ZOJ 1136 Multiple BFS（POJ 1465）

Multiple

---

Time Limit: 10 Seconds      Memory Limit: 32768 KB

---

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).

The input file 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.

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:

Input

22
3
7
0
1

2
1
1

Output

 110

  0

思路 用 m个数能否排出n的倍数
（每个数可以使用无数次） 要把排的的数形成从小到大 ，利用BFS；
然后开个数组，判重 余数相同就是代表不可行


#include <iostream>
#include <stdio.h>
#include <queue>
#include <stack>
#include <set>
#include <vector>
#include <string.h>
#include <algorithm>
using namespace std;
struct node
{
    int digit;
    int pre;
    int yu;
    int id;
}rc[5555];
int d[102];
bool b[5000];
int n,m;
void output(int id)
{
    if(rc[id].pre==-1)
     return;
    output(rc[id].pre);
    printf("%d",rc[id].digit);
}
int f;
void BFS()
{
    f=0;
    memset(b,0,n);
    int cnt=1,i,yu;
    int id;
    rc[0].pre=-1;
    rc[0].yu=0;
    rc[0].id=0;
    queue<int> Q;
    Q.push(0);
    while(!Q.empty())
    {
      id=Q.front();Q.pop();
      for(i=0;i<m;i++)
      {
          if(rc[id].yu==0&&d[i]==0) continue;
          yu=(rc[id].yu*10+d[i])%n;
          if(!b[yu])
          {
              if(yu==0)
               {
                   output(id);
                   printf("%d\n",d[i]);
                   f=1;
                   return;
               }
             b[yu]=1;
             rc[cnt].digit=d[i];
             rc[cnt].pre=id;
             rc[cnt].yu=yu;
             rc[cnt].id=cnt;
             Q.push(cnt++);
          }
      }
    }
}
int main()
{
    int i;
    while(scanf("%d",&n)!=EOF)
    {
       scanf("%d",&m);
       for(i=0;i<m;i++)
        scanf("%d",&d[i]);
       sort(d,d+m);
       if(n==0) {printf("0\n");continue;}
       BFS();
       if(!f) printf("0\n");
    }
    return 0;
}