• <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 - 218040
            • 排名 - 117

            最新評論

            閱讀排行榜

            評論排行榜

            開始時候粗心,狀態轉移時候k寫成k-1了,查了n久.

            The Mailboxes Manufacturers Problem
            Time Limit:1000MS? Memory Limit:65536K
            Total Submit:299 Accepted:227

            Description

            In the good old days when Swedish children were still allowed to blowup their fingers with fire-crackers, gangs of excited kids would plague certain smaller cities during Easter time, with only one thing in mind: To blow things up. Small boxes were easy to blow up, and thus mailboxes became a popular target. Now, a small mailbox manufacturer is interested in how many fire-crackers his new mailbox prototype can withstand without exploding and has hired you to help him. He will provide you with k (1 ≤ k ≤ 10) identical mailbox prototypes each fitting up to m (1 ≤ m ≤ 100) crackers. However, he is not sure of how many firecrackers he needs to provide you with in order for you to be able to solve his problem, so he asks you. You think for a while and then say, “Well,if I blow up a mailbox I can’t use it again, so if you would provide me with only k = 1 mailboxes, I would have to start testing with 1 cracker, then 2 crackers, and so on until it finally exploded. In the worst case, that is if it does not blow up even when filled with m crackers, I would need 1 + 2 + 3 + … + m = m × (m + 1) ? 2 crackers. If m = 100 that would mean more than 5000 fire-crackers!” “That’s too many,” he replies. “What if I give you more than k = 1 mailboxes? Can you find a strategy that requires less crackers?”

            Can you? And what is the minimum number of crackers that you should ask him to provide you with?

            You may assume the following:

            1. If a mailbox can withstand x fire-crackers, it can also withstand x ? 1 fire-crackers.
            2. Upon an explosion, a mailbox is either totally destroyed (blown up) or unharmed, which means that it can be reused in another test explosion.

            Note: If the mailbox can withstand a full load of m fire-crackers, then the manufacturer will of course be satisfied with that answer. But otherwise he is looking for the maximum number of crackers that his mailboxes can withstand.

            Input

            The input starts with a single integer N (1 ≤ N ≤ 10) indicating the number of test cases to follow. Each test case is described by a line containing two integers: k and m, separated by a single space.

            Output

            For each test case print one line with a single integer indicating the minimum number of fire-crackers that is needed, in the worst case, in order to figure out how many crackers the mailbox prototype can withstand.

            Sample Input

            4
            1 10
            1 100
            3 73
            5 100

            Sample Output

            55
            5050
            382
            495

            Source
            Svenskt M?sterskap i Programmering/Norgesmesterskapet 2002

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

            const?int?INF?=?1?<<?28;

            int?d[11][101][101];
            int?sum(int?i,?int?j)?{
            ????
            int?ret?=?0,?k;
            ????
            for?(k=i;?k<=j;?k++)?ret?+=?k;
            ????return?ret;
            }

            int?max(int?a,?int?b)?{
            ????return?a?
            >?b???a?:?b;
            }


            int?main()?{
            ????
            int?caseTime;?
            ????
            int?i,?j,?k,?t,?K,?M,?l;
            ????scanf(
            "%d",?&caseTime);
            ????
            ????
            while?(caseTime--)?{
            ????????scanf(
            "%d%d",?&K,?&M);
            ????????
            for?(i=1;?i<=M;?i++)?{
            ????????????
            for?(j=i;?j<=M;?j++)?{
            ????????????????d[
            1][i][j]?=?sum(i,?j);
            ????????????}
            ????????}
            ????????
            for?(k=2;?k<=K;?k++)?{
            ????????????
            for?(l=0;?l<M;?l++)?{
            ????????????????
            for?(i=1;?i+l<=M;?i++)?{
            ????????????????????j?
            =?i?+?l;
            ????????????????????
            if?(i?==?j)?{
            ????????????????????????d[k][i][j]?
            =?i;
            ????????????????????????continue;
            ????????????????????}
            ????????????????????d[k][i][j]?
            =?INF;
            ????????????????????
            for?(t=i;?t<=j;?t++)?{
            ????????????????????????
            int?tmp;
            ????????????????????????
            if?(t?==?i)?tmp?=?d[k][i+1][j];
            ????????????????????????
            else?if?(t?==?j)?tmp?=?d[k-1][i][j-1];
            ????????????????????????
            else?tmp?=?max(d[k-1][i][t-1],?d[k-1][t+1][j]);
            ????????????????????????tmp?
            =?max(d[k-1][i][t-1],?d[k][t+1][j]);
            ????????????????????????
            if?(d[k][i][j]?>?t?+?tmp)?d[k][i][j]?=?t?+?tmp;
            ????????????????????}
            ????????????????}
            ????????????}
            ????????}
            ????????printf(
            "%d\n",?d[K][1][M]);
            ????}

            ????return?
            0;
            }
            posted on 2007-03-26 00:41 閱讀(2214) 評論(2)  編輯 收藏 引用 所屬分類: ACM題目

            FeedBack:
            # re: pku2904 3維dp 2007-03-27 16:31 litianze
            我是一個剛剛開始做acm題的菜鳥,望大哥幫幫忙,可以介紹一下解決的思想嗎?小弟先謝謝了!  回復  更多評論
              
            # re: pku2904 3維dp 2007-03-27 23:04 
            dp[k][i][j]表示k個郵筒時候放鞭炮數為i..j時候的最優值

            轉移方程為
            dp[k][i][j] = min{t+max(d[k-1][i][t-1],d[k][t+1][j])};

            狀態轉移時候就是考慮選t個鞭炮放時候爆或不爆  回復  更多評論
              
            久久一本综合| 99久久国产热无码精品免费久久久久| 热RE99久久精品国产66热| 亚洲国产天堂久久久久久| 久久国产精品一国产精品金尊| 久久99国产精品二区不卡| 午夜精品久久久久久久无码| 色欲久久久天天天综合网| 91精品国产91久久久久久蜜臀| 国产精品久久久久蜜芽| 日本福利片国产午夜久久| 久久久无码精品午夜| 婷婷综合久久中文字幕蜜桃三电影| 99久久精品免费看国产一区二区三区| 一级做a爰片久久毛片看看 | 久久婷婷是五月综合色狠狠| 996久久国产精品线观看| 一本一道久久综合狠狠老| 久久精品国产一区二区三区不卡| 韩国免费A级毛片久久| 久久久久久精品免费看SSS| 久久精品一区二区影院| 久久99热国产这有精品| av无码久久久久不卡免费网站 | 97久久久久人妻精品专区| 伊人 久久 精品| 久久精品国产72国产精福利| 久久美女人爽女人爽| 91精品国产综合久久婷婷| 久久久久无码精品国产不卡| 久久天天躁狠狠躁夜夜2020一 | 婷婷伊人久久大香线蕉AV| 日韩人妻无码一区二区三区久久99| 精品久久久久久久久久中文字幕 | 久久精品国产只有精品66| 国产激情久久久久影院小草 | 久久人妻无码中文字幕| 色婷婷噜噜久久国产精品12p| 欧美精品一区二区久久| 狠狠色丁香久久婷婷综合_中 | 久久精品国产亚洲精品|