题解 | 矩阵的最小路径和

fmin = lambda x, y: x if x < y else y
fmax = lambda x, y: x if x > y else y

def solve(testcase):
    n, m = MI()
    A = [0 for _ in range(m)]

    for i in range(n):
        B = LII()
        for j in range(m):

            if i > 0 and j > 0:
                B[j] += fmin(A[j], B[j - 1])
            elif i > 0:
                B[j] += A[j]
            elif j > 0:
                B[j] += B[j - 1]
        A = B
    
    print(A[m - 1])

for testcase in range(1):
    solve(testcase)