• <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
            <2010年5月>
            2526272829301
            2345678
            9101112131415
            16171819202122
            23242526272829
            303112345

            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 - 216593
            • 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
              
            亚洲а∨天堂久久精品| 久久精品国产亚洲AV无码娇色| 国产成人精品久久亚洲高清不卡 | 久久综合九色综合网站| 精品久久久无码人妻中文字幕| 99re这里只有精品热久久| 久久99热这里只有精品国产| 亚洲va久久久噜噜噜久久| 国产福利电影一区二区三区久久老子无码午夜伦不 | 久久精品国产一区| 久久综合亚洲鲁鲁五月天| 久久久久久久综合日本亚洲| 2021最新久久久视精品爱| 久久最新精品国产| 亚洲中文字幕久久精品无码APP| 久久青青草原综合伊人| 午夜精品久久久久久毛片| 色综合久久中文字幕综合网| 99久久精品费精品国产一区二区 | 亚洲中文字幕无码久久综合网| 国产精品熟女福利久久AV| 日日躁夜夜躁狠狠久久AV| 亚洲va久久久久| 久久精品国产亚洲Aⅴ香蕉| 成人国内精品久久久久一区| 久久91精品国产91| 久久亚洲中文字幕精品一区| 国产毛片久久久久久国产毛片| 99久久久国产精品免费无卡顿 | 99久久久久| 亚洲国产二区三区久久| 久久亚洲精品成人av无码网站| 久久91精品国产91久| 无码人妻久久一区二区三区蜜桃| 国产综合成人久久大片91| 国产ww久久久久久久久久| 国内精品久久久久久久亚洲| 免费观看成人久久网免费观看| 久久久精品免费国产四虎| 久久综合狠狠综合久久激情 | 青青草原综合久久|