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            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(1917) 評(píng)論(6)  編輯 收藏 引用 所屬分類(lèi): ACM題目

            FeedBack:
            # re: pku(2761)離散化+線(xiàn)段樹(shù) 2007-03-28 18:58 過(guò)客
            怎么注釋這么少?。。?!  回復(fù)  更多評(píng)論
              
            # re: pku(2761)離散化+線(xiàn)段樹(shù) 2007-08-15 19:06 lilu03555
            最近在學(xué)線(xiàn)段樹(shù),今天用遞歸寫(xiě)了一個(gè)線(xiàn)段樹(shù),結(jié)果是錯(cuò)的,實(shí)在是臺(tái)不爽了。。
            看了你寫(xiě)的線(xiàn)段樹(shù),真的是他開(kāi)眼界。。
            佩服佩服。。  回復(fù)  更多評(píng)論
              
            # re: pku(2761)離散化+線(xiàn)段樹(shù) 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(xiàn)段樹(shù) 2007-08-18 21:07 wcn(wengjiaxiang86@163.com)
            呃,是我看錯(cuò)了,不需要變換...
            不過(guò)在判區(qū)間的時(shí)候得稍微改動(dòng)一下,
            不過(guò)超時(shí)了。  回復(fù)  更多評(píng)論
              
            # re: pku(2761)離散化+線(xiàn)段樹(shù) 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(guò)這樣2104還是過(guò)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(xiàn)段樹(shù) 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
              
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