• <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(cè)

            ACM OJ

            My friends

            搜索

            •  

            積分與排名

            • 積分 - 217919
            • 排名 - 117

            最新評(píng)論

            閱讀排行榜

            評(píng)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(píng)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(píng)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(gè)線段樹,結(jié)果是錯(cuò)的,實(shí)在是臺(tái)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(píng)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(duì)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(duì)呀?
            請(qǐng)教一下....  回復(fù)  更多評(píng)論
              
            # re: pku(2761)離散化+線段樹 2007-08-18 21:07 wcn(wengjiaxiang86@163.com)
            呃,是我看錯(cuò)了,不需要變換...
            不過在判區(qū)間的時(shí)候得稍微改動(dòng)一下,
            不過超時(shí)了。  回復(fù)  更多評(píng)論
              
            # re: pku(2761)離散化+線段樹 2007-08-18 21:09 wcn(wengjiaxiang86@163.com)
            呃,是我看錯(cuò)了,
            不需要做變換。只是程序在判區(qū)間重疊的時(shí)候,得再加一個(gè)包含.
            不過這樣2104還是過不去,超時(shí)了,繼續(xù)請(qǐng)教?  回復(fù)  更多評(píng)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(píng)論
              
            色综合久久中文综合网| 日产精品久久久一区二区| 久久精品国产精品亚洲艾草网美妙 | 亚洲国产精品高清久久久| 久久婷婷国产麻豆91天堂| 94久久国产乱子伦精品免费| 久久亚洲国产最新网站| www.久久热.com| 久久热这里只有精品在线观看| 国产91久久精品一区二区| 青青青伊人色综合久久| 一本大道久久香蕉成人网| 香港aa三级久久三级| 四虎影视久久久免费| 狠狠色综合网站久久久久久久| 99久久精品国产麻豆| 久久国产福利免费| 久久婷婷国产综合精品| 国产精品无码久久四虎| 国产成人精品久久一区二区三区 | 亚洲精品乱码久久久久久| 精品久久久久中文字幕日本| 无夜精品久久久久久| 色婷婷久久综合中文久久蜜桃av | 久久精品欧美日韩精品| 久久无码专区国产精品发布| 久久av免费天堂小草播放| 无码人妻久久一区二区三区 | 一级做a爰片久久毛片16| 久久99热这里只有精品66| 99久久精品国产一区二区三区| 伊人久久综合无码成人网| 亚洲中文久久精品无码| 久久国产香蕉一区精品| jizzjizz国产精品久久| 久久国产免费直播| 久久精品中文字幕第23页| 精品国产一区二区三区久久| 久久久中文字幕| 久久精品九九亚洲精品| 99久久精品国产一区二区|