感谢Bloodblaze提供的一道简单的题目。以下是一个使用Fortran90进行DFS遍历的例子。具体题面稍后再说。
感谢Wikibooks和StackOverflow。如果没有他们，我是不可能在一夜之间入门Fortran的;-)
另外，我在网上找到了这些资源可供参考：
- http://www.mathcs.emory.edu/~cheung/Courses/561/Syllabus/6-Fortran/
- https://web.stanford.edu/class/me200c/tutorial_90/
本以为总算可以可以早点去睡觉的，结果发现根本停不下来。这里就不得不提到编程过程中的一大错觉：
再写半个小时应该就能完工了。
我在1点的时候也是这么想的。
我可能是阿华田喝多了。
以及这里的代码高亮似乎并不支持Fortran……
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PROGRAM test IMPLICIT NONE INTEGER, PARAMETER :: STUDENT_COUNT = 7 INTEGER, PARAMETER :: MAX_VACANCY = 3 INTEGER, DIMENSION(STUDENT_COUNT, MAX_VACANCY) :: VACANCY INTEGER, DIMENSION(STUDENT_COUNT) :: schedule = 0 INTEGER :: solutionCount = 0 VACANCY = transpose(reshape( & (/2,4,0, & 1,6,0, & 3,7,0, & 5,0,0, & 1,4,6, & 2,5,0, & 3,6,7/), & (/MAX_VACANCY, STUDENT_COUNT/))) !write(*,*) VACANCY(2,3) CALL assignStudent(1) WRITE(*,*) "Solutions: ", solutionCount CONTAINS RECURSIVE SUBROUTINE assignStudent(studentIndex) INTEGER, INTENT(IN) :: studentIndex INTEGER :: i, j, vc TEST_VACANCY: DO i = 1,MAX_VACANCY vc = VACANCY(studentIndex, i) IF (vc == 0) EXIT TEST_VACANCY ! sanity check DO j = 1,studentIndex - 1 ! DOW is occupied IF (schedule(j) == vc) CYCLE TEST_VACANCY END DO ! so far, so good schedule(studentIndex) = vc IF (studentIndex >= STUDENT_COUNT) THEN ! the solution is complete DO j = 1, STUDENT_COUNT write(*,"(I4)",advance="no") schedule(j) END DO write(*,*) solutionCount = solutionCount + 1 CYCLE TEST_VACANCY END IF ! the solution is not complete ! go to next level CALL assignStudent(studentIndex + 1) END DO TEST_VACANCY END SUBROUTINE assignStudent END PROGRAM test |
运行结果如下
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root@localhost:/usr/local/fortran# gfortran test.f90 root@localhost:/usr/local/fortran# ./a.out 4 1 3 5 6 2 7 4 1 7 5 6 2 3 4 6 3 5 1 2 7 4 6 7 5 1 2 3 Solutions: 4 root@localhost:/usr/local/fortran# |