• <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
            <2006年9月>
            272829303112
            3456789
            10111213141516
            17181920212223
            24252627282930
            1234567

            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 - 217828
            • 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(quán)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(815) 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(shù)據(jù)結(jié)構(gòu)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(fù)  更多評論
              
            # re: pku2985 第一次用兩種數(shù)據(jù)結(jié)構(gòu)解題, 并查集+線段樹 2006-09-22 17:47 
            把區(qū)間劃出來, 節(jié)點(非葉子), 表示該區(qū)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(qū)間(1-10) = 4; 區(qū)間(1-5) = 3;

            就這樣用線段樹動態(tài)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(fù)  更多評論
              
            # re: pku2985 第一次用兩種數(shù)據(jù)結(jié)構(gòu)解題, 并查集+線段樹 2006-09-24 19:53 Optimistic
            偶的第一次呢 靜待。。。  回復(fù)  更多評論
              
            # re: pku2985 第一次用兩種數(shù)據(jù)結(jié)構(gòu)解題, 并查集+線段樹 2006-09-24 22:23 
            +U ^_^  回復(fù)  更多評論
              
            夜夜亚洲天天久久| 嫩草影院久久国产精品| 中文成人久久久久影院免费观看| 伊人久久大香线蕉综合Av| 久久精品国产一区| 99精品国产99久久久久久97 | 99久久精品免费看国产| 久久99精品久久久久久9蜜桃 | 久久久久久毛片免费看| 99久久免费国产特黄| 久久久噜噜噜www成人网| 超级97碰碰碰碰久久久久最新| 国产综合精品久久亚洲| 波多野结衣久久一区二区| 久久电影网一区| 精品伊人久久久| 久久国产视频99电影| 久久99国产综合精品| 97久久精品人妻人人搡人人玩| 精品久久久久久国产免费了| 无码国内精品久久人妻| 国产亚洲精久久久久久无码| 人妻无码久久精品| 国内精品久久久久影院亚洲| 99久久夜色精品国产网站| 久久久久99精品成人片三人毛片| 日韩精品久久久久久免费| 好久久免费视频高清| 性欧美丰满熟妇XXXX性久久久| 99久久综合国产精品免费| 一本色道久久88综合日韩精品 | 久久久噜噜噜www成人网| 久久精品国产久精国产果冻传媒| 国内精品九九久久精品| 久久久久久精品免费看SSS| 日日狠狠久久偷偷色综合免费| 久久精品国产一区二区三区不卡| 久久精品国产亚洲Aⅴ香蕉| 国产一区二区精品久久岳| 欧美精品福利视频一区二区三区久久久精品 | 国产精品国色综合久久|