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            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(816) 評(píng)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(píng)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(diǎn)(非葉子), 表示該區(qū)間里面含有多少個(gè)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(dòng)態(tài)維護(hù)每次集合合并后的集合大小。

            初始化(1-10) = 10;
            因?yàn)殚_始時(shí), 集合大小為1, 1, 1, 1, 1, 1, 1, 1, 1, 1  回復(fù)  更多評(píng)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(píng)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 
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