• <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
            <2007年6月>
            272829303112
            3456789
            10111213141516
            17181920212223
            24252627282930
            1234567

            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 - 216593
            • 排名 - 117

            最新評論

            閱讀排行榜

            評論排行榜

            PKU 3093 Margaritas on the River Walk
                    先對輸入的數組排序,然后類似于01對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[i]加進以前的選擇里面
                                                     else d[i][j]++;//a[i]單獨作為一個選擇(這里需要先對a[i]排序,消除后效性)
                                           }
                                    }
                             }
                     }

            PKU 1037 A decorative fence
                    先dp算出以i為起點的序列的個數,再組合數學
                    td[n][i]和tu[n][i]分別表示個數為n,以i開始的上升和下降的序列個數
                    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
                    雙調歐幾里德旅行商問題(明顯階段dp)
                    動態規劃方程 :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
                    狀態表示: d[i][j][k] (i為13為二進制表示點的狀態, j為當前節點, k為到達j的前驅節點)

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

            FeedBack:
            # re: 對一些DP題目的小結 2007-04-22 08:56 byron
            豪大牛,問一下,這是一些題目嗎????  回復  更多評論
              
            # re: 對一些DP題目的小結 2007-04-24 00:52 
            @byron
            是pku上的題目,我菜菜啊。。。  回復  更多評論
              
            # re: 對一些DP題目的小結 2007-04-26 18:59 oyjpart
            呵呵 就聊上了啊 :)  回復  更多評論
              
            # re: 對一些DP題目的小結 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  更多評論
              
            # re: 對一些DP題目的小結 2007-06-30 22:59 姜雨生
            應該可以更加優化  回復  更多評論
              
            精品久久久久久久久久久久久久久| 国产成人久久精品一区二区三区| 国产一久久香蕉国产线看观看| 97久久精品人妻人人搡人人玩| 丁香狠狠色婷婷久久综合| 亚洲国产精品婷婷久久| 亚洲国产成人精品91久久久 | 99久久99久久久精品齐齐| 九九99精品久久久久久| 色播久久人人爽人人爽人人片aV| 久久天天躁狠狠躁夜夜网站| 国产叼嘿久久精品久久| 人妻无码αv中文字幕久久| 久久久WWW成人免费毛片| 久久久久久久亚洲Av无码| 久久99精品久久久久久野外| 奇米综合四色77777久久| 久久久久这里只有精品| www.久久热.com| 亚洲AV日韩精品久久久久久| 亚洲国产成人久久笫一页| 久久精品嫩草影院| 久久婷婷国产综合精品 | 四虎国产精品成人免费久久| 99久久精品国产一区二区| 久久人爽人人爽人人片AV| 中文字幕久久波多野结衣av| 日本精品一区二区久久久| 999久久久免费国产精品播放| 国内精品久久人妻互换| 久久精品成人欧美大片| 久久久久亚洲av综合波多野结衣| 久久久久亚洲AV成人网人人软件| 91精品观看91久久久久久| 国产精品对白刺激久久久| 丁香五月网久久综合| 久久亚洲精品国产精品| 久久天天躁狠狠躁夜夜躁2O2O | 久久精品国产亚洲av影院| 午夜天堂av天堂久久久| 亚洲国产另类久久久精品|