• <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3月>
            2627281234
            567891011
            12131415161718
            19202122232425
            2627282930311
            2345678

            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 - 217868
            • 排名 - 117

            最新評論

            閱讀排行榜

            評論排行榜

            USE?并查集和線段樹

            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 13:30 閱讀(816) 評論(4)  編輯 收藏 引用 所屬分類: 算法&ACM

            FeedBack:
            # re: 第一次用兩種數據結構解的題目, 紀念一下 2006-09-08 23:01 Optimistic
            哇...偶木了  回復  更多評論
              
            # re: 第一次用兩種數據結構解的題目, 紀念一下 2006-09-08 23:11 
            其實線段樹比較好懂, 但是難在怎么運用-_-個人感覺, 摸索中!~~~  回復  更多評論
              
            # re: 第一次用兩種數據結構解的題目, 紀念一下 2006-09-28 12:21 踏雪赤兔
            進步很快哩~~贊一個!
            P.S.博客手拉手弄好了~  回復  更多評論
              
            # re: 第一次用兩種數據結構解的題目, 紀念一下 2006-09-28 12:57 
            thx!~:)  回復  更多評論
              
            香蕉久久永久视频| 国产国产成人精品久久| 色综合久久天天综线观看| 久久人人爽人人澡人人高潮AV| 欧洲性大片xxxxx久久久| 99久久国产精品免费一区二区| 久久精品国产亚洲av麻豆色欲| 久久久精品国产亚洲成人满18免费网站| 办公室久久精品| 中文字幕久久精品无码| 国产精品综合久久第一页| 久久久久高潮综合影院| 久久这里只有精品首页| 亚洲女久久久噜噜噜熟女| 国内精品欧美久久精品| 久久这里只有精品首页| 国产2021久久精品| 久久国产免费观看精品3| 久久影视综合亚洲| 91久久精品电影| 久久夜色精品国产噜噜麻豆| 久久精品无码av| 77777亚洲午夜久久多喷| 99久久夜色精品国产网站 | 久久影院午夜理论片无码| 欧美丰满熟妇BBB久久久| 久久人妻无码中文字幕| 久久精品二区| 国产精品成人99久久久久| 国产精品福利一区二区久久| 亚洲午夜久久久久久噜噜噜| 国产精品一区二区久久精品涩爱 | 粉嫩小泬无遮挡久久久久久| 亚洲精品视频久久久| 天堂无码久久综合东京热| 久久黄视频| 国产亚洲色婷婷久久99精品91| 久久线看观看精品香蕉国产| 久久精品国产一区| 久久综合九色综合欧美狠狠| 免费观看久久精彩视频|