POJ 3106 Flip and Turn 模拟 分析

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const int maxn=400;
const int maxm=200000;
int n,m;
char a[maxn][maxn];
char b[maxn][maxn];
int stk[10];
int cnt;
char s[maxm];
char cd[7][3]={"1","CX","V","BY","2","AZ","H"};
char sp[7][8]={"1","1H","1H1","1H1H","1H1H1","1H1H1H","H"};
void work(char cmd){
    for (int i=1;i<=n;i++){
        for (int j=1;j<=m;j++){
            if (cmd=='1') b[j][i]=a[i][j];
            if (cmd=='H') b[n-i+1][j]=a[i][j];
        }
    }
    if (cmd=='1') swap(n,m);
    for (int i=1;i<=n;i++){
        for (int j=1;j<=m;j++){
            a[i][j]=b[i][j];
        }
    }
}
void stkPush(char cmd){
    for (int i=0;i<7;i++){
        if (strchr(cd[i],cmd)!=NULL){
            int len=strlen(sp[i]);
            for (int j=0;j<len;j++){
                if (cnt>0&&stk[cnt-1]==sp[i][j]) cnt--;
                else stk[cnt++]=sp[i][j];
                if (cnt>=8) cnt=0;
            }
            break;
        }
    }
}
int main(){
    while (~scanf("%d%d",&n,&m)){
        cnt=0;
        for (int i=1;i<=n;i++) scanf("%s",a[i]+1);
        scanf("%s",s);
        int len=strlen(s);
        for (int i=0;i<len;i++) stkPush(s[i]);
        for (int i=0;i<cnt;i++) work(stk[i]);
        printf("%d %d\n",n,m);
        for (int i=1;i<=n;i++){
            for (int j=1;j<=m;j++){
                printf("%c",a[i][j]);
            }
            printf("\n");
        }
    }
	return 0;
}

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