POJ1465：Multiple（BFS）

reference：http://blog.csdn.net/u013480600/article/details/26458435

Multiple

Time Limit: 1000MS		Memory Limit: 32768K
Total Submissions: 7682		Accepted: 1717

Description

a program that, given a natural number N between 0 and 4999 (inclusively), and M distinct decimal digits X1,X2..XM (at least one), finds the smallest strictly positive multiple of N that has no other digits besides X1,X2..XM (if such a multiple exists).

Input

The input has several data sets separated by an empty line, each data set having the following format: 

On the first line - the number N 
On the second line - the number M 
On the following M lines - the digits X1,X2..XM.

Output

For each data set, the program should write to standard output on a single line the multiple, if such a multiple exists, and 0 otherwise. 

An example of input and output:

Sample Input

22
3
7
0
1

2
1
1

Sample Output

110
0

Source

Southeastern Europe 2000

题意：给你m个数，用这些数字组成最小的可以被n整除的数，不行就输出-1.

思路：m个数排序后直接广搜，剪枝为出现过的余数就跳过，用数组保存节点用于输出结果。

# include <iostream>
# include <cstdio>
# include <queue>
# include <cstring>
# include <algorithm>

using namespace std;
int n, m, a[10], vis[5050];

struct node
{
	int num;//数字 
	int pre;//上一个节点 
	int cal;//取模结果 
}q[5050];

int bfs()
{
	int l=0, r=1;
	node tmp;
	tmp.cal = 0;
	tmp.pre = -1;
	tmp.num = 0;
	q[l] = tmp;
	while(l<r)
	{
		node t = q[l];
		for(int i=0; i<n; ++i)
		{
			int nnum = ((t.cal)*10+a[i]) % m;
			if(!vis[nnum] && (t.pre != -1 || a[i]!= 0))//没有前导0 
			{
				
				vis[nnum] = 1;
				node ne;
				ne.cal = nnum;
				ne.num = a[i];
				ne.pre = l;
				q[r++] = ne;
				if(nnum == 0)
					return r-1;
			}
		}
		++l;
	}
	return -1;
}

void print(int x)
{
	if(q[x].pre != -1)
	{
		print(q[x].pre);
		printf("%d",q[x].num);
	}
}

int main()
{
	while(~scanf("%d%d",&m,&n))
	{
		if(m==0)
		{
			puts("0");
			continue;
		}
		memset(q, 0, sizeof(q));
		memset(vis, 0, sizeof(vis));
		for(int i=0; i<n; ++i)
			scanf("%d",&a[i]);
		sort(a, a+n);
		int ans = bfs();
		if(ans == -1)
			printf("0");
		else
			print(ans);
		puts("");
	}
	return 0;
}