• <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
            <2006年2月>
            2930311234
            567891011
            12131415161718
            19202122232425
            2627281234
            567891011

            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 - 216603
            • 排名 - 117

            最新評(píng)論

            閱讀排行榜

            評(píng)論排行榜

            PKU 3093 Margaritas on the River Walk
                    先對(duì)輸入的數(shù)組排序,然后類似于01對(duì)a[i]做決策,核心代碼加了注釋:
                     for (i=1; i<=n; i++) {
                             for (j=1; j<=maxsum; j++) {
                                    if (j >= sum[i]) d[i][j] = 1; //j比sum[i]大,肯定這時(shí)候d[i][j]=1;
                                    else {
                                            d[i][j] = d[i-1][j];//不考慮a[i]
                                            if (j-a[i]>=0) {//考慮a[i]
                                                     if (d[i-1][j-a[i]] > 0) d[i][j] += d[i-1][j-a[i]];//把a(bǔ)[i]加進(jìn)以前的選擇里面
                                                     else d[i][j]++;//a[i]單獨(dú)作為一個(gè)選擇(這里需要先對(duì)a[i]排序,消除后效性)
                                           }
                                    }
                             }
                     }

            PKU 1037 A decorative fence
                    先dp算出以i為起點(diǎn)的序列的個(gè)數(shù),再組合數(shù)學(xué)
                    td[n][i]和tu[n][i]分別表示個(gè)數(shù)為n,以i開始的上升和下降的序列個(gè)數(shù)
                    易知:
                    td[n][1] = 0;
                    td[n][i] = sigma(tu[n-1][j], j從1..i-1)  = td[n][i-1] + tu[n-1][i-1] ;
                    tu[n][i]  = td[n][n+i-1];

            PKU 2677 Tour
                    雙調(diào)歐幾里德旅行商問題(明顯階段dp)
                    動(dòng)態(tài)規(guī)劃方程 :d[i+1][i] = mint(d[i+1][i], d[i][j]+g[j][i+1]); 
                                                  d[i+1][j] = mint(d[i+1][j], d[i][j]+g[i][i+1]);
                                                   0<=j<i   

            PKU 2288 Islands and Bridges
                    集合DP
                    狀態(tài)表示: d[i][j][k] (i為13為二進(jìn)制表示點(diǎn)的狀態(tài), j為當(dāng)前節(jié)點(diǎn), k為到達(dá)j的前驅(qū)節(jié)點(diǎn))

            posted on 2007-04-20 18:10 閱讀(2119) 評(píng)論(5)  編輯 收藏 引用 所屬分類: 算法&ACM

            FeedBack:
            # re: 對(duì)一些DP題目的小結(jié) 2007-04-22 08:56 byron
            豪大牛,問一下,這是一些題目嗎????  回復(fù)  更多評(píng)論
              
            # re: 對(duì)一些DP題目的小結(jié) 2007-04-24 00:52 
            @byron
            是pku上的題目,我菜菜啊。。。  回復(fù)  更多評(píng)論
              
            # re: 對(duì)一些DP題目的小結(jié) 2007-04-26 18:59 oyjpart
            呵呵 就聊上了啊 :)  回復(fù)  更多評(píng)論
              
            # re: 對(duì)一些DP題目的小結(jié) 2007-06-30 22:55 姜雨生
            Margaritas on the River Walk
            Time Limit:1000MS Memory Limit:65536K
            Total Submit:309 Accepted:132

            Description


            One of the more popular activities in San Antonio is to enjoy margaritas in the park along the river know as the River Walk. Margaritas may be purchased at many establishments along the River Walk from fancy hotels to Joe’s Taco and Margarita stand. (The problem is not to find out how Joe got a liquor license. That involves Texas politics and thus is much too difficult for an ACM contest problem.) The prices of the margaritas vary depending on the amount and quality of the ingredients and the ambience of the establishment. You have allocated a certain amount of money to sampling different margaritas.

            Given the price of a single margarita (including applicable taxes and gratuities) at each of the various establishments and the amount allocated to sampling the margaritas, find out how many different maximal combinations, choosing at most one margarita from each establishment, you can purchase. A valid combination must have a total price no more than the allocated amount and the unused amount (allocated amount – total price) must be less than the price of any establishment that was not selected. (Otherwise you could add that establishment to the combination.)

            For example, suppose you have $25 to spend and the prices (whole dollar amounts) are:

            Vendor A B C D H J
            Price 8 9 8 7 16 5

            Then possible combinations (with their prices) are:

            ABC(25), ABD(24), ABJ(22), ACD(23), ACJ(21), ADJ( 20), AH(24), BCD(24), BCJ(22), BDJ(21), BH(25), CDJ(20), CH(24), DH(23) and HJ(21).

            Thus the total number of combinations is 15.


            Input


            The input begins with a line containing an integer value specifying the number of datasets that follow, N (1 ≤ N ≤ 1000). Each dataset starts with a line containing two integer values V and D representing the number of vendors (1 ≤ V ≤ 30) and the dollar amount to spend (1 ≤ D ≤ 1000) respectively. The two values will be separated by one or more spaces. The remainder of each dataset consists of one or more lines, each containing one or more integer values representing the cost of a margarita for each vendor. There will be a total of V cost values specified. The cost of a margarita is always at least one (1). Input values will be chosen so the result will fit in a 32 bit unsigned integer.


            Output


            For each problem instance, the output will be a single line containing the dataset number, followed by a single space and then the number of combinations for that problem instance.


            Sample Input


            2
            6 25
            8 9 8 7 16 5
            30 250
            1 2 3 4 5 6 7 8 9 10 11
            12 13 14 15 16 17 18 19 20
            21 22 23 24 25 26 27 28 29 30

            Sample Output


            1 15
            2 16509438

            Hint


            Note: Some solution methods for this problem may be exponential in the number of vendors. For these methods, the time limit may be exceeded on problem instances with a large number of vendors such as the second example below.


            Source
            Greater New York 2006
              回復(fù)  更多評(píng)論
              
            # re: 對(duì)一些DP題目的小結(jié) 2007-06-30 22:59 姜雨生
            應(yīng)該可以更加優(yōu)化  回復(fù)  更多評(píng)論
              
            国产成人久久激情91| 色综合久久最新中文字幕| 国产精品久久久久久久午夜片| 久久久久久久97| 中文字幕久久久久人妻| 97精品依人久久久大香线蕉97| 久久一区二区免费播放| 久久精品人妻一区二区三区| 国产高清国内精品福利99久久| 久久免费国产精品一区二区| 青青草原综合久久| 久久精品99无色码中文字幕| 亚洲国产日韩综合久久精品| 国产精品99久久久久久宅男小说| 亚洲美日韩Av中文字幕无码久久久妻妇 | 99久久伊人精品综合观看| 久久精品国产久精国产| 亚洲一区中文字幕久久| 久久99精品久久久久久水蜜桃| 精品久久久久国产免费| 久久一区二区免费播放| 亚洲AV日韩精品久久久久| 国产精品天天影视久久综合网| 国产精品亚洲综合专区片高清久久久 | 麻豆一区二区99久久久久| 91精品国产91久久综合| 久久久中文字幕日本| 亚洲国产另类久久久精品黑人 | 麻豆成人久久精品二区三区免费| 久久精品一区二区| 久久99亚洲综合精品首页| 国内精品人妻无码久久久影院导航| 国产成人精品白浆久久69| 亚洲国产精品嫩草影院久久| 日本强好片久久久久久AAA| 丁香五月综合久久激情| 亚洲欧洲日产国码无码久久99| 88久久精品无码一区二区毛片| 狠狠色噜噜色狠狠狠综合久久 | 中文字幕无码免费久久| Xx性欧美肥妇精品久久久久久|