Question

The challenge

The shortest code by character count to identify and mark water depressions in the ASCII representation of a land from input.

Input will be an ASCII representation of a landscape, having hills, valleys and flat lands. The program should simulate what the landscape would look like if if was flooded – filling all valleys with water (character x).

The landscape will always start and stop with the character _ and will be at least 2 characters long, making the shortest input __.

A hill is defined as a raise, and should not be filled with water:

  __
_/  \_

A valley is defined as a depression and will be filled with water until a flatland is encountered:

_    _
 \__/

Input can be assumed clean and will be composed only from the characters space (), newline (\n), underscore (_), and forward and backward slashes (/ and \). Input can be seen as a continuous line, and any input that contains ambiguous line input such as _/_ or

_   _
 \_/
 / \

Is considered invalid.

Regarding underwater caves, water level should be maintained if cave level goes above water level.

Test cases

Input:
    __/\__
          \__              
             \       ___       ___________
             /      /   \_     \_
             \_____/      \__  _/
                             \/
Output:

    __/\__
          \__              
             \       ___       ___________
             /xxxxxx/   \xxxxxx\_
             \xxxxx/      \xxxxx/
                             \/

---

Input:
                                         __       ___
                                        /  \_____/
                                       / _______
                         ________     /  \     /
                   _____/        \   /__  \    \_
    ____          /               \    /__/   __/
        \_       /                 \     ____/
          \______\                 /____/

Output:
                                         __       ___
                                        /  \xxxxx/
                                       / _______
                         ________     /  \     /
                   _____/        \xxx/__  \xxxx\_
    ____          /               \xxxx/__/xxxxx/
        \xxxxxxxx/                 \xxxxxxxxx/
          \xxxxxx\                 /xxxx/

---

Input:
                                                      __     _
    _       ____                    ____        _____/  \   /
     \     /    \        __________/    \    __/  ___   /___\
      \___/      \       \               \   \___/  /_
                 /________\               \___________\

Output:
                                                      __     _
    _       ____                    ____        _____/  \xxx/
     \xxxxx/    \xxxxxxxxxxxxxxxxxx/    \xxxxxx/  ___   /xxx\
      \xxx/      \xxxxxxx\               \xxx\___/xx/_
                 /xxxxxxxx\               \xxxxxxxxxxx\

Code count includes input/output (i.e full program).

Answer 1

C – 741 621 600 characters (but handles the new cases properly)

$ gcc water.c && ./a.out < test6.txt 
                                     __       ___    
                                    /  \xxxxx/       
                                   / _______         
                     ________     /  \     /         
               _____/        \xxx/__  \xxxx\_        
____          /               \xxxx/__/xxxxx/        
    \xxxxxxxx/                 \xxxxxxxxx/           
      \xxxxxx\                 /xxxx/                

#include<stdio.h>
char d[99][99],*p,*e,*z,*s=d,c,S=' ',D='-',O='.',U='_';n,w,x,N=99,i;
g(y){for(i=0;!i;p+=N,e+=N){i=*p==D;for(z=p;z!=e;z+=y){if(*z!=O&&*z!=
D)break;*z=*z==O?S:U;}}}f(char*n,int r){if(*n==O||*n==D){*n=r>0?'x':
S;int k;for(k=0;k<9;k++)f(n+k/3*N-N+k%3-1,r+k/3-1);}}main(){for(p=s;
gets(p);p+=N,n++){x=strlen(p)-1;w=x>w?x:w;}for(p=s,e=d[N];p<s+N;p++)
{for(i=1,z=p;z<e;z+=N)c=*z,c==0?*z=c=S:0,i?c==S?*z=O:c==U?*z=D:0:0,(
c=='/'&&z[1]!=U)||(c=='\\'&&z[-1]!=D)||c==U?i=1-i:0;}p=s;e=s+w;g(1);
p=s+w;e=s;g(-1);for(p=s;p<s+w;p++){for(z=p;*z==S;z+=N);f(z,1);}for(i
=0;i<n;i++)printf("%.*s\n",w+1,d[i]);}