    }
    int getMoneyAmount(int n) {
        vector> data(n+1,vector(n+1,0));
        return getMinCost(1,n,data);

    }
};

思路参考 http://www.1point3acres.com/bbs/thread-197552-1-1.html