• <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 - 217869
            • 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
              
            99久久免费只有精品国产| 久久中文字幕人妻熟av女| 99麻豆久久久国产精品免费| 久久国产乱子伦免费精品| 久久精品aⅴ无码中文字字幕不卡| 久久久久亚洲Av无码专| 国产精品99久久精品爆乳| 国产精品99久久久久久宅男小说| 久久久av波多野一区二区| 51久久夜色精品国产| 国产精品乱码久久久久久软件| 久久国产欧美日韩精品| 久久夜色精品国产亚洲av| 色婷婷综合久久久中文字幕| 精品人妻伦九区久久AAA片69| 伊人久久无码中文字幕| 久久青青草原综合伊人| 久久久久久亚洲AV无码专区| 久久精品无码免费不卡| 久久婷婷五月综合色高清| 亚洲国产精品无码久久九九| 久久精品男人影院| 69久久精品无码一区二区| 伊人伊成久久人综合网777| 国产三级精品久久| 久久棈精品久久久久久噜噜| 欧美精品乱码99久久蜜桃| 精品无码久久久久久久动漫| www久久久天天com| 久久精品国产亚洲AV嫖农村妇女 | 久久人人爽人人爽人人片AV不| 91精品国产91久久久久久青草| 久久婷婷五月综合97色| 久久精品成人免费国产片小草| 国产精品久久久久久久 | 久久婷婷国产麻豆91天堂| 人妻无码αv中文字幕久久琪琪布 人妻无码久久一区二区三区免费 人妻无码中文久久久久专区 | 久久精品国产亚洲av麻豆小说| 久久无码专区国产精品发布| 久久精品国产男包| 久久这里只有精品首页|