题目描述：

LeetCode 799. Champagne Tower

We stack glasses in a pyramid, where the first row has 1 glass, the second row has 2 glasses, and so on until the 100th row.  Each glass holds one cup (250ml) of champagne.

Then, some champagne is poured in the first glass at the top.  When the top most glass is full, any excess liquid poured will fall equally to the glass immediately to the left and right of it.  When those glasses become full, any excess champagne will fall equally to the left and right of those glasses, and so on.  (A glass at the bottom row has it's excess champagne fall on the floor.)

For example, after one cup of champagne is poured, the top most glass is full.  After two cups of champagne are poured, the two glasses on the second row are half full.  After three cups of champagne are poured, those two cups become full - there are 3 full glasses total now.  After four cups of champagne are poured, the third row has the middle glass half full, and the two outside glasses are a quarter full, as pictured below.

Now after pouring some non-negative integer cups of champagne, return how full the j-th glass in the i-th row is (both i and j are 0 indexed.)

Example 1:
Input: poured = 1, query_glass = 1, query_row = 1
Output: 0.0
Explanation: We poured 1 cup of champange to the top glass of the tower (which is indexed as (0, 0)). There will be no excess liquid so all the glasses under the top glass will remain empty.

Example 2:
Input: poured = 2, query_glass = 1, query_row = 1
Output: 0.5
Explanation: We poured 2 cups of champange to the top glass of the tower (which is indexed as (0, 0)). There is one cup of excess liquid. The glass indexed as (1, 0) and the glass indexed as (1, 1) will share the excess liquid equally, and each will get half cup of champange.

Note:

• poured will be in the range of [0, 10 ^ 9].
• query_glass and query_row will be in the range of [0, 99].

题目大意：

用等差数列的杯子堆叠成香槟塔，从第一层的杯子倒入香槟，当杯子倒满时，多余的香槟会等量地流入下方的左右两个杯子中，以此类推。

求倒入poured杯香槟后，第query_row行的第query_glass杯香槟的体积。

解题思路：

动态规划（Dynamic Programming）

状态转移方程详见代码

Python代码：

class Solution(object):
    def champagneTower(self, poured, query_row, query_glass):
        """
        :type poured: int
        :type query_row: int
        :type query_glass: int
        :rtype: float
        """
        dp = [[0] * (query_row + 2) for x in range(query_row + 2)]
        dp[0][0] += poured
        for x in range(query_row + 1):
            for y in range(x + 1):
                if dp[x][y] > 1:
                    dp[x + 1][y] += 0.5 * (dp[x][y] - 1)
                    dp[x + 1][y + 1] += 0.5 * (dp[x][y] - 1)
                    dp[x][y] = 1
        return dp[query_row][query_glass]

 

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