• <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>
            隨筆 - 87  文章 - 279  trackbacks - 0
            <2025年7月>
            293012345
            6789101112
            13141516171819
            20212223242526
            272829303112
            3456789

            潛心看書研究!

            常用鏈接

            留言簿(19)

            隨筆分類(81)

            文章分類(89)

            相冊(cè)

            ACM OJ

            My friends

            搜索

            •  

            積分與排名

            • 積分 - 217940
            • 排名 - 117

            最新評(píng)論

            閱讀排行榜

            評(píng)論排行榜

            Two Ends
            Time Limit:1000MS? Memory Limit:65536K
            Total Submit:625 Accepted:271

            Description
            In the two-player game "Two Ends", an even number of cards is laid out in a row. On each card, face up, is written a positive integer. Players take turns removing a card from either end of the row and placing the card in their pile. The player whose cards add up to the highest number wins the game. Now one strategy is to simply pick the card at the end that is the largest -- we'll call this the greedy strategy. However, this is not always optimal, as the following example shows: (The first player would win if she would first pick the 3 instead of the 4.)
            3 2 10 4
            You are to determine exactly how bad the greedy strategy is for different games when the second player uses it but the first player is free to use any strategy she wishes.

            Input
            There will be multiple test cases. Each test case will be contained on one line. Each line will start with an even integer n followed by n positive integers. A value of n = 0 indicates end of input. You may assume that n is no more than 1000. Furthermore, you may assume that the sum of the numbers in the list does not exceed 1,000,000.

            Output
            For each test case you should print one line of output of the form:
            In game m, the greedy strategy might lose by as many as p points.
            where m is the number of the game (starting at game 1) and p is the maximum possible difference between the first player's score and second player's score when the second player uses the greedy strategy. When employing the greedy strategy, always take the larger end. If there is a tie, remove the left end.

            Sample Input

            4 3 2 10 4
            8 1 2 3 4 5 6 7 8
            8 2 2 1 5 3 8 7 3
            0

            Sample Output

            In game 1, the greedy strategy might lose by as many as 7 points.
            In game 2, the greedy strategy might lose by as many as 4 points.
            In game 3, the greedy strategy might lose by as many as 5 points.

            Source
            East Central North America 2005

            #include? < iostream >
            #include?
            < cmath >
            using ? namespace ?std;

            const ? int ?MAXN? = ? 1001 ;
            int ?dp[MAXN][MAXN];

            int ?main()
            {
            ????
            int ?i,?j,?l,?n;
            ????
            int ?t? = ? 1 ;
            ????
            int ?tmp;
            ????
            int ?a[MAXN];
            ????
            while ?(scanf( " %d " ,? & n),?n > 0 )
            ????
            {
            ????????memset(dp,?
            0 ,? sizeof (dp));
            ????????
            for ?(i = 1 ;?i <= n;?i ++ )
            ????????????scanf(
            " %d " ,? & a[i]);
            ????????????
            ????????
            for ?(i = 1 ;?i < n;?i ++ )
            ????????????dp[i][i
            + 1 ]? = ?abs(a[i]? - ?a[i + 1 ]);
            ????????
            for ?(l = 4 ;?l <= n;?l += 2 )
            ????????
            {
            ????????????
            for ?(i = 1 ;?i <= n - l + 1 ;?i ++ )
            ????????????
            {
            ????????????????j?
            = ?i? + ?l? - ? 1 ;
            ????????????????
            if ?(a[j]? <= ?a[i + 1 ])
            ????????????????
            {
            ????????????????????tmp?
            = ?dp[i + 2 ][j]? + ?a[i]? - ?a[i + 1 ];
            ????????????????????
            if ?(dp[i][j]? < ?tmp)
            ????????????????????????dp[i][j]?
            = ?tmp;
            ????????????????}

            ????????????????
            if ?(a[i]? < ?a[j - 1 ])
            ????????????????
            {
            ????????????????????tmp?
            = ?dp[i][j - 2 ]? + ?a[j]? - ?a[j - 1 ];
            ????????????????????
            if ?(dp[i][j]? < ?tmp)
            ????????????????????????dp[i][j]?
            = ?tmp;
            ????????????????}

            ????????????????
            if ?(a[i]? >= ?a[j - 1 ])
            ????????????????
            {
            ????????????????????tmp?
            = ?dp[i + 1 ][j - 1 ]? + ?a[j]? - ?a[i];
            ????????????????????
            if ?(dp[i][j]? < ?tmp)
            ????????????????????????dp[i][j]?
            = ?tmp;
            ????????????????}

            ????????????????
            if ?(a[j]? > ?a[i + 1 ])
            ????????????????
            {
            ????????????????????tmp?
            = ?dp[i + 1 ][j - 1 ]? + ?a[i]? - ?a[j];
            ????????????????????
            if ?(dp[i][j]? < ?tmp)
            ????????????????????????dp[i][j]?
            = ?tmp;
            ????????????????}

            ????????????}

            ????????}

            ????????printf(
            " In?game?%d,?the?greedy?strategy?might?lose?by?as?many?as?%d?points.\n " ,?t ++ ,?dp[ 1 ][n]);
            ????}

            ????system(
            " pause " );
            ????
            return ? 0 ;
            }

            posted on 2006-08-28 16:00 閱讀(572) 評(píng)論(0)  編輯 收藏 引用 所屬分類: ACM題目
            久久久久亚洲av成人网人人软件| 韩国三级大全久久网站| 久久一区二区三区免费| 久久综合一区二区无码| 久久精品aⅴ无码中文字字幕不卡| 色综合久久综合中文综合网| 久久久国产乱子伦精品作者| 精品99久久aaa一级毛片| 亚洲婷婷国产精品电影人久久| 人妻无码久久一区二区三区免费 | 久久黄色视频| 性高湖久久久久久久久| 久久精品成人免费观看97| 丁香色欲久久久久久综合网| 久久亚洲国产中v天仙www| 亚洲美日韩Av中文字幕无码久久久妻妇 | 精品综合久久久久久88小说| 国产A级毛片久久久精品毛片| 久久艹国产| 国产精品久久久久无码av| 亚洲精品综合久久| 久久久久99精品成人片牛牛影视| 国产精品禁18久久久夂久 | 91精品婷婷国产综合久久| 精产国品久久一二三产区区别| 精品久久久久久亚洲| 久久综合亚洲色一区二区三区| 麻豆久久| 久久久久亚洲av毛片大| 91精品日韩人妻无码久久不卡| 精品久久香蕉国产线看观看亚洲| 亚洲国产精品久久电影欧美| 色婷婷狠狠久久综合五月| 品成人欧美大片久久国产欧美...| 久久久久久久尹人综合网亚洲| 久久久无码精品亚洲日韩蜜臀浪潮| 无夜精品久久久久久| 欧美一级久久久久久久大| 欧美性猛交xxxx免费看久久久| 久久人人爽人人爽人人片AV东京热| 久久强奷乱码老熟女|