• <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>

            oyjpArt ACM/ICPC算法程序設計空間

            // I am new in programming, welcome to my blog
            I am oyjpart(alpc12, 四城)
            posts - 224, comments - 694, trackbacks - 0, articles - 6

            SRM406 PTS500 FoldThePaper

            Posted on 2008-06-18 11:29 oyjpart 閱讀(1560) 評論(0)  編輯 收藏 引用 所屬分類: ACM/ICPC或其他比賽程序設計

            Problem Statement

                 You have a rectangular piece of paper that's divided into 1x1 cells, each of which has an integer value. The paper will be described by a vector <string> paper. The ith element of paper will be a space delimited list of integers, where the jth integer of the ith element of paper represents the value of the jth cell of the ith row of the paper.



            You want to perform a sequence of folds on the paper, where you may fold anywhere along an axis that is in between two rows or columns of the paper. After performing a fold, we wish to model the folded paper as a new, flat piece of paper. We will do this by considering two overlapping cells as a single cell, with a value that is the sum of the individual cells.



            You wish to perform a sequence of folds such that the value of some single cell in the resulting piece of paper is as large as possible. Return this value.

            Definition

                
            Class: FoldThePaper
            Method: getValue
            Parameters: vector <string>
            Returns: int
            Method signature: int getValue(vector <string> paper)
            (be sure your method is public)
                

            Constraints

            - paper will contain between 1 and 12 elements, inclusive.
            - Each element of paper will be a single-space delimited list of integers with no leading or trailing spaces.
            - Each element of paper will contain between 1 and 12 integers, inclusive.
            - Each element of paper will contain the same number of integers.
            - Each element of paper will contain between 1 and 50 characters, inclusive.
            - Each integer in paper will be between -100 and 100, inclusive.
            - Each integer in paper will have no leading zeros.
            - An integer in paper equal to zero will not have a preceding negative sign.

            Examples

            0)
                
            {
            "1 1 1",
            "1 1 1"
            }
            Returns: 6
            We can collapse every cell onto the upper-left cell.
            1)
                
            {
            "1 -1",
            "1 -1"
            }
            Returns: 2
            We should perform only the fold between the two rows, and take the resulting left column.
            2)
                
            {
            "1 -1 -1 1",
            "-1 -1 -1 -1",
            "-1 -1 -1 -1",
            "1 -1 -1 1"
            }
            Returns: 4
            Folding between the middle rows then the middle columns allows us to combine the four corner cells.
            3)
                
            {
            "20 13 -2 100",
            "-12 0 4 -3",
            "4 1 -36 21"
            }
            Returns: 131

            4)
                
            {
            "0"
            }
            Returns: 0

            This problem statement is the exclusive and proprietary property of TopCoder, Inc. Any unauthorized use or reproduction of this information without the prior written consent of TopCoder, Inc. is strictly prohibited. (c)2003, TopCoder, Inc. All rights reserved.


            題目大意是有一個12*12的矩陣,現在可以對這個矩陣橫向或縱向折疊,出在重疊位置的數相加。
            求折疊過程中任意位置產生的最大數。

            很多大牛fail了,我一個DFS+剪枝也超時了,一共32人pass sys test,1000pts無人ac,此套題難度還是很大的。

            基本思路是狀態壓縮DP,橫向(1<<12)*縱向(1<<12)*加和。

            但是這樣會超時。關鍵是沒有利用到折疊的信息。

            預先生成某個位置的狀態(由那些位置疊加而來),就可以減少檢查量,就可以ac了。

            如何生成這些狀態呢?沒錯,又是一個DP. 呵呵。


            久久久久综合网久久| 久久99精品久久久久子伦| 久久成人18免费网站| 国产免费久久精品丫丫| 国产精品欧美久久久久天天影视| 狠狠精品干练久久久无码中文字幕| 久久久久久A亚洲欧洲AV冫| 久久亚洲精品成人无码网站| 亚洲AV无码久久| 免费国产99久久久香蕉| 日韩十八禁一区二区久久 | 国产欧美久久久精品| 99国内精品久久久久久久| 人妻中文久久久久| 国产精品久久久天天影视| 久久综合视频网站| 国产精品久久99| 久久亚洲sm情趣捆绑调教 | 久久精品二区| 久久久久99精品成人片试看| 久久久久国产视频电影| 丰满少妇高潮惨叫久久久| 一本大道久久香蕉成人网| 国产精品美女久久久久av爽 | 99久久国产宗和精品1上映| 国产成人精品久久综合| 国产成人精品久久| 人妻丰满?V无码久久不卡| 国产成人久久精品二区三区| 精品无码久久久久久尤物| 无码人妻久久一区二区三区蜜桃 | 香蕉久久一区二区不卡无毒影院 | 久久精品国产精品青草| 久久午夜伦鲁片免费无码| 亚洲午夜精品久久久久久app| 久久本道久久综合伊人| 大美女久久久久久j久久| 久久久久久久综合日本亚洲| 久久99精品久久久久久| 国产精品视频久久久| 色综合久久精品中文字幕首页 |