• <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 - 218108
            • 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(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 13:30 閱讀(818) 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(shù)據(jù)結(jié)構(gòu)解的題目, 紀念一下 2006-09-08 23:01 Optimistic
            哇...偶木了  回復(fù)  更多評論
              
            # re: 第一次用兩種數(shù)據(jù)結(jié)構(gòu)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(fù)  更多評論
              
            # re: 第一次用兩種數(shù)據(jù)結(jié)構(gòu)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(fù)  更多評論
              
            # re: 第一次用兩種數(shù)據(jù)結(jié)構(gòu)解的題目, 紀念一下 2006-09-28 12:57 
            thx!~:)  回復(fù)  更多評論
              
            久久天天躁狠狠躁夜夜不卡| 亚洲精品乱码久久久久久久久久久久 | 国内精品久久久久| 国产精品一久久香蕉产线看| 久久天堂AV综合合色蜜桃网| 久久精品国产亚洲av日韩| 久久国产免费观看精品| 久久久噜噜噜久久| 精品久久久久久久| 久久亚洲精品成人无码网站| 成人久久综合网| 国产精品成人久久久| 青青草原1769久久免费播放| 人妻无码精品久久亚瑟影视 | 中文精品久久久久人妻不卡| 久久ZYZ资源站无码中文动漫| 中文字幕亚洲综合久久2| 中文字幕精品无码久久久久久3D日动漫| 模特私拍国产精品久久| 久久香蕉国产线看观看精品yw| 久久久精品人妻一区二区三区蜜桃 | 99久久国产主播综合精品| 人妻无码精品久久亚瑟影视| 久久se精品一区二区| 国产成人精品久久| 色悠久久久久久久综合网 | 久久久精品久久久久久| 精品久久久久久久无码 | 久久国产精品免费一区| 久久人人添人人爽添人人片牛牛| 一本色道久久88—综合亚洲精品 | 国产三级久久久精品麻豆三级 | 91精品国产色综久久| 国产精品一区二区久久不卡| 久久久久国产精品人妻| 亚洲国产视频久久| 久久久久亚洲AV无码去区首| 久久亚洲国产成人影院网站| 66精品综合久久久久久久| 夜夜亚洲天天久久| 久久亚洲国产成人影院网站|