Small Multiple———图论最短路

6616: Small Multiple

时间限制: 1 Sec  内存限制: 512 MB
提交: 459  解决: 98
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题目描述

Find the smallest possible sum of the digits in the decimal notation of a positive multiple of K.
Constraints
2≤K≤105
K is an integer.

输入

Input is given from Standard Input in the following format:
K

输出

Print the smallest possible sum of the digits in the decimal notation of a positive multiple of K.

样例输入

6

样例输出

3

提示

12=6×2 yields the smallest sum.

来源/分类

ABC077&ARC084 

思路：从0开始对当前数进行加一建边并将权值变成1，然后再乘十建边权值为0。然后最短路搜索从1-0

# include<bits/stdc++.h>
using namespace std;
const int INF=1<<30;
struct edge
{
    int to,w,nxt;
};
edge e[200005];  //前向星建边
int head[100005],dis[100005];
int t,s,d,cnt,sta[100005],ed[100005];
void add(int a,int b,int c)
{
    e[cnt].to=b;
    e[cnt].w=c;
    e[cnt].nxt=head[a];
    head[a]=cnt++;
}
void spfa(int from)
{
    fill(dis,dis+100004,INF);
    dis[from]=0;
    queue<int>q;
    q.push(from);
    while(!q.empty()){     
        int u=q.front();
        q.pop();
        for(int i=head[u];i!=-1;i=e[i].nxt){
            if(dis[e[i].to]>dis[u]+e[i].w){
                dis[e[i].to]=dis[u]+e[i].w;
                q.push(e[i].to);
            }
        }
    }
}
int main()
{
    int i,j,a,b,c;
    int k;
    cin>>k;
        cnt=0;
        memset(head,-1,sizeof(head));
        for(i=0;i<k;++i){
            add(i,(i+1)%k,1);
            add(i,(i*10)%k,0);
        }
 
        int ans=INF;
        spfa(1);
         printf("%d\n",dis[0]+1);
 
    return 0;
}