• <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

            Space Ant
            Time Limit:1000MS? Memory Limit:10000K
            Total Submit:113 Accepted:84

            Description
            The most exciting space discovery occurred at the end of the 20th century. In 1999, scientists traced down an ant-like creature in the planet Y1999 and called it M11. It has only one eye on the left side of its head and just three feet all on the right side of its body and suffers from three walking limitations:

            1. It can not turn right due to its special body structure.
            2. It leaves a red path while walking.
            3. It hates to pass over a previously red colored path, and never does that.

            The pictures transmitted by the Discovery space ship depicts that plants in the Y1999 grow in special points on the planet. Analysis of several thousands of the pictures have resulted in discovering a magic coordinate system governing the grow points of the plants. In this coordinate system with x and y axes, no two plants share the same x or y.
            An M11 needs to eat exactly one plant in each day to stay alive. When it eats one plant, it remains there for the rest of the day with no move. Next day, it looks for another plant to go there and eat it. If it can not reach any other plant it dies by the end of the day. Notice that it can reach a plant in any distance.
            The problem is to find a path for an M11 to let it live longest.
            Input is a set of (x, y) coordinates of plants. Suppose A with the coordinates (xA, yA) is the plant with the least y-coordinate. M11 starts from point (0,yA) heading towards plant A. Notice that the solution path should not cross itself and all of the turns should be counter-clockwise. Also note that the solution may visit more than two plants located on a same straight line.

            Input
            The first line of the input is M, the number of test cases to be solved (1 <= M <= 10). For each test case, the first line is N, the number of plants in that test case (1 <= N <= 50), followed by N lines for each plant data. Each plant data consists of three integers: the first number is the unique plant index (1..N), followed by two positive integers x and y representing the coordinates of the plant. Plants are sorted by the increasing order on their indices in the input file. Suppose that the values of coordinates are at most 100.

            Output
            Output should have one separate line for the solution of each test case. A solution is the number of plants on the solution path, followed by the indices of visiting plants in the path in the order of their visits.

            Sample Input

            2
            10
            1 4 5
            2 9 8
            3 5 9
            4 1 7
            5 3 2
            6 6 3
            7 10 10
            8 8 1
            9 2 4
            10 7 6
            14
            1 6 11
            2 11 9
            3 8 7
            4 12 8
            5 9 20
            6 3 2
            7 1 6
            8 2 13
            9 15 1
            10 14 17
            11 13 19
            12 5 18
            13 7 3
            14 10 16
            

            Sample Output

            10 8 7 3 4 9 5 6 2 1 10
            14 9 10 11 5 12 8 7 6 13 4 14 1 3 2
            

            Source
            Tehran 1999

            ?

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

            const ? int ?MAXN? = ? 1000 ;
            const ? int ?INF? = ? 2000000000 ;

            struct ?XYDATA
            {
            ????
            int ?x,?y,?index;
            }
            ;

            int ?inorder(XYDATA?a,?XYDATA?b)
            {
            ????
            return ?a.y? < ?b.y? || ?(a.y? == ?b.y? && ?a.x? < ?b.x);
            }


            int ?cross(XYDATA?a,?XYDATA?b)
            {
            ????
            return ?a.x? * ?b.y? - ?a.y? * ?b.x;
            }


            double ?Cos(XYDATA?a,?XYDATA?b)
            {
            ????
            return ?(a.x * b.x + a.y * b.y)? / ?(sqrt(a.x * a.x + a.y * a.y)? * ?sqrt(b.x * b.x + b.y * b.y));
            }

            void ?Solve()
            {
            ????
            int ?n;
            ????
            int ?i,?j;
            ????
            int ? out [MAXN];
            ????
            int ?outSize? = ? 0 ?;
            ????
            bool ?mp[ 101 ][ 101 ];
            ????
            int ?t;
            ????
            double ?tmp;
            ????XYDATA?a[MAXN];
            ????XYDATA?v,?u;
            ????
            int ?beg;

            ????scanf(
            " %d " ,? & n);
            ????
            for ?(i = 0 ;?i < n;?i ++ )
            ????????scanf(
            " %d%d%d " ,? & a[i].index,? & a[i].x,? & a[i].y);
            ????
            ????sort(a,?a
            + n,?inorder);
            ????
            ????
            for ?(i = 0 ;?i < 101 ;?i ++ )
            ????????
            for ?(j = 0 ;?j < 101 ;?j ++ )
            ????????????mp[i][j]?
            = ? false ;

            ????v.x?
            = ?a[ 0 ].x;
            ????v.y?
            = ? 0 ;
            ????mp[a[
            0 ].x][a[ 0 ].y]? = ? true ;
            ????beg?
            = ? 0 ;
            ????
            out [outSize ++ ]? = ?a[ 0 ].index;
            ????
            for ?(i = 1 ;?i < n;?i ++ )
            ????
            {
            ????????
            bool ?isFind? = ? false ;
            ????????tmp?
            = ? - 2 ;
            ????????
            for ?(j = 0 ;?j < n;?j ++ )
            ????????????
            if ?(beg? != ?j? && ? ! mp[a[j].x][a[j].y])
            ????????????
            {
            ????????????????u.x?
            = ?a[j].x? - ?a[beg].x;
            ????????????????u.y?
            = ?a[j].y? - ?a[beg].y;
            ????????????????
            int ?cro? = ?cross(v,?u);
            ????????????????
            if ?(cro? >= ? 0 )????
            ????????????????
            {
            ????????????????????isFind?
            = ? true ;
            ????????????????????
            if ?(Cos(v,?u)? > ?tmp)?
            ????????????????????
            {
            ????????????????????????tmp?
            = ?Cos(v,?u);
            ????????????????????????t?
            = ?j;
            ????????????????????}

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

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

            ????????
            if ?(isFind)
            ????????
            {
            ????????????v.x?
            = ?a[t].x? - ?a[beg].x;
            ????????????v.y?
            = ?a[t].y? - ?a[beg].y;
            ????????????mp[a[t].x][a[t].y]?
            = ? true ;
            ????????????beg?
            = ?t;
            ????????????
            out [outSize ++ ]? = ?a[t].index;
            ????????}

            ????????
            else ?
            ????????
            {
            ????????????
            break ;
            ????????}

            ????}

            ????printf(
            " %d " ,?outSize);
            ????
            for ?(i = 0 ;?i < outSize;?i ++ )
            ????????printf(
            " ?%d " ,? out [i]);
            ????printf(
            " \n " );
            }



            int ?main()
            {
            ????
            int ?n;

            ????scanf(
            " %d " ,? & n);

            ????
            while ?(n -- ? != ? 0 )
            ????
            {
            ????????Solve();
            ????}

            ????
            return ? 0 ;
            }
            posted on 2006-09-27 13:54 閱讀(544) 評論(0)  編輯 收藏 引用 所屬分類: ACM題目
            久久精品女人天堂AV麻| 久久久精品2019免费观看| 亚洲AV无码久久精品蜜桃| 久久久久亚洲AV片无码下载蜜桃| 伊人色综合久久天天人手人婷 | 欧美日韩久久中文字幕| 久久99精品久久久久久动态图 | 亚洲精品乱码久久久久久不卡| 午夜精品久久久久久久| 91精品国产综合久久四虎久久无码一级 | 久久久久18| 久久久久亚洲精品无码蜜桃| 精品久久久久久久久久中文字幕| 伊人久久大香线蕉亚洲五月天 | 久久精品亚洲福利| 91久久精一区二区三区大全| 亚洲AV无码久久精品色欲| 久久亚洲国产欧洲精品一| 久久精品aⅴ无码中文字字幕不卡 久久精品aⅴ无码中文字字幕重口 | 午夜久久久久久禁播电影| 久久996热精品xxxx| 国产午夜精品理论片久久影视| 久久亚洲中文字幕精品有坂深雪 | 99久久精品国产一区二区三区| 久久天天婷婷五月俺也去| 久久亚洲精品无码观看不卡| 久久99精品九九九久久婷婷| 久久久精品人妻一区二区三区蜜桃| 欧美亚洲日本久久精品| 久久免费视频1| 亚洲欧洲精品成人久久曰影片 | 国产午夜免费高清久久影院 | 国产午夜精品久久久久九九| 激情伊人五月天久久综合| 亚洲狠狠婷婷综合久久久久| 久久成人小视频| 亚洲精品无码久久久影院相关影片| 天天影视色香欲综合久久| 老男人久久青草av高清| 国产午夜精品理论片久久| 国产免费久久精品99久久|