Sticks
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 115683 | Accepted: 26614 |
Description
George took sticks of the same length and cut them randomly until all parts became at most 50 units long. Now he wants to return sticks to the original state, but he forgot how many sticks he had originally and how long they were originally. Please help him and design a program which computes the smallest possible original length of those sticks. All lengths expressed in units are integers greater than zero.
Input
The input contains blocks of 2 lines. The first line contains the number of sticks parts after cutting, there are at most 64 sticks. The second line contains the lengths of those parts separated by the space. The last line of the file contains zero.
Output
The output should contains the smallest possible length of original sticks, one per line.
Sample Input
9 5 2 1 5 2 1 5 2 1 4 1 2 3 4 0
Sample Output
6 5
Source
题解：深度优先搜索经典题目。需要配合各种剪枝。
1.把木棍长度从大到小排序，从长的开始搜极大减少搜索次数。
2.长度a[i]的不能匹配，因为有序，和它等长的全部剪掉。
代码：
1 #include<stdio.h> 2 #include<string.h> 3 #include<stdbool.h> 4 int i,j,n,m,maxi,sum,ans, 5 a[100]; 6 7 bool vis[100]; 8 9 int 10 pre() 11 { 12 sum=0;ans=0; 13 memset(a,0,sizeof(a)); 14 memset(vis,false,sizeof(vis)); 15 16 return 0; 17 } 18 19 bool 20 dfs(int num,int pos,int len) 21 { 22 int i; 23 if (num==n) return true; 24 for(i=pos;i<=n;i++) 25 { 26 if(vis[i]) continue; 27 if(a[i]+len<ans) 28 { 29 vis[i]=true; 30 if (dfs(num+1,i+1,len+a[i])) return true;//如果dfs(used+1,i+1,len+a[i])返回false，则第i根不可用，标记为false 31 vis[i]=false; 32 while((a[i]==a[i+1])&&((i+1)<n)) //长度为a[i]的不能匹配，则和它同样长的不需要再匹配。 33 { 34 i++; 35 } 36 if (len==0) return false;//如果当前长度为0，证明没有合适的，return false 37 } else 38 if(a[i]+len==ans) 39 { 40 vis[i]=true; 41 if (dfs(num+1,1,0)) return true; 42 vis[i]=false; 43 return false; 44 } 45 } 46 return false; 47 48 } 49 50 void 51 qsort(int head,int tail) 52 { 53 int i,j,x; 54 i=head;j=tail; 55 x=a[head]; 56 while(i<j) 57 { 58 while((i<j)&&(a[j]<=x)) j--; 59 a[i]=a[j]; 60 while((i<j)&&(a[i]>=x)) i++; 61 a[j]=a[i]; 62 } 63 a[i]=x; 64 if (head<(i-1)) qsort(head,i-1); 65 if ((i+1)<tail) qsort(i+1,tail); 66 } 67 68 int 69 main() 70 { 71 72 while(~scanf("%d ",&n)) 73 { 74 pre(); 75 if(n==0) break; 76 for(i=1;i<=n;i++) 77 { 78 scanf("%d",&a[i]); 79 sum+=a[i]; 80 } 81 82 qsort(1,n); //按木棍长度从大到小排序。 83 maxi=a[n]; 84 for(i=n;i>=1;i--) 85 { 86 if ((sum%i==0)&&((sum/i)>=maxi)) 87 { 88 ans=sum/i; 89 memset(vis,false,sizeof(vis)); 90 if (dfs(0,1,0)) 91 { 92 printf("%d ",ans); 93 break; 94 } 95 } 96 } 97 } 98 return 0; 99 }