POJ2248 Addition Chains

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概述

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5122		Accepted: 2766		Special Judge

Description

An addition chain for n is an integer sequence <a0, a1,a2,...,am="">with the following four properties:

a0 = 1
am = n
a0 < a1 < a2 < ... < a_m-1 < am
For each k (1<=k<=m) there exist two (not necessarily different) integers i and j (0<=i, j<=k-1) with ak=ai+aj

You are given an integer n. Your job is to construct an addition chain for n with minimal length. If there is more than one such sequence, any one is acceptable.
For example, <1,2,3,5> and <1,2,4,5> are both valid solutions when you are asked for an addition chain for 5.

Input

The input will contain one or more test cases. Each test case consists of one line containing one integer n (1<=n<=100). Input is terminated by a value of zero (0) for n.

Output

For each test case, print one line containing the required integer sequence. Separate the numbers by one blank.
Hint: The problem is a little time-critical, so use proper break conditions where necessary to reduce the search space.

Sample Input

Sample Output

1 2 4 5
1 2 4 6 7
1 2 4 8 12
1 2 4 5 10 15
1 2 4 8 9 17 34 68 77

Source

Ulm Local 1997

DFS迭代加深。

每次优先取数列后面的大数凑成新数，如果数过小或者大于目标就剪掉。

 1 /*by SilverN*/
 2 #include<algorithm>
 3 #include<iostream>
 4 #include<cstring>
 5 #include<cstdio>
 6 #include<cmath>
 7 using namespace std;
 8 const int mxn=300;
 9 int n;
10 int a[mxn];
11 int dep[mxn];
12 bool v[mxn];
13 bool dfs(int now,int limit){
14     if(now>limit)return 0;
15     int i,j;
16     for(i=now-1;i;i--){
17         for(j=i;j;j--){
18             a[now]=a[i]+a[j];
19             if(a[now]>n)continue;
20             if(a[now]<=a[now-1])break;
21             if(a[now]==n) return 1; 
22             if(dfs(now+1,limit))return 1;
23         }
24     }
25     return 0;
26 }
27 int main(){
28     a[1]=1;
29     a[2]=2;
30     while(scanf("%d",&n) && n){
31         if(n==1){
32             printf("1 n");
33             continue;
34         }
35         if(n==2){
36             printf("1 2 n");
37             continue;
38         }
39         memset(v,0,sizeof v);
40         int i,j;
41         int ans=1000;
42         for(i=1;i<=100;i++){
43             if(dfs(3,i)){
44                 ans=i;
45                 for(j=1;j<=i;j++)printf("%d ",a[j]);
46                 printf("n");
47                 break;    
48             }
49         }
50     }
51     return 0;
52 }