UVa - 227 - Puzzle(gets(),scanf("%c",&x))

思路：简单模拟，按照题目意思进行编码即可。

#include<cstdio>

using namespace std;
char m[5][5];
char cmd[100];
int main(){
// freopen("input.txt","r",stdin);

int b_x ,b_y ,cases=0;

while(gets(m[0])){
if(m[0][0]=='Z'){
break;
}else{

for(int i=1 ;i<5 ;i++){
gets(m[i]);
}

for(int i=0 ;i<5 ;i++){
for(int j=0 ;j<5 ;j++){
if(m[i][j] == ' '){
b_x = i; b_y = j;
break;
}
}
}
}
int count = 0;

while(scanf("%c",&cmd[count])!=EOF){
if(cmd[count]!='0'){
count++;
}else {
break;
}
}

cmd[count] = 0;
getchar();

int x=b_x ,y=b_y ,flag = 0;
for(int i=0 ;cmd[i] ;i++){
switch(cmd[i]){
case 'A':x = b_x-1; y = b_y; break;
case 'B':x = b_x+1; y = b_y; break;
case 'L':x = b_x; y = b_y-1; break;
case 'R':x = b_x; y = b_y+1; break;
}

if(x<0||x>4||y<0||y>4){
flag = 1;
break;
}else{
m[b_x][b_y] = m[x][y];
m[x][y] = ' ';
b_x = x; b_y = y;
}
}

if(cases++){
printf("\nPuzzle #%d:\n",cases);
}else{
printf("Puzzle #%d:\n",cases);
}

if(flag){
printf("This puzzle has no final configuration.\n");
}else{
for(int i=0 ;i<5 ;i++){
printf("%c",m[i][0]);
for(int j=1 ;j<5 ;j++){
printf(" %c",m[i][j]);
}
printf("\n");
}
}

}

return 0;
}