1132.txt

北大ACM题目例程 详细的解答过程 程序实现 算法分析

#define debug 0
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<stdlib.h>
#define NMAX 33
#define INF 1000000001
int xp[4]={0,1,0,-1},yp[4]={1,0,-1,0};
int dx[4]={0,0,-1,-1},dy[4]={0,-1,-1,0};
int stx,sty;
int map[NMAX][NMAX];
char in[NMAX*NMAX];

int set(char t)
{
	switch(t)
	{
	case 'N':return 0;
	case 'E':return 1;
	case 'S':return 2;
	case 'W':return 3;
	}
	return -1;
}
void solve()
{
	int i=0;
	int cx=stx,cy=sty;
	int direct;
	while(in[i]!='.')
	{
		direct=set(in[i]);
		map[cy+dy[direct]][cx+dx[direct]]=1;
		cx+=xp[direct];
		cy+=yp[direct];
		i++;
	}
	int j;
	for(i=31;i>=0;i--)
	{
		for(j=0;j<32;j++)
			if(map[i][j]==1)
				printf("X");
			else
				printf(".");
		printf("\n");
	}
	printf("\n");
}
main()
{
#if debug 	
	freopen("in.txt","r",stdin);
	freopen("out.txt","w",stdout);
#endif
	int t,i;
	scanf("%d",&t);
	for(i=1;i<=t;i++)
	{
		printf("Bitmap #%d\n",i);
		memset(map,0,sizeof(map));
		scanf("%d%d",&stx,&sty);
		
		scanf("%s",in);
		solve();
	}
#if debug
	fclose(stdin);
	fclose(stdout);
#endif;
	return 1;
}