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            Dain

            寫出一個(gè)可以工作的程序并不夠

            統(tǒng)計(jì)

            留言簿(3)

            積分與排名

            良師益友

            閱讀排行榜

            評(píng)論排行榜

            最大的子序列和問題

            求解該問題的四種算法:
            時(shí)間O(N3),算法一
            int ?MaxSubsequenceSum( const ? int ?A[], int ?N)
            {
            ????
            int
            ?ThisSum,MaxSum,i,j,k;
            ????
            ????MaxSum?
            = ? 0
            ;
            ????
            for (i? = ? 0 ;i? < ?N;i ++
            )
            ????????
            for (j? = ?i;j? < ?N;j ++
            )
            ????????
            {
            ????????????ThisSum?
            = ? 0
            ;
            ????????????
            for (k? = ?i;k? <= ?j;k ++ )????ThisSum? +=
            ?A[k];????????????????
            ????????????
            if (ThisSum? > ?MaxSum)????MaxSum? =
            ?ThisSum;
            ????????}

            ????????
            ????
            return ?MaxSum;
            }
            時(shí)間O(N2),算法二
            int ?MaxSubsequenceSum( const ? int ?A[], int ?N)
            {
            ????
            int
            ?ThisSum,MaxSum,i,j;
            ????
            ????MaxSum?
            = ? 0
            ;
            ????
            for (i? = ? 0 ;i? < ?N;i ++
            )
            ????
            {
            ????????ThisSum?
            = ? 0
            ;
            ????????
            for (j? = ?i;j? < ?N;j ++
            )
            ????????
            {
            ????????????ThisSum?
            +=
            ?A[k];????????????????
            ????????????
            if (ThisSum? > ?MaxSum)????MaxSum? =
            ?ThisSum;
            ????????}

            ????}

            ????????
            ????
            return ?MaxSum;
            }
            時(shí)間O(NlogN),算法三
            static ? int ?MaxSubSum( const ? int ?A[], int ?Left, int ?Right)
            {
            ????
            int
            ?MaxLeftSum,MaxRightSum;
            ????
            int
            ?MaxLeftBorderSum,MaxRightBorderSum;
            ????
            int
            ?LeftBorderSum,RightBorderSum;
            ????
            int
            ?Center,i;
            ????
            ????
            if (Left? ==
            ?Right)
            ????????
            if (A[left]? > ? 0 )???? return
            ?A[left];
            ????????
            else ???? return ? 0
            ;
            ????????????
            ????Center?
            = ?(Left? + ?Right)? / ? 2
            ;
            ????MaxLeftSum?
            =
            ?MaxSubSum(A,Left,Center);
            ????MaxRightSum?
            = ?MaxSubSum(A,Center? + ? 1
            ,Right);
            ????
            ????MaxLeftBorderSum?
            = ? 0
            ;
            ????LeftBorderSum?
            = ? 0
            ;
            ????
            for (i? = ?Center;i? >= ?Left;i --
            )
            ????
            {
            ????????LeftBorderSum?
            +=
            ?A[i];
            ????????
            if (LeftBorderSum? > ?MaxLeftBorderSum)????MaxLeftBorderSum? =
            ?LeftBorderSum;
            ????}

            ????
            ????MaxRightBorderSum?
            = ? 0 ;
            ????RightBorderSum?
            = ? 0
            ;
            ????
            for (i? = ?Center? + ? 1 ;i? <= ?Right;i ++
            )
            ????
            {
            ????????RightBorderSum?
            +=
            ?A[i];
            ????????
            if (RightBorderSum? > ?MaxRightBorderSum)????MaxRightBorderSum? =
            ?RightBorderSum;
            ????}

            ????
            ????
            return ?Max3(MaxLeftSum,MaxRightSum,MaxLeftBorderSum? + ?MaxRightBorderSum);
            }


            int ?MaxSubsequenceSum( const int??A[],int ?N)
            {
            ????
            return ?MaxSubSum(A, 0 ,N? - ? 1
            );????
            }
            時(shí)間O(N),算法四
            intMaxSubsequenceSum( const int ?A[], int ?N)
            {
            ????
            int ?ThisSum,MaxSum,i;
            ????
            ????ThisSum?
            = ?MaxSum? = ? 0 ;
            ????
            for (i? = ? 0 ;i? < ?N;i ++ )
            ????
            {
            ????????ThisSum?
            += ?A[i];
            ????????
            if (ThisSum? > ?MaxSum)
            ????????????MaxSum?
            = ?ThisSum;
            ????????
            else
            ????????????ThisSum?
            = ? 0 ;
            ????}

            ????
            ????
            return ?MaxSum;
            }


            參考《數(shù)據(jù)結(jié)構(gòu)與算法分析》

            posted on 2007-02-07 10:52 Dain 閱讀(1067) 評(píng)論(7)  編輯 收藏 引用 所屬分類: 算法筆記

            評(píng)論

            # re: 最大的子序列和問題[未登錄] 2007-02-08 01:36

            呵呵, dp是王道  回復(fù)  更多評(píng)論   

            # re: 最大的子序列和問題 2007-02-08 16:53 Dain

            嗯,dp
            對(duì)于這個(gè)問題,算法三和四都是不錯(cuò)的思路@豪
              回復(fù)  更多評(píng)論   

            # re: 最大的子序列和問題 2007-02-09 16:24 alai04

            樓主開什么玩笑,算法四明顯是不對(duì)的嘛,是考一下我們的眼力嗎?  回復(fù)  更多評(píng)論   

            # re: 最大的子序列和問題 2007-03-25 20:50 felix

            四不對(duì)
            如:
            1 5 -3 2 4
              回復(fù)  更多評(píng)論   

            # re: 最大的子序列和問題 2007-03-25 20:51 felix

            三用的是分治,不是DP吧?  回復(fù)  更多評(píng)論   

            # re: 最大的子序列和問題 2007-03-30 11:12 dqchen

            第四個(gè)
            MaxSum = -((static_cast<unsigned int>(~0)) >> 1) - 1;

            ThisSum = 0;
            for (int i = 0;i < A.size();i++)
            {
            ThisSum += A[i];
            if(ThisSum > MaxSum)
            MaxSum = ThisSum;
            else
            ThisSum = 0;
            }
              回復(fù)  更多評(píng)論   

            # 第四個(gè)算法的實(shí)現(xiàn)是錯(cuò)誤的 2008-07-04 17:53 wetwoo

            輸入如下:-2 11 -4 13 -5 -2
            輸出如下:13
            正確輸出應(yīng)該為:20
            修改如下:
            intMaxSubsequenceSum( const int A[], int N)
            {
            int ThisSum,MaxSum,i;

            ThisSum = MaxSum = 0 ;
            for (i = 0 ;i < N;i ++ )
            {
            ThisSum += A[i];
            if (ThisSum > MaxSum)
            MaxSum = ThisSum;
            else if(ThisSum < 0)
            ThisSum = 0 ;
            }

            return MaxSum;
            }   回復(fù)  更多評(píng)論   

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