• <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
            <2007年4月>
            25262728293031
            1234567
            891011121314
            15161718192021
            22232425262728
            293012345

            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 - 217832
            • 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
              
            久久综合给合久久狠狠狠97色 | 久久99久国产麻精品66| 久久男人AV资源网站| 欧美色综合久久久久久| 99久久99久久精品国产片果冻 | 国产精品久久久久免费a∨| 精品久久久久香蕉网| 久久精品三级视频| 久久99精品国产99久久6男男| 亚洲国产成人精品女人久久久 | 国色天香久久久久久久小说| 色综合色天天久久婷婷基地| 国产aⅴ激情无码久久| 青青青国产成人久久111网站| 久久综合鬼色88久久精品综合自在自线噜噜 | 国产精品无码久久综合| 午夜精品久久久久久影视777| 精品乱码久久久久久久| 欧美国产成人久久精品| 中文字幕久久欲求不满| 亚洲天堂久久精品| 亚洲国产一成人久久精品| 一本色道久久88精品综合| 久久人妻少妇嫩草AV无码蜜桃| AV狠狠色丁香婷婷综合久久| 久久人人爽人人爽人人片av麻烦| 国产三级观看久久| 九九热久久免费视频| 久久久久亚洲精品无码网址| 国产精品久久网| 久久国产精品免费一区| 伊人久久大香线蕉影院95| 久久精品中文字幕久久| 秋霞久久国产精品电影院| 久久青青草原精品影院| 国产精品久久永久免费| 欧美亚洲另类久久综合| 久久国产福利免费| 亚洲国产一成久久精品国产成人综合 | 亚洲熟妇无码另类久久久| 无码超乳爆乳中文字幕久久|