• <ins id="pjuwb"></ins>
    <blockquote id="pjuwb"><pre id="pjuwb"></pre></blockquote>
    <noscript id="pjuwb"></noscript>
          <sup id="pjuwb"><pre id="pjuwb"></pre></sup>
            <dd id="pjuwb"></dd>
            <abbr id="pjuwb"></abbr>

            coreBugZJ

            此 blog 已棄。

            生成全排列的非回溯方法(TopCoder SRM 591 DIV 2)

            問題來自 TopCoder SRM 591 DIV 2 的第二題:

            Problem Statement
               
            Let X and Y be two strings of equal length, consisting of uppercase English letters only. The two strings are called convertible if there is a permutation P of the English alphabet with the following property: if each character c in the string X is replaced by the character P(c), we obtain the string Y. (In other words, X and Y are convertible iff the following holds: whenever two letters of X are equal, the corresponding two letters of Y must be equal, and vice versa.)  For example, consider the string "ABCA". We can choose to replace each 'A' by a 'F', each 'B' by a 'B', and each 'C' by a 'G', obtaining the string "FBGF". Thus the strings "ABCA" and "FBGF" are convertible. The strings "ABCA" and "EFGH" are not convertible, because the two 'A's in the first string must correspond to the same letter in the second string. The strings "ABCA" and "EEEE" are not convertible, because different letters in the first string must correspond to different letters in the second string.  You are given two strings A and B of the same length. These strings only contain English letters from 'A' to 'I', inclusive. (That is, only the first 9 letters of the alphabet are used.)  You want to change A and B into two strings that are convertible. The only allowed change is to choose some indices (possibly none) and to remove the characters at those indices from each of the strings. (I.e., the removed characters must be at the same positions in both strings. For example, it is not allowed to only remove character 0 of A and character 3 of B.) For example, if A="ABAC", B="AHHA" and the chosen indices are 0 and 2, then the resulting strings will be "BC" and "HA". Our goal is to choose as few indices as possible, given that after the erasing we want to obtain two convertible strings. Compute and return the smallest possible number of chosen indices.
            Definition
               
            Class:
            ConvertibleStrings
            Method:
            leastRemovals
            Parameters:
            string, string
            Returns:
            int
            Method signature:
            int leastRemovals(string A, string B)
            (be sure your method is public)
               

            Constraints
            -
            A will contain between 1 and 50 characters, inclusive.
            -
            A and B will be of the same length.
            -
            A will contain characters from 'A' to 'I' only.
            -
            B will contain characters from 'A' to 'I' only.

            我的思路是窮舉A中字母與B中字母的對應關系,看哪種對應需要刪除的字母最少,這一最少值即是答案。
            窮舉對應關系,就是生成全排列。
            我生成全排列的方式是回溯。

            之后看其他人的代碼,發現一個由給定排列求出其下一個排列的函數,于是學習一下,自己實現如下:

            // 生成下一字典序排列
            // 假設a中元素互不相同
            // 若已經是最后一個字典序排列,則返回0,否則返回1
            int next_permutation( int a[], int n ) {
                    
            int i, j;
                    
            for ( i = n-1; (0 < i) && (a[i-1> a[i]); --i ) {
                    }
                    
            if ( 0 >= i ) {
                            
            return 0;
                    }
                    
            for ( j = n-1; j >= i; --j ) {
                            
            if ( a[ j ] > a[ i-1 ] ) {
                                    
            int tmp = a[ i-1 ];
                                    a[ i
            -1 ] = a[ j ];
                                    a[ j ] 
            = tmp;
                                    j 
            = n - 1;
                                    
            while ( i < j ) {
                                            tmp 
            = a[ i ];
                                            a[ i ] 
            = a[ j ];
                                            a[ j ] 
            = tmp;
                                            
            ++i; --j;
                                    }
                                    
            break;
                            }
                    }
                    
            return 1;
            }

            還有人使用的是C++的 <algorithm> 中 next_permutation 函數,功能一樣。


            posted on 2013-09-28 17:03 coreBugZJ 閱讀(910) 評論(0)  編輯 收藏 引用 所屬分類: ACM 、Algorithm

            久久精品无码专区免费| 久久久精品国产免大香伊 | 久久久久亚洲AV片无码下载蜜桃| 国产伊人久久| 91精品国产91久久久久久蜜臀| 国产一级做a爰片久久毛片| 国产成人精品久久二区二区| 狠狠色婷婷久久一区二区三区| 99久久婷婷国产综合亚洲| 久久夜色精品国产噜噜亚洲AV| 日本一区精品久久久久影院| 狠狠人妻久久久久久综合蜜桃| 久久97精品久久久久久久不卡| 99麻豆久久久国产精品免费| 91精品国产综合久久精品| 国产精品免费福利久久| 99久久777色| 久久久久国产精品嫩草影院| 色播久久人人爽人人爽人人片aV | 2020国产成人久久精品| 国产91色综合久久免费| 日本一区精品久久久久影院| 精品久久久久久国产三级| 久久一区二区三区免费| 久久久久人妻一区精品性色av| 久久精品aⅴ无码中文字字幕重口| 国产精品久久久久9999| 久久午夜无码鲁丝片午夜精品| 久久国产AVJUST麻豆| 狠狠色婷婷久久一区二区三区| 成人精品一区二区久久| 久久亚洲精品国产精品| 99国内精品久久久久久久| 久久精品极品盛宴观看| 女人香蕉久久**毛片精品| 女同久久| 久久精品国产福利国产秒| 久久人人添人人爽添人人片牛牛 | 久久亚洲精品无码播放| 亚洲av日韩精品久久久久久a| 日韩亚洲国产综合久久久|