POJ 1017 -- Packets


http://poj.org/problem?id=1017
A factory produces products packed in square packets of the same height h and of the sizes 1*1, 2*2, 3*3, 4*4, 5*5, 6*6. These products are always delivered to customers in the square parcels of the same height h as the products have and of the size 6*6. Because of the expenses it is the interest of the factory as well as of the customer to minimize the number of parcels necessary to deliver the ordered products from the factory to the customer. A good program solving the problem of finding the minimal number of parcels necessary to deliver the given products according to an order would save a lot of money. You are asked to make such a program.
Input
The input file consists of several lines specifying orders. Each line specifies one order. Orders are described by six integers separated by one space representing successively the number of packets of individual size from the smallest size 1*1 to the biggest size 6*6. The end of the input file is indicated by the line containing six zeros.
Sample Input
0 0 4 0 0 1 
7 5 1 0 0 0 
0 0 0 0 0 0
http://www.2cto.com/kf/201310/249520.html
首先6*6,5*5,4*4的一次只能放一个,这样有多少个,就要用多少个包装袋,3*3的最多可以放4个,所以需要(num[3]+3)/4个包装袋,然后统计出空余部分2*2的个数。显然当放4*4时空出5个2*2,放3个3*3时,空出1个2*2;放2个3*3时,空出3个2*2;放1个3*3时,空出5个2*2;然后将2*2的全部搞定。最后用总的空间减去已放的空间就可以得到剩余的1*1的空间。
a/b的向上取整可以写为(a+(b-1))/b.
int main() 
    int n,a,b,c,d,e,f,x,y; 
    int u[4]={0,5,3,1}; //当放i个3*3时,空出来的2*2的个数 
   
    while(1) 
    
        scanf("%d%d%d%d%d%d",&a,&b,&c,&d,&e,&f); 
        if(a==0&&b==0&&c==0&&d==0&&e==0&&f==0) 
            break
        n=d+e+f+(c+3)/4;//必须要这么多个包装袋 
        y=5*d+u[c%4];//在已有n个的情况下,能装下y个2*2的 
        if(b>y) 
            n+=(b-y+8)/9;//把多的2*2的弄进来 
        x=36*n-36*f-25*e-16*d-9*c-4*b; 
        if(a>x) 
            n+=(a-x+35)/36;//把1*1的弄进来 
        printf("%d\n",n); 
    
    return 0; 
}
http://www.hankcs.com/program/cpp/poj-1017-packets.html
有 1 * 1 到 6 * 6 的产品,最少用几个 6 * 6 的箱子装它们。
贪心策略是先装大的,再装小的,看我专门画的图就全明白了。生活常识,桶里先放碎石,再放沙,最后还可以装不少水。 6 * 6 和 5 * 5 以及 4 * 4 肯定独享一个箱子(不可能再放得下同规格的产品)。装了 4 * 4 和 3 * 3 的箱子还可以放 2 * 2 的产品.
int main(int argc, char *argv[])
{
#ifndef ONLINE_JUDGE
    freopen("in.txt""r", stdin);
    freopen("out.txt""w", stdout);
#endif
    int n, p1, p2, p3, p4, p5, p6, x , y;
    int space[4] = {0, 5, 3, 1}; // 一个箱子放入几个 3 * 3 后留下的缝隙可以放入几个 2 * 2
    while(true)
    {
        cin >> p1 >> p2 >> p3 >> p4 >> p5 >> p6;
        if(p1 == 0 && p2 == 0 && p3 == 0 && p4 == 0 && p5 == 0 && p6 == 0)
        {
            break;
        }
        // 第一次放“放不下第二个该型号”的型号,需要消耗n个箱子
        n = p4 + p5 + p6 + ceil(p3 / 4.0); // 对 3 * 3 的 package 来讲,四个刚好,否则向上取整
        // 计算 4 * 4 的型号和 3 * 3 的型号之间的缝隙可以放下多少个 2 * 2 的型号
        y = 5 * p4 + space[p3 % 4];
        // 2 * 2 的型号填入这些缝隙,如果填不下,再多消耗几个箱子
        if(p2 > y)
        {
            n += ceil((p2 - y) / 9.0); // 多出来的每9个刚好一个箱子
        }
        // 计算现在空了多少个 1 * 1 的位置
        x = 36 * n - 36 * p6 - 25 * p5 - 16 * p4 - 9 * p3 - 4 * p2;
        // 1 * 1 的型号填入这些缝隙,如果填不下,再多消耗几个箱子
        if(p1>x)
        {
            n += ceil((p1 - x) / 36.0);
        }
        cout << n << endl;
    }
#ifndef ONLINE_JUDGE
    fclose(stdin);
    fclose(stdout);
    system("out.txt");
#endif
    return 0;
}

首先可分析出,要放 6*6、5*5、4*4 规格的货物时都各需要一个箱子。
在放入了一个5*5货物的箱子中还留有可放11个 1*1 货物的空位。
在放入了一个4*4货物的箱子中还留有可放5个 2*2 货物的空位,或20个 1*1 的空位。
对于3*3货物的情况见下表:
一个箱子中3*3货物数量        剩余2*2的空间       剩余1*1的空间
       4                      0                 0
       3                      1                 5
       2                      3                 6
       1                      5                 7
对于2*2的货物,一个箱子最多可放9个 2*2 的货物,每减少一个 2*2 的货物课增加4个 1*1 的货物。每个箱子可放36个1*1的货物。
显然6*6,5*5,4*4的产品每次只能放一个,且放完后只能放1*1的产品。对于3*3的格子,设置数组lim[i][j]表示放了i个j*j的产品后最多还能放多少个2*2的。
显然lim[4][3]=0,lim[3][3]=1,lim[2][3]=3,lim[1][3]=5,lim[0][3]=9;  lim[1][4]=5;
Max[i]表示当放i*i的产品时最多能放多少个。
然后每次从大往小放,有多的空间就尽可能的放2*2和1*1.注意有时2*2的产品数量比较少,此时可以放更多的1*1。
#define Maxn 10 
int num[Maxn]; 
int lim[Maxn][Maxn],Max[Maxn]; 
   
int main() 
   
    memset(lim,0,sizeof(lim)); 
    lim[4][3]=0,lim[3][3]=1,lim[2][3]=3,lim[1][3]=5,lim[0][3]=9; 
    lim[1][4]=5; 
    Max[6]=1,Max[3]=4,Max[4]=1,Max[5]=1,Max[2]=9,Max[1]=36; 
   
    while(scanf("%d",&num[1])) 
    
        int flag=num[1]; 
   
        for(int i=2;i<=6;i++) 
        
            scanf("%d",&num[i]); 
            flag+=num[i]; 
        
        if(!flag) 
            break
   
        int ans=0; 
        for(int i=6;i>=2;i--) 
        
            if(!num[i]) 
                continue
            if(num[i]>=Max[i]) //比能放的最多的还多,不止放一个 
            
                int tmp=num[i]/Max[i]; //放的个数 
                ans+=tmp; 
                num[i]%=Max[i]; //剩下的个数 
   
                int tt=tmp*lim[Max[i]][i];//可以填补的2*2的个数 
   
                if(num[2]) 
                
                    if(num[2]>=tt) //2*2的产品比较多的话,可以全部添进去 
                        num[2]-=tt; 
                    else 
                    
                        tt=num[2]; //比较少的话只添一部分 
                        num[2]=0; 
                    
   
                
                if(num[1]) //根据2*2的占用情况,添加1*1 
                
                    num[1]-=36*tmp-tt*4-i*i*Max[i]*tmp; 
                    if(num[1]<0) 
                        num[1]=0; 
                
   
            
            if(num[i]) //剩余的 
            
                ans++; //再添加一个 
                int tt=lim[num[i]][i]; //2*2的个数 
   
                if(num[2]) 
                
                    if(tt>=num[2]) 
                    
                        tt=num[2]; 
                        num[2]=0; 
                    
                    else 
                        num[2]-=tt; 
                
                if(num[1]) //1*1的可添加的个数 
                
                    num[1]-=36-tt*4-num[i]*i*i; 
                    if(num[1]<0) 
                        num[1]=0; 
                
                num[i]=0; 
            
        } //1*1的单独处理 
        if(num[1]) 
                ans+=(num[1]%36)?(num[1]/36+1):(num[1]/36); 
        printf("%d\n",ans); 
    
   
   return 0; 
}


Read full article from 1017 -- Packets

Labels

LeetCode (1432) GeeksforGeeks (1122) LeetCode - Review (1067) Review (882) Algorithm (668) to-do (609) Classic Algorithm (270) Google Interview (237) Classic Interview (222) Dynamic Programming (220) DP (186) Bit Algorithms (145) POJ (141) Math (137) Tree (132) LeetCode - Phone (129) EPI (122) Cracking Coding Interview (119) DFS (115) Difficult Algorithm (115) Lintcode (115) Different Solutions (110) Smart Algorithm (104) Binary Search (96) BFS (91) HackerRank (90) Binary Tree (86) Hard (79) Two Pointers (78) Stack (76) Company-Facebook (75) BST (72) Graph Algorithm (72) Time Complexity (69) Greedy Algorithm (68) Interval (63) Company - Google (62) Geometry Algorithm (61) Interview Corner (61) LeetCode - Extended (61) Union-Find (60) Trie (58) Advanced Data Structure (56) List (56) Priority Queue (53) Codility (52) ComProGuide (50) LeetCode Hard (50) Matrix (50) Bisection (48) Segment Tree (48) Sliding Window (48) USACO (46) Space Optimization (45) Company-Airbnb (41) Greedy (41) Mathematical Algorithm (41) Tree - Post-Order (41) ACM-ICPC (40) Algorithm Interview (40) Data Structure Design (40) Graph (40) Backtracking (39) Data Structure (39) Jobdu (39) Random (39) Codeforces (38) Knapsack (38) LeetCode - DP (38) Recursive Algorithm (38) String Algorithm (38) TopCoder (38) Sort (37) Introduction to Algorithms (36) Pre-Sort (36) Beauty of Programming (35) Must Known (34) Binary Search Tree (33) Follow Up (33) prismoskills (33) Palindrome (32) Permutation (31) Array (30) Google Code Jam (30) HDU (30) Array O(N) (29) Logic Thinking (29) Monotonic Stack (29) Puzzles (29) Code - Detail (27) Company-Zenefits (27) Microsoft 100 - July (27) Queue (27) Binary Indexed Trees (26) TreeMap (26) to-do-must (26) 1point3acres (25) GeeksQuiz (25) Merge Sort (25) Reverse Thinking (25) hihocoder (25) Company - LinkedIn (24) Hash (24) High Frequency (24) Summary (24) Divide and Conquer (23) Proof (23) Game Theory (22) Topological Sort (22) Lintcode - Review (21) Tree - Modification (21) Algorithm Game (20) CareerCup (20) Company - Twitter (20) DFS + Review (20) DP - Relation (20) Brain Teaser (19) DP - Tree (19) Left and Right Array (19) O(N) (19) Sweep Line (19) UVA (19) DP - Bit Masking (18) LeetCode - Thinking (18) KMP (17) LeetCode - TODO (17) Probabilities (17) Simulation (17) String Search (17) Codercareer (16) Company-Uber (16) Iterator (16) Number (16) O(1) Space (16) Shortest Path (16) itint5 (16) DFS+Cache (15) Dijkstra (15) Euclidean GCD (15) Heap (15) LeetCode - Hard (15) Majority (15) Number Theory (15) Rolling Hash (15) Tree Traversal (15) Brute Force (14) Bucket Sort (14) DP - Knapsack (14) DP - Probability (14) Difficult (14) Fast Power Algorithm (14) Pattern (14) Prefix Sum (14) TreeSet (14) Algorithm Videos (13) Amazon Interview (13) Basic Algorithm (13) Codechef (13) Combination (13) Computational Geometry (13) DP - Digit (13) LCA (13) LeetCode - DFS (13) Linked List (13) Long Increasing Sequence(LIS) (13) Math-Divisible (13) Reservoir Sampling (13) mitbbs (13) Algorithm - How To (12) Company - Microsoft (12) DP - Interval (12) DP - Multiple Relation (12) DP - Relation Optimization (12) LeetCode - Classic (12) Level Order Traversal (12) Prime (12) Pruning (12) Reconstruct Tree (12) Thinking (12) X Sum (12) AOJ (11) Bit Mask (11) Company-Snapchat (11) DP - Space Optimization (11) Dequeue (11) Graph DFS (11) MinMax (11) Miscs (11) Princeton (11) Quick Sort (11) Stack - Tree (11) 尺取法 (11) 挑战程序设计竞赛 (11) Coin Change (10) DFS+Backtracking (10) Facebook Hacker Cup (10) Fast Slow Pointers (10) HackerRank Easy (10) Interval Tree (10) Limited Range (10) Matrix - Traverse (10) Monotone Queue (10) SPOJ (10) Starting Point (10) States (10) Stock (10) Theory (10) Tutorialhorizon (10) Kadane - Extended (9) Mathblog (9) Max-Min Flow (9) Maze (9) Median (9) O(32N) (9) Quick Select (9) Stack Overflow (9) System Design (9) Tree - Conversion (9) Use XOR (9) Book Notes (8) Company-Amazon (8) DFS+BFS (8) DP - States (8) Expression (8) Longest Common Subsequence(LCS) (8) One Pass (8) Quadtrees (8) Traversal Once (8) Trie - Suffix (8) 穷竭搜索 (8) Algorithm Problem List (7) All Sub (7) Catalan Number (7) Cycle (7) DP - Cases (7) Facebook Interview (7) Fibonacci Numbers (7) Flood fill (7) Game Nim (7) Graph BFS (7) HackerRank Difficult (7) Hackerearth (7) Inversion (7) Kadane’s Algorithm (7) Manacher (7) Morris Traversal (7) Multiple Data Structures (7) Normalized Key (7) O(XN) (7) Radix Sort (7) Recursion (7) Sampling (7) Suffix Array (7) Tech-Queries (7) Tree - Serialization (7) Tree DP (7) Trie - Bit (7) 蓝桥杯 (7) Algorithm - Brain Teaser (6) BFS - Priority Queue (6) BFS - Unusual (6) Classic Data Structure Impl (6) DP - 2D (6) DP - Monotone Queue (6) DP - Unusual (6) DP-Space Optimization (6) Dutch Flag (6) How To (6) Interviewstreet (6) Knapsack - MultiplePack (6) Local MinMax (6) MST (6) Minimum Spanning Tree (6) Number - Reach (6) Parentheses (6) Pre-Sum (6) Probability (6) Programming Pearls (6) Rabin-Karp (6) Reverse (6) Scan from right (6) Schedule (6) Stream (6) Subset Sum (6) TSP (6) Xpost (6) n00tc0d3r (6) reddit (6) AI (5) Abbreviation (5) Anagram (5) Art Of Programming-July (5) Assumption (5) Bellman Ford (5) Big Data (5) Code - Solid (5) Code Kata (5) Codility-lessons (5) Coding (5) Company - WMware (5) Convex Hull (5) Crazyforcode (5) DFS - Multiple (5) DFS+DP (5) DP - Multi-Dimension (5) DP-Multiple Relation (5) Eulerian Cycle (5) Graph - Unusual (5) Graph Cycle (5) Hash Strategy (5) Immutability (5) Java (5) LogN (5) Manhattan Distance (5) Matrix Chain Multiplication (5) N Queens (5) Pre-Sort: Index (5) Quick Partition (5) Quora (5) Randomized Algorithms (5) Resources (5) Robot (5) SPFA(Shortest Path Faster Algorithm) (5) Shuffle (5) Sieve of Eratosthenes (5) Strongly Connected Components (5) Subarray Sum (5) Sudoku (5) Suffix Tree (5) Swap (5) Threaded (5) Tree - Creation (5) Warshall Floyd (5) Word Search (5) jiuzhang (5)

Popular Posts