uva 529 Addition Chains(迭代深搜）

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Addition Chains

An addition chain for n is an integer sequence $<a_0, a_1, a_2, \dots, a_m>$ with the following four properties:

a₀ = 1
a_m = n
a₀<a₁<a₂<...<a_m-1<a_m
For each k ( $1 \le k \le m$ ) there exist two (not neccessarily different) integers i and j ( $0 \le i, j \le k-1$ ) with a_k =a_i +a_j

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 integer n ( $1 \le n \le 10000$ ). 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

571215770

Sample Output

1 2 4 51 2 4 6 71 2 4 8 121 2 4 5 10 151 2 4 8 9 17 34 68 77

题目大意：给出一个数n，要求找出一个集合，集合的末尾数必须为n,并且集合中的任意一个数必须可以用集合中另外两个数组成,找出集合数最小的，并输出。

解题思路：迭代深搜的方法，先确定数的基数，就是说组成这个数最少需要几个数，例如1个数最大组成1， 2个数最大组成2，而3个数最大为4（每增加一个数，最大值 * 2）

一个很重要的剪枝，当当前第i个数，为k， k的在剩余的位数中以倍数形式增长后任小于目标值。

#include <stdio.h>#include <string.h>#define N 20int order, num[N], ok;void DFS(int cur, int sum){    if (cur >= sum + 1){if (num[sum] == order){    ok = 1;    return ;}    }    else{if (num[cur - 1] << sum - cur + 1 < order)    return;for (int i = cur - 1; i >= 0; i--)    for (int j = cur - 1; j >= 0; j--){num[cur] = num[i] + num[j];DFS(cur + 1, sum);if (ok) return ;    }    }}int main(){    while (scanf("%d", &order), order){int k, n;ok = 0;for (k = 0; ; k++)    if (1 << k >= order)break;for (n = k; ; n++){    memset(num, 0, sizeof(num));    num[1] = 1;    DFS(2, n);    if (ok) break;}for (int i = 1; i < n; i++)    printf("%d ", num[i]);printf("%d\n", num[n]);    }    return 0;}