• <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>

            Google code jam 2008 R1A - Milkshakes

            Problem

            You own a milkshake shop. There are N different flavors that you can prepare, and each flavor can be prepared "malted" or "unmalted". So, you can make 2N different types of milkshakes.

            Each of your customers has a set of milkshake types that they like, and they will be satisfied if you have at least one of those types prepared. At most one of the types a customer likes will be a "malted" flavor.

            You want to make N batches of milkshakes, so that:

            • There is exactly one batch for each flavor of milkshake, and it is either malted or unmalted.
            • For each customer, you make at least one milkshake type that they like.
            • The minimum possible number of batches are malted.

            Find whether it is possible to satisfy all your customers given these constraints, and if it is, what milkshake types you should make.

            If it is possible to satisfy all your customers, there will be only one answer which minimizes the number of malted batches.

            Input

            • One line containing an integer C, the number of test cases in the input file.

            For each test case, there will be:

            • One line containing the integer N, the number of milkshake flavors.
            • One line containing the integer M, the number of customers.
            • M lines, one for each customer, each containing:
              • An integer T >= 1, the number of milkshake types the customer likes, followed by
              • T pairs of integers "X Y", one for each type the customer likes, where X is the milkshake flavor between 1 and N inclusive, and Y is either 0 to indicate unmalted, or 1 to indicated malted. Note that:
                • No pair will occur more than once for a single customer.
                • Each customer will have at least one flavor that they like (T >= 1).
                • Each customer will like at most one malted flavor. (At most one pair for each customer has Y = 1).
              All of these numbers are separated by single spaces.

            Output

            • C lines, one for each test case in the order they occur in the input file, each containing the string "Case #X: " where X is the number of the test case, starting from 1, followed by:
              • The string "IMPOSSIBLE", if the customers' preferences cannot be satisfied; OR
              • N space-separated integers, one for each flavor from 1 to N, which are 0 if the corresponding flavor should be prepared unmalted, and 1 if it should be malted.

            Limits

            Small dataset

            C = 100
            1 <= N <= 10
            1 <= M <= 100

            Large dataset

            C = 5
            1 <= N <= 2000
            1 <= M <= 2000

            The sum of all the T values for the customers in a test case will not exceed 3000.

            Sample


            Input
             

            Output
             
            2
            5
            3
            1 1 1
            2 1 0 2 0
            1 5 0
            1
            2
            1 1 0
            1 1 1
            Case #1: 1 0 0 0 0
            Case #2: IMPOSSIBLE

            In the first case, you must make flavor #1 malted, to satisfy the first customer. Every other flavor can be unmalted. The second customer is satisfied by getting flavor #2 unmalted, and the third customer is satisfied by getting flavor #5 unmalted.

            In the second case, there is only one flavor. One of your customers wants it malted and one wants it unmalted. You cannot satisfy them both.

            Analysis
            On the surface, this problem appears to require solving the classic problem "Satisfiability," the canonical example of an NP-complete problem. The customers represent clauses, the milkshake flavors represent variables, and malted and unmalted flavors represent whether the variable is negated.

            We are not evil enough to have chosen a problem that hard! The restriction that makes this problem easier is that the customers can only like at most one malted flavor (or equivalently, the clauses can only have at most one negated variable.)

            Using the following steps, we can quickly find whether a solution exists, and if so, what the solution is.

            1. Start with every flavor unmalted and consider the customers one by one.
            2. If there is an unsatisfied customer who only likes unmalted flavors, and all those flavors have been made malted, then no solution is possible.
            3. If there is an unsatisfied customer who has one favorite malted flavor, then we must make that flavor malted. We do this, then go back to step 2.
            4. If there are no unsatisfied customers, then we already have a valid solution and can leave the remaining flavors unmalted.

            Notice that whenever we made a flavor malted, we were forced to do so. Therefore, the solution we got must have the minimum possible number of malted flavors.

            With clever data structures, the above algorithm can be implemented to run in linear time.

            More information:

            The Satisfiability problem - Horn clauses


            Source Code
            #include <iostream>

            using namespace std;

            #define Rep(i,n) for (int i(0),_n(n); i<_n; ++i)

            struct Flavor{
                
            int X;
                
            char Y;
            }
            ;

            struct Customer{
                
            int T;
                Flavor
            * F;
                Customer() 
            {
                    F 
            = NULL;
                }

                
            ~Customer() {
                    
            if(NULL!=F) {
                        delete[] F;
                        F 
            = NULL;
                    }

                }

                
            void Init(int t) {
                    T 
            = t;
                    F 
            = new Flavor[T];
                }

                
            void SetFlavor(int i, int X, int Y) {
                    F[i].X 
            = X;
                    F[i].Y 
            = Y;
                }

                
            int GetFlavorX(int i) {
                    
            return F[i].X;
                }

                
            int GetFlavorY(int i) {
                    
            return F[i].Y;
                }

                
            bool IsSatisfied() {
                    
            return T==0;
                }

                
            void Satisfy() {
                    T
            =0;
                    
            if(NULL!=F) {
                        delete[] F;
                        F 
            = NULL;
                    }

                }

                
            bool IsSatisfing(int i, int *f) {
                    
            return f[F[i].X]==F[i].Y;
                }

                
            void SetMalted(int i, int *f) {
                    f[F[i].X] 
            = 1;
                }

            }
            ;

            int main()
            {
                
            int C;
                FILE 
            *fp = fopen("A.out""w");
                scanf(
            "%d"&C);
                Rep(c, C) 
            {
                    
            int N;
                    scanf(
            "%d"&N);
                    
            int* f = new int[N+1];
                    Rep(i ,N
            +1{
                        f[i]
            =0;
                    }

                    
            int M;
                    scanf(
            "%d"&M);
                    Customer
            * customer = new Customer[M];
                    Rep(m, M) 
            {
                        
            int T;
                        scanf(
            "%d"&T);
                        customer[m].Init(T);
                        Rep(t, T) 
            {
                            
            int X, Y;
                            scanf(
            "%d%d"&X,&Y);
                            customer[m].SetFlavor(t,X,Y);
                        }

                    }

                    
            bool findSolution = true;
                    
            int m = 0;
                    
            while(m<M) {
                        
            if(customer[m].IsSatisfied()) {
                            m
            ++;
                            
            continue;
                        }

                        
            bool malted = false;
                        
            int idx;
                        
            bool satisfied = false;
                        Rep(t, customer[m].T) 
            {
                            
            if(customer[m].GetFlavorY(t)==1{
                                malted 
            = true;
                                idx 
            = t;
                            }

                            
            if(customer[m].IsSatisfing(t,f)) {
                                satisfied 
            = true;
                            }

                        }

                        
            if(!satisfied) {
                            
            if(malted) {
                                customer[m].SetMalted(idx,f);
                                customer[m].Satisfy();
                                m
            =0;
                            }
             else {
                                findSolution 
            = false;
                                
            break;
                            }

                        }
             else {
                            m
            ++;
                        }

                    }

                    fprintf(fp,
            "Case #%d: ", c+1);
                    
            if(findSolution) {
                        Rep(i ,N) 
            {
                            fprintf(fp,
            "%d ", f[i+1]);
                        }

                        fprintf(fp,
            "\n");
                    }
             else {
                        fprintf(fp,
            "IMPOSSIBLE\n");
                    }

                    delete[] customer;
                    delete[] f;

                }

                fclose(fp);
            }

            posted on 2009-08-12 21:17 Chauncey 閱讀(409) 評論(0)  編輯 收藏 引用

            導航

            <2009年8月>
            2627282930311
            2345678
            9101112131415
            16171819202122
            23242526272829
            303112345

            統計

            常用鏈接

            留言簿

            隨筆檔案(4)

            文章檔案(3)

            搜索

            最新評論

            閱讀排行榜

            評論排行榜

            亚洲国产精品久久电影欧美| 色综合久久综精品| 亚洲精品无码久久千人斩| 亚洲精品国产美女久久久| 国产人久久人人人人爽| 久久精品国产第一区二区| 欧美伊人久久大香线蕉综合| 国产精品国色综合久久| 久久久久亚洲精品天堂久久久久久 | 国产精品亚洲综合久久| 久久久精品人妻一区二区三区四| 精品久久香蕉国产线看观看亚洲 | 18禁黄久久久AAA片| 久久这里只有精品久久| 久久99精品久久久大学生| 国产福利电影一区二区三区久久老子无码午夜伦不 | 久久国产综合精品五月天| 久久青青草原精品国产| 午夜精品久久影院蜜桃| 久久伊人精品青青草原高清| 国产激情久久久久久熟女老人| 狠狠久久综合伊人不卡| 精品久久人妻av中文字幕| 久久精品一本到99热免费| 国产—久久香蕉国产线看观看 | 一本久久a久久精品综合香蕉 | 久久国产美女免费观看精品| 国产精品99久久免费观看| 国产成人精品三上悠亚久久| 久久久精品波多野结衣| 91久久九九无码成人网站| 久久国产精品久久国产精品| 国内精品人妻无码久久久影院| 色婷婷综合久久久久中文| 狠狠色噜噜色狠狠狠综合久久| 久久久久久久91精品免费观看| 亚洲欧洲精品成人久久曰影片 | 东方aⅴ免费观看久久av| 国产精品一区二区久久精品涩爱| 久久乐国产综合亚洲精品| 国产69精品久久久久久人妻精品|