/*
ID: lvxiaol3
LANG: C++
TASK: maze1
*/
#include<iostream>
#include<fstream>
#include<string.h>
using namespace std;
#define MAXWID 38
#define MAXHT 100
typedef struct Point Point;
struct Point
{
int r, c;
};
int wid, ht;
char maze[MAXHT*2+1][MAXWID*2+1+2];
int dist[MAXHT*2+1][MAXWID*2+1];
Point
Pt(int r, int c)
{
Point p;
p.r = r;
p.c = c;
return p;
}
typedef struct Queue Queue;
struct Queue
{
Point p;
int d;
};
Queue floodq[MAXHT*MAXWID];
int bq, eq;
/* if no wall between point p and point np, add np to queue with distance d+1 */
void
addqueue(int d, Point p, Point np)
{
if(maze[(p.r+np.r)/2][(p.c+np.c)/2] == ' ' && maze[np.r][np.c] == ' ')
{
maze[np.r][np.c] = '*';
floodq[eq].p = np;
floodq[eq].d = d+1;
eq++;
}
}
/* if there is an exit at point exitp, plug it and record a start point
* at startp */
void lookexit(Point exitp, Point startp)
{
if(maze[exitp.r][exitp.c] == ' ')
{
addqueue(0, startp, startp);
maze[exitp.r][exitp.c] = '#';
}
}
int main()
{
FILE *fin, *fout;
Point p;
int i, r, c, m, d;
fin = fopen("maze1.in", "r");
fout = fopen("maze1.out", "w");
//assert(fin != NULL && fout != NULL);
fscanf(fin, "%d %d\n", &wid, &ht);
wid = 2*wid+1;
ht = 2*ht+1;
for(i=0; i<ht; i++)
fgets(maze[i], sizeof(maze[i]), fin);
/* find exits */
for(i=1; i<wid; i+=2)
{
lookexit(Pt(0, i), Pt(1, i));
lookexit(Pt(ht-1, i), Pt(ht-2, i));
}
for(i=1; i<ht; i+=2)
{
lookexit(Pt(i, 0), Pt(i, 1));
lookexit(Pt(i, wid-1), Pt(i, wid-2));
}
/* must have found at least one square with an exit */
/* since two exits might open onto the same square, perhaps eq == 1 */
for(bq = 0; bq < eq; bq++)
{
p = floodq[bq].p;
d = floodq[bq].d;
dist[p.r][p.c] = d;
addqueue(d, p, Pt(p.r-2, p.c));
addqueue(d, p, Pt(p.r+2, p.c));
addqueue(d, p, Pt(p.r, p.c-2));
addqueue(d, p, Pt(p.r, p.c+2));
}
/* find maximum distance */
m = 0;
for(r=0; r<ht; r++)
for(c=0; c<wid; c++)
if(dist[r][c] > m)
m = dist[r][c];
fprintf(fout, "%d\n", m);
return 0;
}
很基礎的一個BFS或者是Floodfill問題，直接上標程吧。。。寫得太挫了。。