# Résolution de problèmes algorithmiques

Given a description of a landscape, describe how the rain water will drain.

# Problem statement

You are given a rectangular grid. Grid cells represent squares of a landscape. Every grid cell c has an integer height, denoted height[c]. It is raining, and water can drain to the border of the grid, following a path of grid cells with non-increasing height. Compute for each grid cell, how much water is trapped above it. (The actual problem asks only for the total amount.)

# Reduction to a shortest path problem

When you see the word “flood”, you might think that this is a flow problem. But it can be reduced to a variant of a shortest path problem.

Consider a path P from some cell c to a border cell. Water above c can drain to the border along P. Formally, if the water height above c exceeds stricty the maximum height[d] over all grid cells d in P (call it the height of the path), then this surplus can drain along P. This means that for every grid cell we wish to compute a path to the border minimizing the maximum height of the traversed cells. If this height is k, then the trapped water level above c is precisely k.

This can be done by a variant of Dijkstra’s algorithm, where instead of the addition of edge costs we use the maximum operator, and instead of having weights on the edges, we have weights on the edges. The algorithm computes for every grid cell c, the trapped water level above it in a variable water[c]. It uses a priority queue called heap, storing triplets. Each triplet consists of a cell coordinate and the maximum height of a discovered path from the border to this cell.

from heapq import heappop, heappush, heapify

class Solution:

def sum_matrix(self, M):
return sum(sum(row) for row in M)

def trapRainWater(self, heightMap):
rows = len(heightMap)
cols = len(heightMap[0])
R = range(rows)
C = range(cols)
water = [[None for j in C] for i in R]
# initialize with border cells
heap = [(heightMap[i][j], i, j) for i in R for j in C if
i == 0 or i == rows - 1 or j == 0 or j == cols - 1]
heapify(heap)
while heap:
h, i, j = heappop(heap)
if water[i][j] is None:            # not yet visited
water[i][j] = h
for i2, j2 in [(i-1,j),(i+1,j),(i,j-1),(i,j+1)]:
if 0 <= i2 < rows and 0 <= j2 < cols:
priority = max(h, heightMap[i2][j2])
heappush(heap, (priority, i2, j2))
return self.sum_matrix(water) - self.sum_matrix(heightMap)


# Complexity

The complexity is $O(n m \log (nm))$ for an n by m grid.