• <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 - 217972
            • 排名 - 117

            最新評論

            閱讀排行榜

            評論排行榜

            The k-th Largest Group
            Time Limit:2000MS? Memory Limit:131072K
            Total Submit:1222 Accepted:290

            Description

            Newman likes playing with cats. He possesses lots of cats in his home. Because the number of cats is really huge, Newman wants to group some of the cats. To do that, he first offers a number to each of the cat (1, 2, 3, …, n). Then he occasionally combines the group cat i is in and the group cat j is in, thus creating a new group. On top of that, Newman wants to know the size of the k-th biggest group at any time. So, being a friend of Newman, can you help him?

            Input

            1st line: Two numbers N and M (1 ≤ N, M ≤ 200,000), namely the number of cats and the number of operations.

            2nd to (m + 1)-th line: In each line, there is number C specifying the kind of operation Newman wants to do. If C = 0, then there are two numbers i and j (1 ≤ i, jn) following indicating Newman wants to combine the group containing the two cats (in case these two cats are in the same group, just do nothing); If C = 1, then there is only one number k (1 ≤ k ≤ the current number of groups) following indicating Newman wants to know the size of the k-th largest group.

            Output

            For every operation “1” in the input, output one number per line, specifying the size of the kth largest group.

            Sample Input

            10 10
            0 1 2
            1 4
            0 3 4
            1 2
            0 5 6
            1 1
            0 7 8
            1 1
            0 9 10
            1 1

            Sample Output

            1
            2
            2
            2
            2

            Hint

            When there are three numbers 2 and 2 and 1, the 2nd largest number is 2 and the 3rd largest number is 1.

            Source
            POJ Monthly--2006.08.27, zcgzcgzcg

            #include? < iostream >
            using ? namespace ?std;
            const ? int ?MAXN? = ? 200001 ;

            class ?UFset
            {
            public :
            ????
            int ?parent[MAXN];
            ????UFset();
            ????
            int ?Find( int );
            ????
            void ?Union( int ,? int );
            }
            ;

            UFset::UFset()
            {
            ????memset(parent,?
            - 1 ,? sizeof (parent));
            }


            int ?UFset::Find( int ?x)
            {
            ????
            if ?(parent[x]? < ? 0 )
            ????????
            return ?x;
            ????
            else
            ????
            {
            ????????parent[x]?
            = ?Find(parent[x]);
            ????????
            return ?parent[x];
            ????}
            // ?壓縮路徑
            }


            void ?UFset::Union( int ?x,? int ?y)
            {
            ????
            int ?pX? = ?Find(x);
            ????
            int ?pY? = ?Find(y);
            ????
            int ?tmp;
            ????
            if ?(pX? != ?pY)
            ????
            {
            ????????tmp?
            = ?parent[pX]? + ?parent[pY];? // ?加權合并
            ???????? if ?(parent[pX]? > ?parent[pY])
            ????????
            {
            ????????????parent[pX]?
            = ?pY;
            ????????????parent[pY]?
            = ?tmp;
            ????????}

            ????????
            else
            ????????
            {
            ????????????parent[pY]?
            = ?pX;
            ????????????parent[pX]?
            = ?tmp;
            ????????}

            ????}

            }


            int ?f[(MAXN + 1 ) * 3 ]? = ? { 0 } ;
            int ?n,?m;

            void ?initTree()
            {
            ????
            int ?l? = ? 1 ,?r? = ?n;
            ????
            int ?c? = ? 1 ;
            ????
            while ?(l? < ?r)
            ????
            {
            ????????f[c]?
            = ?n;
            ????????c?
            = ?c? * ? 2 ;
            ????????r?
            = ?(l? + ?r)? / ? 2 ;
            ????}

            ????f[c]?
            = ?n; // 葉子初始化
            }


            void ?insertTree( int ?k)
            {
            ????
            int ?l? = ? 1 ,?r? = ?n;
            ????
            int ?c? = ? 1 ;
            ????
            int ?mid;

            ????
            while ?(l? < ?r)
            ????
            {
            ????????f[c]
            ++ ;
            ????????mid?
            = ?(r? + ?l)? / ? 2 ;
            ????????
            if ?(k? > ?mid)
            ????????
            {
            ????????????l?
            = ?mid? + ? 1 ;
            ????????????c?
            = ?c? * ? 2 ? + ? 1 ;
            ????????}

            ????????
            else
            ????????
            {
            ????????????r?
            = ?mid;
            ????????????c?
            = ?c? * ? 2 ;
            ????????}

            ????}

            ????f[c]
            ++ ; // 葉子增加1
            }


            void ?delTree( int ?k)
            {
            ????
            int ?l? = ? 1 ,?r? = ?n;
            ????
            int ?c? = ? 1 ;
            ????
            int ?mid;

            ????
            while ?(l? < ?r)
            ????
            {
            ????????f[c]
            -- ;
            ????????mid?
            = ?(r? + ?l)? / ? 2 ;
            ????????
            if ?(k? > ?mid)
            ????????
            {
            ????????????l?
            = ?mid? + ? 1 ;
            ????????????c?
            = ?c? * ? 2 ? + ? 1 ;
            ????????}

            ????????
            else
            ????????
            {
            ????????????r?
            = ?mid;
            ????????????c?
            = ?c? * ? 2 ;
            ????????}

            ????}

            ????f[c]
            -- ; // 葉子減少1
            }


            int ?searchTree( int ?k)
            {
            ????
            int ?l? = ? 1 ,?r? = ?n;
            ????
            int ?c? = ? 1 ;
            ????
            int ?mid;

            ????
            while ?(l? < ?r)
            ????
            {
            ????????mid?
            = ?(l? + ?r)? / ? 2 ;
            ????????
            if ?(k? <= ?f[ 2 * c + 1 ])
            ????????
            {
            ????????????l?
            = ?mid? + ? 1 ;
            ????????????c?
            = ?c? * ? 2 ? + ? 1 ;
            ????????}

            ????????
            else
            ????????
            {
            ????????????k?
            -= ?f[ 2 * c + 1 ];
            ????????????r?
            = ?mid;
            ????????????c?
            = ?c? * ? 2 ;
            ????????}

            ????}

            ????
            return ?l;
            }


            int ?main()
            {
            ????
            int ?i,?j;
            ????
            int ?x,?y;
            ????
            int ?k;
            ????
            int ?l,?r;
            ????
            int ?cmd;
            ????
            int ?px,?py;
            ????
            int ?tx,?ty,?tz;
            ????UFset?UFS;

            ????
            ????scanf(
            " %d%d " ,? & n,? & m);
            ????initTree();
            ????
            for ?(i = 0 ;?i < m;?i ++ )
            ????
            {
            ????????scanf(
            " %d " ,? & cmd);
            ????????
            if ?(cmd? == ? 0 )
            ????????
            {
            ????????????scanf(
            " %d%d " ,? & x,? & y);
            ????????????px?
            = ?UFS.Find(x);
            ????????????py?
            = ?UFS.Find(y);
            ????????????
            if ?(px? != ?py)
            ????????????
            {
            ????????????????tx?
            = ? - UFS.parent[px];
            ????????????????ty?
            = ? - UFS.parent[py];
            ????????????????tz?
            = ?tx? + ?ty;
            ????????????????UFS.Union(x,?y);
            ????????????????insertTree(tz);
            ????????????????delTree(tx);
            ????????????????delTree(ty);
            ????????????}

            ????????}

            ????????
            else
            ????????
            {
            ????????????scanf(
            " %d " ,? & k);
            ????????????printf(
            " %d\n " ,?searchTree(k));
            ????????}

            ????}

            ????
            return ? 0 ;
            }
            posted on 2006-09-06 13:28 閱讀(816) 評論(4)  編輯 收藏 引用 所屬分類: ACM題目

            FeedBack:
            # re: pku2985 第一次用兩種數據結構解題, 并查集+線段樹 2006-09-22 13:24 A3
            可否講解一下線段樹部分  回復  更多評論
              
            # re: pku2985 第一次用兩種數據結構解題, 并查集+線段樹 2006-09-22 17:47 
            把區間劃出來, 節點(非葉子), 表示該區間里面含有多少個元素。
            如果 n = 10;
            而集合大小分別是 1, 1, 2, 6;

            則 區間(1-10) = 4; 區間(1-5) = 3;

            就這樣用線段樹動態維護每次集合合并后的集合大小。

            初始化(1-10) = 10;
            因為開始時, 集合大小為1, 1, 1, 1, 1, 1, 1, 1, 1, 1  回復  更多評論
              
            # re: pku2985 第一次用兩種數據結構解題, 并查集+線段樹 2006-09-24 19:53 Optimistic
            偶的第一次呢 靜待。。。  回復  更多評論
              
            # re: pku2985 第一次用兩種數據結構解題, 并查集+線段樹 2006-09-24 22:23 
            +U ^_^  回復  更多評論
              
            欧美一区二区三区久久综合 | 青青青青久久精品国产h| 99999久久久久久亚洲| 7国产欧美日韩综合天堂中文久久久久| 久久国产亚洲精品无码| 久久精品国产亚洲欧美| 欧美性大战久久久久久| 久久久久亚洲Av无码专| 久久九九久精品国产免费直播| 成人综合久久精品色婷婷| 久久人人爽人人爽人人AV东京热| 99久久免费只有精品国产| 久久久久亚洲AV成人网人人网站| 超级碰久久免费公开视频| 欧美日韩精品久久久久| 九九久久精品国产| 91精品国产乱码久久久久久| 欧美亚洲国产精品久久高清| 成人精品一区二区久久久| 欧美一区二区三区久久综合| 久久影院亚洲一区| 精品国产青草久久久久福利| 亚洲AV日韩精品久久久久久久| 久久精品国产亚洲7777| 1000部精品久久久久久久久| 中文字幕乱码久久午夜| 久久中文字幕人妻熟av女| 久久精品无码免费不卡| 久久99精品久久久久久水蜜桃| 精品久久久久久中文字幕人妻最新| 久久久久久久久久久| 久久久久国产精品人妻| 国产精品久久久久久五月尺| 伊人色综合九久久天天蜜桃 | 久久精品国产久精国产果冻传媒| 精品无码久久久久久久动漫| 亚洲午夜久久影院| 99久久精品国产一区二区| 品成人欧美大片久久国产欧美| 狠狠干狠狠久久| 精品久久久久中文字幕一区|