• <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
            <2025年7月>
            293012345
            6789101112
            13141516171819
            20212223242526
            272829303112
            3456789

            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 - 217940
            • 排名 - 117

            最新評論

            閱讀排行榜

            評論排行榜

            Feed the dogs
            Time Limit:6000MS? Memory Limit:65536K
            Total Submit:1387 Accepted:222

            Description
            Wind loves pretty dogs very much, and she has n pet dogs. So Jiajia has to feed the dogs every day for Wind. Jiajia loves Wind, but not the dogs, so Jiajia use a special way to feed the dogs. At lunchtime, the dogs will stand on one line, numbered from 1 to n, the leftmost one is 1, the second one is 2, and so on. In each feeding, Jiajia choose an inteval[i,j], select the k-th pretty dog to feed. Of course Jiajia has his own way of deciding the pretty value of each dog. It should be noted that Jiajia do not want to feed any position too much, because it may cause some death of dogs. If so, Wind will be angry and the aftereffect will be serious. Hence any feeding inteval will not contain another completely, though the intervals may intersect with each other.

            Your task is to help Jiajia calculate which dog ate the food after each feeding.

            Input
            The first line contains n and m, indicates the number of dogs and the number of feedings.

            The second line contains n integers, describe the pretty value of each dog from left to right. You should notice that the dog with lower pretty value is prettier.

            Each of following m lines contain three integer i,j,k, it means that Jiajia feed the k-th pretty dog in this feeding.

            You can assume that n<100001 and m<50001.

            Output
            Output file has m lines. The i-th line should contain the pretty value of the dog who got the food in the i-th feeding.

            Sample Input

            7 2
            1 5 2 6 3 7 4
            1 5 3
            2 7 1
            

            Sample Output

            3
            2
            

            Source
            POJ Monthly--2006.02.26,zgl & twb

            #include? < iostream >
            #include?
            < algorithm >
            using ? namespace ?std;

            const ? int ?MAXN? = ? 100001 ;
            const ? int ?MAXM? = ? 50001 ;

            struct ?PosData
            {
            ????
            int ?b,?e;
            ????
            int ?k;
            ????
            int ?index;
            }
            ;

            struct ?Adata
            {
            ????
            int ?index;
            ????
            int ?incode;
            ????
            int ?data;
            }
            ;

            Adata?a[MAXN];
            int ?aa[MAXN];
            PosData?d[MAXM];
            int ?f[ 3 * (MAXN + 1 )]? = ? { 0 } ;
            int ?n,?m;
            int ? out [MAXM];

            inline?
            int ?inorder(PosData?x,?PosData?y)
            {
            ????
            return ?x.b? < ?y.b;
            }


            inline?
            int ?inorder2(Adata?x,?Adata?y)
            {
            ????
            return ?x.data? < ?y.data;
            }


            inline?
            int ?inorder3(Adata?x,?Adata?y)
            {
            ????
            return ?x.index? < ?y.index;
            }


            void ?initTree()
            {
            ????memset(f,?
            0 ,? sizeof (f));
            }


            void ?insertTree( int ?v)
            {
            ????
            int ?l? = ? 1 ,?r? = ?MAXN;
            ????
            int ?c? = ? 1 ;
            ????
            int ?mid;
            ????
            while ?(l? < ?r)
            ????
            {
            ????????mid?
            = ?(l? + ?r)? / ? 2 ;
            ????????f[c]
            ++ ;
            ????????
            if ?(v? > ?mid)
            ????????
            {
            ????????????c?
            = ?c? * ? 2 ? + ? 1 ;
            ????????????l?
            = ?mid? + ? 1 ;
            ????????}

            ????????
            else
            ????????
            {
            ????????????c?
            = ?c? * ? 2 ;
            ????????????r?
            = ?mid;
            ????????}

            ????}

            ????f[c]
            ++ ;
            }


            void ?delTree( int ?v)
            {
            ????
            int ?l? = ? 1 ,?r? = ?MAXN;
            ????
            int ?c? = ? 1 ;
            ????
            int ?mid;
            ????
            while ?(l? < ?r)
            ????
            {
            ????????mid?
            = ?(l? + ?r)? / ? 2 ;
            ????????f[c]
            -- ;
            ????????
            if ?(v? > ?mid)
            ????????
            {
            ????????????c?
            = ?c? * ? 2 ? + ? 1 ;
            ????????????l?
            = ?mid? + ? 1 ;
            ????????}

            ????????
            else
            ????????
            {
            ????????????c?
            = ?c? * ? 2 ;
            ????????????r?
            = ?mid;
            ????????}

            ????}

            ????f[c]
            -- ;
            }


            int ?searchTree( int ?k)
            {
            ????
            int ?l? = ? 1 ,?r? = ?MAXN;
            ????
            int ?c? = ? 1 ;
            ????
            int ?mid;
            ????
            while ?(l? < ?r)
            ????
            {
            ????????mid?
            = ?(l? + ?r)? / ? 2 ;
            ????????
            if ?(k? <= ?f[ 2 * c])
            ????????
            {
            ????????????c?
            = ? 2 ? * ?c;
            ????????????r?
            = ?mid;
            ????????}

            ????????
            else
            ????????
            {
            ????????????k?
            -= ?f[ 2 * c];
            ????????????c?
            = ? 2 ? * ?c? + ? 1 ;
            ????????????l?
            = ?mid? + ? 1 ;
            ????????}

            ????}

            ????
            return ?l;
            }


            int ?main()
            {
            ????
            int ?i,?j;

            ????scanf(
            " %d%d " ,? & n,? & m);

            ????
            for ?(i = 1 ;?i <= n;?i ++ )
            ????
            {
            ????????scanf(
            " %d " ,? & a[i].data);
            ????????a[i].index?
            = ?i;
            ????}

            ????sort(a
            + 1 ,?a + n + 1 ,?inorder2);
            ????
            // 離散化
            ???? for ?(i = 1 ;?i <= n;?i ++ )
            ????
            {
            ????????a[i].incode?
            = ?i;
            ????????aa[i]?
            = ?a[i].data;
            ????}

            ????sort(a
            + 1 ,?a + n + 1 ,?inorder3);

            ????
            for ?(i = 0 ;?i < m;?i ++ )
            ????
            {
            ????????scanf(
            " %d%d%d " ,? & d[i].b,? & d[i].e,? & d[i].k);
            ????????d[i].index?
            = ?i;
            ????}

            ????sort(d,?d
            + m,?inorder);

            ????
            for ?(j = d[ 0 ].b;?j <= d[ 0 ].e;?j ++ )
            ????????insertTree(a[j].incode);
            ????
            out [d[ 0 ].index]? = ?aa[searchTree(d[ 0 ].k)];

            ????
            for ?(i = 1 ;?i < m;?i ++ )
            ????
            {
            ????????
            if ?(d[i - 1 ].e? < ?d[i].b)
            ????????
            {
            ????????????
            for ?(j = d[i - 1 ].b;?j <= d[i - 1 ].e;?j ++ )
            ????????????????delTree(a[j].incode);
            ????????????
            for ?(j = d[i].b;?j <= d[i].e;?j ++ )
            ????????????????insertTree(a[j].incode);
            ????????}

            ????????
            else
            ????????
            {
            ????????????
            for ?(j = d[i - 1 ].b;?j < d[i].b;?j ++ )
            ????????????
            {
            ????????????????delTree(a[j].incode);
            ????????????}

            ????????????
            for ?(j = d[i - 1 ].e + 1 ;?j <= d[i].e;?j ++ )
            ????????????
            {
            ????????????????insertTree(a[j].incode);
            ????????????}

            ????????}

            ????????
            ????????
            out [d[i].index]? = ?aa[searchTree(d[i].k)];
            ????}


            ????
            for ?(i = 0 ;?i < m;?i ++ )
            ????????printf(
            " %d\n " ,? out [i]);
            ????

            ????
            return ? 0 ;
            }
            posted on 2006-09-13 23:40 閱讀(1924) 評論(6)  編輯 收藏 引用 所屬分類: ACM題目

            FeedBack:
            # re: pku(2761)離散化+線段樹 2007-03-28 18:58 過客
            怎么注釋這么少!!!!  回復(fù)  更多評論
              
            # re: pku(2761)離散化+線段樹 2007-08-15 19:06 lilu03555
            最近在學(xué)線段樹,今天用遞歸寫了一個線段樹,結(jié)果是錯的,實(shí)在是臺不爽了。。
            看了你寫的線段樹,真的是他開眼界。。
            佩服佩服。。  回復(fù)  更多評論
              
            # re: pku(2761)離散化+線段樹 2007-08-18 11:42 wcn(wengjiaxiang86@163.com)
            pku2104差不多的一道題,
            對k進(jìn)行變換,
            d[i].k=d[i].e-d[i]-b+2-d[i].k;
            變成求第k大的元素.你的程序似乎連sample都不對呀?
            請教一下....  回復(fù)  更多評論
              
            # re: pku(2761)離散化+線段樹 2007-08-18 21:07 wcn(wengjiaxiang86@163.com)
            呃,是我看錯了,不需要變換...
            不過在判區(qū)間的時候得稍微改動一下,
            不過超時了。  回復(fù)  更多評論
              
            # re: pku(2761)離散化+線段樹 2007-08-18 21:09 wcn(wengjiaxiang86@163.com)
            呃,是我看錯了,
            不需要做變換。只是程序在判區(qū)間重疊的時候,得再加一個包含.
            不過這樣2104還是過不去,超時了,繼續(xù)請教?  回復(fù)  更多評論
              
            # re: pku(2761)離散化+線段樹 2008-03-19 20:36 l-y-p
            強(qiáng)大,向牛人學(xué)習(xí)了……  回復(fù)  更多評論
              
            久久亚洲精品国产亚洲老地址| 久久精品国产99国产精偷| 久久国产精品一区| 久久人人爽人人爽人人片av麻烦| 亚洲精品美女久久777777| 一级做a爱片久久毛片| 午夜肉伦伦影院久久精品免费看国产一区二区三区 | 精品国产青草久久久久福利| 欧美色综合久久久久久| 欧美一区二区三区久久综合| 久久国产精品一区| 国产精品久久久天天影视| 久久午夜免费视频| 国产高潮久久免费观看| 久久无码人妻一区二区三区午夜 | 一本久道久久综合狠狠躁AV| 国产午夜免费高清久久影院| 亚洲国产精品无码久久青草| 99久久免费只有精品国产| 亚洲精品无码久久一线| 久久99这里只有精品国产| 国产69精品久久久久9999| 久久久久成人精品无码中文字幕 | 久久午夜夜伦鲁鲁片免费无码影视 | 国产精品久久久久久久久久影院 | 久久精品国产亚洲AV忘忧草18| 亚洲成人精品久久| 久久精品国产精品青草app| 人妻无码αv中文字幕久久| 久久只有这精品99| 亚洲国产成人精品久久久国产成人一区二区三区综 | 日本免费久久久久久久网站| 久久综合噜噜激激的五月天| 天堂无码久久综合东京热| 久久亚洲精品无码观看不卡| 999久久久免费国产精品播放| 狠狠88综合久久久久综合网| 久久精品无码专区免费东京热| 五月丁香综合激情六月久久 | 国产精品久久久久久久久久免费| 国产精品久久久久国产A级|