POJ2248Addition Chains（dfs+剪枝）

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Addition Chains

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 4388		Accepted: 2406		Special Judge

Description

An addition chain for n is an integer sequence 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

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
using namespace std;
//Accepted	200K	360MS	C++	965B	2013-04-03 21:19:09
const int N = 100;
int num[N], ans[N];
int len = 11;//len为自己定义的搜索深度的上限。
int n;
//int length;

void dfs(int dep) {
    if(dep > len) return ;
    if(num[dep-1] == n) {
        if(dep < len) {
            //length = dep; Time Limit Exceeded
            len = dep;  //用搜索到的结果对回溯时可能的结果剪枝
            for(int i = 0; i <= dep-1; i++)
                ans[i] = num[i];
        }
        return;
    }
    for(int i = dep-1; i >= 0; i--) { //从后向往前搜，搜索的的技巧。
        num[dep] = num[i] + num[dep-1];
        if(num[dep] > n) continue;//剪枝
        dfs(dep+1);
    }
}

void init() {
    memset(num, 0,sizeof(num));
    memset(ans, 0, sizeof(ans));
    len = 11;//每次都会改变。
}

int main()
{
   while((scanf("%d", &n) != EOF) && n) {
        num[0] = 1;///the beginning of the search
        dfs(1);
        for(int i = 0; i < len-1; i++) {
            cout << ans[i] << ' ';
        }
        cout << ans[len-1] << endl;
        init();
   }
}