• <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>

            GoogleCodejam2009 Round 1C Bribe the Prisoners

            Bribe the Prisoners

            Problem

            In a kingdom there are prison cells (numbered 1 to P) built to form a straight line segment. Cells number i and i+1 are adjacent, and prisoners in adjacent cells are called "neighbours." A wall with a window separates adjacent cells, and neighbours can communicate through that window.

            All prisoners live in peace until a prisoner is released. When that happens, the released prisoner's neighbours find out, and each communicates this to his other neighbour. That prisoner passes it on to his other neighbour, and so on until they reach a prisoner with no other neighbour (because he is in cell 1, or in cell P, or the other adjacent cell is empty). A prisoner who discovers that another prisoner has been released will angrily break everything in his cell, unless he is bribed with a gold coin. So, after releasing a prisoner in cell A, all prisoners housed on either side of cell A - until cell 1, cell P or an empty cell - need to be bribed.

            Assume that each prison cell is initially occupied by exactly one prisoner, and that only one prisoner can be released per day. Given the list of Q prisoners to be released in Q days, find the minimum total number of gold coins needed as bribes if the prisoners may be released in any order.

            Note that each bribe only has an effect for one day. If a prisoner who was bribed yesterday hears about another released prisoner today, then he needs to be bribed again.

            Input

            The first line of input gives the number of cases, N. N test cases follow. Each case consists of 2 lines. The first line is formatted as

            P Q
            where P is the number of prison cells and Q is the number of prisoners to be released.
            This will be followed by a line with Q distinct cell numbers (of the prisoners to be released), space separated, sorted in ascending order.

             

            Output

            For each test case, output one line in the format

            Case #X: C
            where X is the case number, starting from 1, and C is the minimum number of gold coins needed as bribes.

             

            Limits

            1 ≤ N ≤ 100
            QP
            Each cell number is between 1 and P, inclusive.

            Small dataset

            1 ≤ P ≤ 100
            1 ≤ Q ≤ 5

            Large dataset

            1 ≤ P ≤ 10000
            1 ≤ Q ≤ 100

            Sample


            Input
             

            Output
             
            2
            8 1
            3
            20 3
            3 6 14
            Case #1: 7
            Case #2: 35

            Note

            In the second sample case, you first release the person in cell 14, then cell 6, then cell 3. The number of gold coins needed is 19 + 12 + 4 = 35. If you instead release the person in cell 6 first, the cost will be 19 + 4 + 13 = 36.


            題目分析:
            從直覺上來說,這是一道動態規劃的題目,關鍵是如何建立狀態遞推。有一個很明顯的規律是,釋放一個牢房的犯人,只能影響到左邊第一個空的牢房和右邊第一個空的牢房,而與其它的無關。所以,釋放了一個囚犯,整個連續的牢房被分成了2段,而這兩段又都可以看成是單獨的兩個互不影響的一段,這樣,腦子里有一種很直覺的想法就是,這是一棵類似于二叉樹的結構,這顆二叉樹最后的形態就決定了最終的結果。
            我們從第二個Case為例子介紹算法:
            1.首先對該題所提到的監獄增加兩個cell,0號和P + 1號, 這兩個cell是不存在的,只是為了增加程序的可編寫和公式的一致,我們用一個vector v 來存儲釋放囚犯的監獄號,v按照從小到達的順序排列,如第二個例子中,v(0...4) = (0,3, 6, 14, 21)
            2. 我們用F[i][j]表示 從v[i]到v[j]這一段中所需要的賄賂金最小值,那么,F[0][4]就是最后需要的結果, 代表從v[0]到v[4]也就是從(0, 21)這之間的牢房中釋放囚犯所需要的錢(不包括邊界,實際需要錢賄賂的囚犯在1到20號房子中)
            3. 例如第三個例子,F[0][4] = Min(F[0][i] + F[i][3]) + (v[4] - v[0] - 2) , 其可以中i = 1, 2, 3
                可以從第三個例子中看到,無論你選擇釋放哪一個囚犯,所需的金額都是一定的,正好是這段之間住人的牢房數 - 1
            4.更加抽象出最終的公式, 我們用L代表左邊界, R代表右邊界
                F[L][R] = Min(F[L][i] + F[i][R]) + v[R] - V[L] - 2,   i = L + 1, L + 2, ……R- 1,  當L + 1 != R時
                F[L][R] = 0, 當L + 1 == R時, 這個就是上面提到的二叉樹的葉子節點
               通過這個公式就可以得到最終的結果了。

             1#include <iostream>
             2#include <vector>
             3#include <queue>
             4#include <string>
             5#include <algorithm>
             6#include <set>
             7#include <map>
             8using namespace std;
             9
            10#define ONLINEJUDGE
            11#define MAXN 11000
            12#define MAXQ 110
            13#define MAXP 110
            14#define MIN(a, b) ((a) < (b) ? (a):(b))
            15
            16int P, Q;
            17vector<int> v;
            18vector<int> Ret, buf;
            19int F[MAXP][MAXP];
            20
            21int Find(int l, int r)
            22{
            23    if(l + 1 == r)
            24    {
            25        F[l][r] = 0;
            26        return F[l][r];
            27    }

            28    if(F[l][r] >= 0return F[l][r];
            29
            30    int i, j;
            31    int iMin;
            32    iMin = 999999999;
            33    for(i = l + 1; i < r; i++)
            34    {
            35        iMin = MIN(iMin, Find(l, i) + Find(i, r));
            36    }

            37    F[l][r] = iMin + v[r] - v[l] - 2;
            38    return F[l][r];
            39}

            40
            41int main()
            42{
            43#ifdef ONLINEJUDGE
            44    freopen("C-large.in""r", stdin);
            45    freopen("C-large.out""w", stdout);
            46#endif
            47
            48    int iCaseTimes, i, j;
            49    int iBuf;
            50    int iMax, iRet, iMin;
            51
            52    scanf("%d"&iCaseTimes);
            53    for(int k = 0; k < iCaseTimes; k++)
            54    {
            55        printf("Case #%d: ", k + 1);
            56        scanf("%d%d"&P, &Q);
            57        v.clear();
            58        v.push_back(0);
            59        for(i = 0; i < Q; i++)
            60        {
            61            scanf("%d"&iBuf);
            62            v.push_back(iBuf);
            63        }

            64        v.push_back(P + 1);
            65
            66        iMin = 0;
            67        memset(F, -1sizeof(F));
            68        sort(v.begin(), v.end());
            69        
            70        iMin = 999999999;
            71        for(i = 1; i < v.size() - 1; i++)
            72        {
            73            iMin = MIN(iMin, Find(0, i) + Find(i, v.size() - 1));
            74        }

            75        F[0][v.size() - 1= iMin + v[v.size() - 1- v[0- 2;
            76        printf("%d\n", F[0][v.size() - 1]);
            77    }

            78    return 0;
            79}

            posted on 2009-09-14 19:28 Philip85517 閱讀(1435) 評論(2)  編輯 收藏 引用 所屬分類: GoogleCodeJam

            評論

            # re: GoogleCodejam2009 Round 1C Bribe the Prisoners[未登錄] 2009-09-15 16:09 vincent

            googlecodejam啊 ..貌似是9月2日?
            當天有事情沒參加..orz  回復  更多評論   

            # re: GoogleCodejam2009 Round 1C Bribe the Prisoners[未登錄] 2009-09-15 23:32 Philip85517

            @vincent
            9月2號開始的是資格賽,這個是上個星期天的比賽。
            估計Round2 是死活進不去了,太弱了……
              回復  更多評論   

            導航

            <2009年9月>
            303112345
            6789101112
            13141516171819
            20212223242526
            27282930123
            45678910

            統計

            常用鏈接

            留言簿

            隨筆分類

            隨筆檔案

            文章分類

            文章檔案

            搜索

            最新評論

            閱讀排行榜

            評論排行榜

            久久久久亚洲精品中文字幕| 久久久久久夜精品精品免费啦| 久久久久久久久久久久中文字幕 | 精品人妻伦九区久久AAA片69| 久久无码精品一区二区三区| 欧美久久综合九色综合| 亚洲伊人久久精品影院| 久久无码av三级| 久久久久亚洲国产| 日本免费一区二区久久人人澡| 欧美久久一级内射wwwwww.| 久久亚洲精品人成综合网| 精品欧美一区二区三区久久久| 久久久久国产精品人妻| 精品国产91久久久久久久| 久久国产亚洲精品| 国产精品美女久久久久av爽| 奇米综合四色77777久久| 中文字幕无码av激情不卡久久| 国产精品18久久久久久vr| 久久久久se色偷偷亚洲精品av| 国产成人精品久久亚洲高清不卡 | 国产成人久久777777| 久久精品无码一区二区无码| 香蕉久久AⅤ一区二区三区| 久久综合九色综合精品| 色欲av伊人久久大香线蕉影院| 久久久久九九精品影院| 亚洲国产精品久久久久婷婷老年 | 久久99国产精品一区二区| 久久久久青草线蕉综合超碰| 久久成人精品| 国产—久久香蕉国产线看观看| 久久久久se色偷偷亚洲精品av| 亚洲一级Av无码毛片久久精品| 久久精品成人欧美大片| 国产精品久久久99| 国产亚洲精午夜久久久久久| 国产精品美女久久久免费| 94久久国产乱子伦精品免费| 久久se精品一区精品二区|