ZOJ 1136 Multiple (BFS）

Multiple

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Time Limit: 10 Seconds      Memory Limit: 32768 KB

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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

题意：给m个数随意拼成一个最小n的倍数。

思路：bfs 输出的时候相当于遍历路径一边，所以有pre.

#include <cstdio>
#include <iostream>
#include <cstdlib>
#include <algorithm>
#include <ctime>
#include <cmath>
#include <string>
#include <cstring>
#include <stack>
#include <queue>
#include <list>
#include <vector>
#include <map>
#include <set>
using namespace std;

const int INF=0x3f3f3f3f;
const double eps=1e-10;
const double PI=acos(-1.0);
#define maxn 5000

struct Node
{
    int num, pre, id, yu;
};
Node node[maxn];
int vis[maxn];
int n, m;
int a[maxn];
void output(int id)
{
    if(node[id].pre == -1)
        return;
    output(node[id].pre);
    printf("%d", node[id].num);
}
void bfs()
{
    memset(vis, 0, sizeof vis);
    node[0].id = 0;
    node[0].pre = -1;
    node[0].yu = 0;
    int cnt = 1;
    queue<int> q;
    q.push(0);
    while(!q.empty())
    {
        int id = q.front();
        q.pop();
        for(int i = 0; i < m; i++)
        {
            if(node[id].yu == 0 && a[i] == 0)
                continue;
            int yu = (node[id].yu*10+ a[i])%n;
            if(!vis[yu])
            {
                if(yu == 0)
                {
                    output(id);
                    printf("%d\n", a[i]);
                    return;
                }
                vis[yu] = 1;
                node[cnt].pre = id;
                node[cnt].num = a[i];
                node[cnt].yu = yu;
                node[cnt].id = cnt;
                q.push(cnt++);
            }
        }
    }
    printf("%d\n", 0);
}
int main()
{
    while(~scanf("%d",&n))
    {
        scanf("%d", &m);
        for(int i = 0; i < m; i++) scanf("%d", &a[i]);
        sort(a, a + m);
        if(n == 0)
            printf("%d\n", 0);
        else
            bfs();
    }
    return 0;
}