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            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

            A decorative fence
            Time Limit:1000MS  Memory Limit:10000K
            Total Submit:1548 Accepted:440

            Description
            Richard just finished building his new house. Now the only thing the house misses is a cute little wooden fence. He had no idea how to make a wooden fence, so he decided to order one. Somehow he got his hands on the ACME Fence Catalogue 2002, the ultimate resource on cute little wooden fences. After reading its preface he already knew, what makes a little wooden fence cute.
            A wooden fence consists of N wooden planks, placed vertically in a row next to each other. A fence looks cute if and only if the following conditions are met:
            ?The planks have different lengths, namely 1, 2, . . . , N plank length units.
            ?Each plank with two neighbors is either larger than each of its neighbors or smaller than each of them. (Note that this makes the top of the fence alternately rise and fall.)
            It follows, that we may uniquely describe each cute fence with N planks as a permutation a1, . . . , aN of the numbers 1, . . . ,N such that (any i; 1 < i < N) (ai − ai−1)*(ai − ai+1) > 0 and vice versa, each such permutation describes a cute fence.
            It is obvious, that there are many di erent cute wooden fences made of N planks. To bring some order into their catalogue, the sales manager of ACME decided to order them in the following way: Fence A (represented by the permutation a1, . . . , aN) is in the catalogue before fence B (represented by b1, . . . , bN) if and only if there exists such i, that (any j < i) aj = bj and (ai < bi). (Also to decide, which of the two fences is earlier in the catalogue, take their corresponding permutations, find the first place on which they differ and compare the values on this place.) All the cute fences with N planks are numbered (starting from 1) in the order they appear in the catalogue. This number is called their catalogue number.


            After carefully examining all the cute little wooden fences, Richard decided to order some of them. For each of them he noted the number of its planks and its catalogue number. Later, as he met his friends, he wanted to show them the fences he ordered, but he lost the catalogue somewhere. The only thing he has got are his notes. Please help him find out, how will his fences look like.

             

            Input
            The first line of the input file contains the number K (1 <= K <= 100) of input data sets. K lines follow, each of them describes one input data set.
            Each of the following K lines contains two integers N and C (1 <= N <= 20), separated by a space. N is the number of planks in the fence, C is the catalogue number of the fence.
            You may assume, that the total number of cute little wooden fences with 20 planks fits into a 64-bit signed integer variable (long long in C/C++, int64 in FreePascal). You may also assume that the input is correct, in particular that C is at least 1 and it doesn抰 exceed the number of cute fences with N planks.

            Output
            For each input data set output one line, describing the C-th fence with N planks in the catalogue. More precisely, if the fence is described by the permutation a1, . . . , aN, then the corresponding line of the output file should contain the numbers ai (in the correct order), separated by single spaces.

            Sample Input

            2
            2 1
            3 3

             

            Sample Output

            1 2
            2 3 1

             

            Source
            CEOI 2002


            也算是DP+分段統計中的經典題了 呵呵
            DP的狀態表示如下
            dp[style][n][i][j] :
            style 代表走向 0 代表向上(也就是下次要向下) 1代表向下
            n代表總共的fence數
            i代表比當前選擇的fence高的fence數(注意 當前fence是一個隱藏的參數 因為該狀態不需要知道當前fence是哪個 只需要知道有多少比這個fence高 多少比這個fence低 就可以代表整個狀態)
            j代表比當前選擇的fence低的fence數

            這樣很直觀的得到了一個DP方程

                dp[0][i][j][k] += dp[1][i-1][j-m][k+m-1]; (m = 1 ... j (inclusive))

                dp[1][i][j][k] += dp[0][i-1][j+m-1][k-m];   (m = 1... k(inclusive))

            具體請參考源代碼

             1#include <stdio.h>
             2#include <string.h>
             3
             4const int N = 21;
             5__int64 dp[2][N][N][N];
             6int n;
             7__int64 idx;
             8bool chk[N];
             9
            10void pre() {
            11    int i, j, m;
            12
            13    memset(dp, 0, sizeof(dp));
            14
            15    dp[0][1][0][0= 1;
            16    dp[1][1][0][0= 1;
            17
            18    for(i = 2; i <= 20++i) {
            19        for(j = 0; j < i; ++j) {
            20            int k = i - j - 1;
            21            for(m = 1; m <= j; ++m) 
            22                dp[0][i][j][k] += dp[1][i-1][j-m][k+m-1];
            23            for(m = 1; m <= k; ++m)
            24                dp[1][i][j][k] += dp[0][i-1][j+m-1][k-m];
            25        }
            26    }
            27}
            28
            29void DFS(int nowint last, int style, __int64 idx) {
            30    if(now <= 0) return;
            31    int i, j;
            32    for(i = 0; i < n; ++i) if(!chk[i]) {
            33        if(style == 0 && i < last) continue;
            34        if(style == 1 && i > last) return;
            35
            36        chk[i] = true;
            37        int big = 0, small = 0;
            38        for(j = 0; j < n; ++j) if(!chk[j]) {
            39            if(j < i) small++;
            40            if(j > i) big++;
            41        }
            42
            43        if(style == 0 || style == -1) {
            44            if(idx > dp[1][now][big][small]) idx -= dp[1][now][big][small];
            45            else {
            46                printf("%d ", i+1);
            47                DFS(now-1, i, 1, idx);    return;
            48            }
            49        }
            50
            51        if(style == 1 || style == -1) {
            52            if(idx > dp[0][now][big][small]) idx -= dp[0][now][big][small];
            53            else {
            54                printf("%d ", i+1);
            55                DFS(now-1, i, 0, idx);    return;
            56            }
            57        }
            58        chk[i] = false;
            59    }
            60}
            61
            62int main() {
            63    int ntc;
            64    pre();
            65    scanf("%d "&ntc);
            66    while(ntc--) {
            67        scanf("%d %I64d"&n, &idx);
            68        memset(chk, false, sizeof(chk));
            69        DFS(n, -1-1, idx);
            70        putchar('\n');
            71    }
            72    return 0;
            73}

            Feedback

            # re: PKU1037 A decorative fence DP+分段統計  回復  更多評論   

            2007-08-18 15:57 by deoxyz
            你的最后一維根本不需要啊.......這樣空間復雜度會下降不少啊...

            # re: PKU1037 A decorative fence DP+分段統計  回復  更多評論   

            2007-08-18 16:27 by oyjpart
            真的么?
            那你怎么寫的呢?

            # re: PKU1037 A decorative fence DP+分段統計  回復  更多評論   

            2007-08-19 10:08 by deoxyz
            就你的這個程序的話 直接把所有有關最后一維的東西全部刪掉就行了,第三維完全可以用前兩維表示~而且輸出可以不用遞歸會快點~ 我ACM也剛學1年多點 有空交流交流 我QQ120148455

            # re: PKU1037 A decorative fence DP+分段統計  回復  更多評論   

            2007-08-19 11:05 by oyjpart
            恩 是的

            # re: PKU1037 A decorative fence DP+分段統計  回復  更多評論   

            2007-10-10 13:28 by floyd635
            的確只要三維就可以完成...
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