• <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

            ACM OJ

            My friends

            搜索

            •  

            積分與排名

            • 積分 - 217940
            • 排名 - 117

            最新評論

            閱讀排行榜

            評論排行榜

            Giftbox
            Time Limit:4000MS? Memory Limit:131072K
            Total Submit:1210 Accepted:218

            Description

            Bobobo and Bottle are good friends, and the birthday of Bottle is coming. So Bobobo is considering preparing Bottle an unexpected gift for his birthday. As a common sense that when a person at his or her birthday party, he or she will open the gift with the presence of the person who gives the gift and then expresses the appreciation. Bobobo knows that the characteristic of Bottle is rush when he receives something important, so he wants to play a joke on him in this way…

            Bobobo comes into a gift shop, and there are a lot of different kinds of gift boxes. Bobobo intends to choose various boxes with different size and choose one of the gift boxes to contain the precious gift and place this box into another bigger box and place this bigger box into another bigger one… So Bottle will not see the gift until he opens the innermost box. Imagine the process of his opening the boxes, how rush Bottle will be ^_^ !

            The gift boxes are n-dimensional. An n-dimensional box with dimensions ( X1, X2, …, Xn ) can be put into another box with dimensions ( Y1, Y2, …, Yn ) if there exists a permutation π on { 1, 2, …, n } such that Xπ1 < Y1, Xπ2 < Y2, …, Xπn < Yn. The gift is also n-dimensional and it can be put into a box if it satisfies the criterion above. And Bobobo must try his best to choose as many as boxes to contain the gift.

            Input

            The input file contains multiple test cases. The first line of each test case contains two numbers. The first one is a positive integer number N (1 ≤ N ≤ 500), the number of boxes in the gift shop, and the second one is a positive number d (3 ≤ d ≤ 1 000), the dimension of all the boxes. The next one line contains d positive integers ( G1, G2, …, Gd ) representing the dimensions of the gift. And the subsequent N lines each contain d positive integers ( X1, X2, …, Xd ) representing the dimensions of each box. You may assume that all the numbers you encounter are positive integers and less than 231. The input data is terminated by EOF.

            Output

            The output of each test case will contain only one line. Output the maximum number of the boxes that Bobobo can choose. If Bobobo can not find any box which can contain the gift, output “Please look for another gift shop!”

            Sample Input

            5 7
            4 6 8 2 7 5 3
            2 8 13 6 10 9 4
            80 70 12 3 6 8 2
            8 7 4 6 9 10 12
            100 200 300 400 500 600 700
            800 800 800 800 800 800 800

            Sample Output

            3

            Source
            POJ Monthly--2006.09.29, sza

            ?

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

            const ? int ?MAXN? = ? 510 ;
            const ? int ?MAXM? = ? 1010 ;

            int ?n,?m;
            int ?data[MAXN][MAXM];
            int ?map[MAXN][MAXN];
            int ?i,?j,?k;
            int ?f[MAXN];
            int ?d[MAXN];
            int ?ans;

            int ?DP( int ?b)
            {
            ????
            if ?(d[b]? > ? 0 )? return ?d[b];
            ????
            int ?i;
            ????
            int ?t? = ? 0 ;
            ????
            for ?(i = 1 ;?i <= n;?i ++ )
            ????
            {
            ????????
            if ?(f[i]? && ?b? != ?i? && ?map[b][i]? && ?t? < ?DP(i)? + ? 1 )?t? = ?DP(i)? + ? 1 ;
            ????}

            ????d[b]?
            = ?t;
            ????
            return ?d[b];
            }


            int ?main()
            {???
            ????
            while ?(scanf( " %d%d " ,? & n,? & m)? != ?EOF)
            ????
            {
            ????????
            for ?(i = 0 ;?i <= n;?i ++ )
            ????????
            {
            ????????????
            for ?(j = 0 ;?j < m;?j ++ )
            ????????????????scanf(
            " %d " ,? & data[i][j]);
            ????????????sort(data[i],?data[i]
            + m);
            ????????}

            ????????
            ????????memset(f,?
            1 ,? sizeof (f));
            ????????
            ????????
            for ?(i = 1 ;?i <= n;?i ++ )
            ????????????
            for ?(j = 0 ;?j < m;?j ++ )
            ????????????????
            if ?(data[i][j]? <= ?data[ 0 ][j])
            ????????????????
            {
            ????????????????????f[i]?
            = ? 0 ;
            ????????????????????
            break ;
            ????????????????}

            ????????????????
            ????????
            for ?(i = 0 ;?i <= n;?i ++ )
            ????????????
            for ?(j = 1 ;?j <= n;?j ++ )
            ????????????
            {
            ????????????????map[i][j]?
            = ? 1 ;
            ????????????????
            for ?(k = 0 ;?k < m;?k ++ )
            ????????????????
            {
            ????????????????????
            if ?(data[i][k]? >= ?data[j][k])
            ????????????????????
            {
            ????????????????????????map[i][j]?
            = ? 0 ;
            ????????????????????????
            break ;
            ????????????????????}

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

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

            ????????????
            ????????memset(d,?
            0 ,? sizeof (d));

            ????????ans?
            = ?DP( 0 );

            ????????
            if ?(ans? == ? 0 )
            ????????????printf(
            " Please?look?for?another?gift?shop!\n " );
            ????????
            else
            ????????????printf(
            " %d\n " ,?ans);
            ???????????????????????
            ????}

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

            posted on 2006-09-30 01:46 閱讀(521) 評論(0)  編輯 收藏 引用 所屬分類: ACM題目
            日韩va亚洲va欧美va久久| 久久只有这精品99| 久久免费国产精品一区二区| 久久免费小视频| 久久久久久久久久久| 欧美伊香蕉久久综合类网站| 久久夜色撩人精品国产小说| 亚洲精品无码久久久久sm| 亚洲国产精品婷婷久久| 国产精品成人久久久| 嫩草影院久久99| 久久永久免费人妻精品下载| 久久婷婷人人澡人人| 99久久精品九九亚洲精品| 久久丫精品国产亚洲av| 2021国产精品午夜久久| 久久精品无码一区二区三区日韩| 亚洲AV日韩AV天堂久久| 午夜精品久久久久| 久久久久女教师免费一区| 久久亚洲精品中文字幕三区| 久久精品国产亚洲av影院| 综合久久一区二区三区 | 综合久久久久久中文字幕亚洲国产国产综合一区首 | 久久综合噜噜激激的五月天| 久久婷婷五月综合色99啪ak| 99久久99这里只有免费的精品| 亚洲国产精品狼友中文久久久 | 久久青青草原精品国产不卡| 久久久久久免费一区二区三区| 97久久婷婷五月综合色d啪蜜芽| 四虎国产精品免费久久| 久久国产香蕉一区精品| 久久久久99精品成人片三人毛片| 久久久精品午夜免费不卡| 高清免费久久午夜精品| 久久久精品午夜免费不卡| 久久综合欧美成人| 久久精品免费大片国产大片| 久久天天躁狠狠躁夜夜2020老熟妇| 国产精品gz久久久|