[HDU 4119]Isabella's Message[字符串模拟]

题意:

给出一个表格和一块蒙版, 露出的字母按一定顺序读出, 蒙版旋转, 初始位置不定.

只认识给定的一些词. 问可辨识的短语. 多种情况输出字典序最小的.

思路:

字符串处理的模拟题.

主要是有几步:

1. 读入.

2. 得到蒙版四个方向读出的串.

3. 将这四个串以四种方式串联.

4. 消除前导空格, 合并中间连续的多个空格.

5. 判断是否是认识的词.

6. 将字典序最小的输出.

模拟恶心题, 写清步骤很重要. 磨刀不误砍柴工, 不然后面很容易乱套!

其中, 2 中分出两个函数, 分别收集 和 旋转.

3 中可以用一个矩阵来显示4个循环. 比直接for 要清晰些...

4 是最麻烦的!

仔细分析各个标志改变状态的时间点 和 各个标志发挥作用以及清除的时间点!!!

5 查找直接用set 中的find 函数, 但是要认识到 判断是恰好发生在遇到空格之后,且判断完毕tmp清空 .

#include <set>#include <string>#include <cstdio>#include <cstdlib>#include <iostream>using namespace std;char code[55][55],mark[55][55],s[25];int t,n,m,nCase = 0;set<string> dic,ans;string getstring(){    string ret("");    for(int i = 0; i < n; ++i)for(int j = 0; j < n; ++j)if(mark[i][j] == '*') ret += code[i][j];///dia-circle + judge    return ret;}void rotate(){    char tmp[55][55];    for(int i = 0; i < n; ++i)for(int j = 0; j < n; ++j)tmp[i][j] = mark[i][j];///copy    for(int i = 0; i < n; ++i)for(int j = 0; j < n; ++j)mark[j][n-1-i] = tmp[i][j];///rotate}string preprocess(string s)///deal with blanks{    string ret("");    int len = s.length();    bool flag = true;///ignore blank    int i = 0;    while(s[i] == '.')i++;    for(; i < len; ++i)    {        if(s[i] == '.')        {            flag = false;///blank delay!!! only mark it when encounter blank, because you may encounter blanks!!!        }//the blank after the first word activated the flag         else//while the second word set the first blank and at the same time eliminated the flag        {            if(!flag)            {                ret += ' ';                flag = true;            }            ret += s[i];        }    }    return ret;}bool check(string s){    int len = s.length();    string key;    for(int i = 0; i < len; ++i)    {        if(s[i] == ' ')        {            if(dic.find(key) == dic.end()) return false;            key.clear();        }        else key += s[i];    }    if(dic.find(key) == dic.end()) return false;    return true;}int r[4][4] ={    {0,1,2,3},    {1,2,3,0},    {2,3,0,1},    {3,0,1,2}};//circle shiftvoid solve(){    scanf("%d",&n);    dic.clear();    ans.clear();    for(int i = 0; i < n; ++i) scanf("%s",code[i]);    for(int i = 0; i < n; ++i) scanf("%s",mark[i]);    scanf("%d",&m);    for(int i = 0; i < m; ++i)    {        scanf("%s",s);        string d(s);///char to string        dic.insert(d);///set    }    string tostring[4];    for(int i = 0; i < 4; ++i)    {        tostring[i] = getstring();        rotate();    }///get four string    for(int i = 0; i < 4; ++i)    {        string tocheck("");        for(int j = 0; j < 4; ++j) tocheck += tostring[r[i][j]];///concatenate string        string st = preprocess(tocheck);        if(check(st)) ans.insert(st);    }    printf("Case #%d: ",++nCase);    if(ans.size() == 0) printf("FAIL TO DECRYPT\n");    else printf("%s\n",(*ans.begin()).c_str());///automatically lexicographically ordered in set}int main(){    scanf("%d",&t);    while(t--)    {        solve();    }    return 0;}