uva 529 - Addition Chains
2012-08-15 16:16
399 查看
Addition Chains |
with the following four properties:
a0 = 1
am = n
a0<a1<a2<...<am-1<am
For each k (
) there exist two (not neccessarily different) integers
i and j (
) 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 Specification
The input file will contain one or more test cases. Each test case consists of one line containing one integern (
). Input is terminated by a value of zero (0) for
n.
Output Specification
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
5 7 12 15 77 0
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
迭代加深dfs很适合n=10000,17个数字,虽然感觉上迭代本身存在大量重复计算,但是重复的搜索树的深度小;
迭代加深搜索算法就是仿广度优先搜索的深度优先搜索。既能满足深度优先搜索的线性存储要求,又能保证发现一个最小深度的目标结点。
#include <string.h> #include <stdio.h> int a[100],best[100],n,min,f; int minlen(int num,int len) {int x=num,y=len; while (x<n) {x=x<<1;++y;} return y; } void dfs(int len) { int i,j,time,x; if ((a[len]==n)&&(len<min)) {f=0; min=len; for (i=1;i<=len;i++) best[i]=a[i]; } if ((a[len]>=n)||(len>=min)) return ; if (minlen(a[len],len)>=min) return ; for (i=1;i<=len;i++) for (j=i;j<=len;j++) { a[len+1]=a[i]+a[j]; if (a[len+1]>a[len]) dfs(len+1); } } int main() { int i; a[1]=1; a[2]=2; best[1]=1; best[2]=2; while (scanf("%d",&n),n) { f=1; min=minlen(2,2)-1; if (n<=2) {min=n;f=0;} while (f) {++min; dfs(2); } for (i=1;i<min;i++) printf("%d ",best[i]); printf("%d\n",n); } return 0; }
[/code]
相关文章推荐
- UVA 529 Addition Chains(迭代搜索)
- UVA 529 - Addition Chains,迭代加深搜索+剪枝
- UVA - 529 Addition Chains(迭代+dfs)
- uva 529 Addition Chains(迭代深搜)
- UVA 529 Addition Chains(迭代搜索)
- UVA 529 - Addition Chains
- uva 529 Addition Chains
- UVA 529 Addition Chains
- uva529 Addition Chains
- UVa 529 - Addition Chains ,迭代加深搜索+减枝
- UVA529- Addition Chains(迭代+DFS)
- poj 2248--Addition Chains (uva 529--Addition Chains)
- uva529
- UVa 529 POJ 2248 - Addition Chains ,迭代加深搜索+减枝
- uva 529(暴力求解)
- UVA 529 - Addition Chains 剪枝+迭代深搜
- UVA529加深迭代搜索,起点为1,终点为n,数列中的每个数由前面俩个数的和组成的最短数列
- uva 529 迭代DFS
- uva529 迭代加深+必要剪枝
- UVA529