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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