标签：tom return minimum cow hal capacity nbsp pair sam
Farmer John must measure Q (1 <= Q <= 20,000) quarts of his finest milk and deliver it in one big bottle to a customer. He fills that bottle with exactly the number of quarts that the customer orders.
Farmer John has always been frugal. He is at the cow hardware store where he must purchase a set of pails with which to measure out Q quarts of milk from his giant milk tank. Since the pails each cost the same amount, your task is to figure out a minimal set of pails Farmer John can purchase in order to fill a bottle with exactly Q quarts of milk. Additionally, since Farmer John has to carry the pails home, given two minimal sets of pails he should choose the "smaller" one as follows: Sort the sets in ascending order. Compare the first pail in each set and choose the set with the smallest pail. If the first pails match, compare the second pails and choose from among those, else continue until the two sets differ. Thus the set {3, 5, 7, 100} should be chosen over {3, 6, 7, 8}.
To measure out milk, FJ may completely fill a pail from the tank and pour it into the bottle. He can never remove milk from the bottle or pour milk anywhere except into the bottle. With a one-quart pail, FJ would need only one pail to create any number of quarts in a bottle. Other pail combinations are not so convenient.
Determine the optimally small number of pails to purchase, given the guarantee that at least one solution is possible for all contest input data.
| Line 1: | The single integer Q |
| Line 2: | A single integer P (1 <= P <= 100) which is the number of pails in the store |
| Lines 3..P+2: | Each line contains a single integer pail_value (1 <= pail_value <= 10000), the number of quarts a pail holds |
16 3 3 5 7
The output is a single line of space separated integers that contains:
2 3 5
——————————————————————————————————————————————————————题解
这道题是迭代深搜，但是以为是个记录状态的背包，死活没写出来
这道题记忆化搜索判断可行比纯dp要快，因为要用的状态都是跳跃幅度较大的，记忆化搜索反而用的状态比较少
思路和ax+by=1方程解然后不断得到新的差值为1的序列如果这个序列长度==最小的数，那么就回得到后面所有的数
http://www.cnblogs.com/ivorysi/p/6279637.html
所以这道题的个数不会太多，用DFSID+记搜
1 /* 2 ID: ivorysi 3 LANG: C++ 4 PROG: milk4 5 */ 6 #include <iostream> 7 #include <cstdio> 8 #include <cstring> 9 #include <algorithm> 10 #include <queue> 11 #include <set> 12 #include <vector> 13 #include <string.h> 14 #include <cmath> 15 #define siji(i,x,y) for(int i=(x);i<=(y);++i) 16 #define gongzi(j,x,y) for(int j=(x);j>=(y);--j) 17 #define xiaosiji(i,x,y) for(int i=(x);i<(y);++i) 18 #define sigongzi(j,x,y) for(int j=(x);j>(y);--j) 19 #define inf 0x3f3f3f3f 20 //#define ivorysi 21 #define mo 97797977 22 #define hash 974711 23 #define base 47 24 #define pss pair<string,string> 25 #define MAXN 5000 26 #define fi first 27 #define se second 28 #define pii pair<int,int> 29 #define esp 1e-8 30 typedef long long ll; 31 using namespace std; 32 int q,p; 33 int f[20005]; 34 int ld,used[105],w[105]; 35 int flag=0,cnt; 36 bool calc(int tmp) { 37 if(f[tmp]!=-1) return f[tmp]; 38 f[tmp]=0; 39 siji(i,1,p) { 40 if(used[i] && tmp%w[i]==0) return f[tmp]=1; 41 } 42 siji(i,1,p) { 43 if(used[i] && tmp>=w[i]) { 44 f[tmp]=(f[tmp] || calc(tmp-w[i])); 45 if(f[tmp]==1) return f[tmp]; 46 } 47 } 48 return f[tmp]; 49 } 50 void dfs(int d,int pr) { 51 if(flag) return; 52 if(d==ld){ 53 siji(i,1,q) f[i]=-1; 54 f[0]=1; 55 if(calc(q)) { 56 flag=1; 57 printf("%d ",ld); 58 siji(i,1,p) { 59 if(used[i]) { 60 ++cnt; 61 printf("%d%c",w[i]," \n"[cnt==ld]); 62 } 63 } 64 } 65 return; 66 } 67 siji(i,pr+1,p) { 68 used[i]=1; 69 dfs(d+1,i); 70 if(flag) return; 71 used[i]=0; 72 } 73 } 74 void solve() { 75 scanf("%d%d",&q,&p); 76 siji(i,1,p) { 77 scanf("%d",&w[i]); 78 } 79 sort(w+1,w+p+1); 80 f[0]=1; 81 siji(i,1,p) { 82 ld=i; 83 dfs(0,0); 84 if(flag) break; 85 } 86 } 87 int main(int argc, char const *argv[]) 88 { 89 #ifdef ivorysi 90 freopen("milk4.in","r",stdin); 91 freopen("milk4.out","w",stdout); 92 #else 93 freopen("f1.in","r",stdin); 94 #endif 95 solve(); 96 return 0; 97 }
标签：tom return minimum cow hal capacity nbsp pair sam
原文地址：http://www.cnblogs.com/ivorysi/p/6391907.html
