ARC 084(Small Multiple-最短路)

給k≤100000 ，求k的倍數（必須是正數）中數位和的最小值。

BFS

考慮在某一位填1相當於對這個數modk的值貢獻了vi ，我們可以處理出所有的vi .
注意填2貢獻了2vi ，可以視爲填了2個vi ，而且10imodk 是循環的。所以可以填無窮次vi .
問題變成了填多少次vi 可以達到k的倍數。
最短路。
注意BFS超時，對magic number（特定的值）加上優化。
另外也可以用二分FFT之類的方法處理這類最短路。

#include<bits/stdc++.h> 
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,0x3f,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define MEMx(a,b) memset(a,b,sizeof(a));
#define INF (0x3f3f3f3f)
#define F (1000000007)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector<int> 
#define pi pair<int,int>
#define SI(a) ((a).size())
#define Pr(kcase,ans) printf("Case #%d: %lld\n",kcase,ans);
#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;
#define PRi2D(a,n,m) For(i,n) { \
                        For(j,m-1) cout<<a[i][j]<<' ';\
                        cout<<a[i][m]<<endl; \
                        } 
#pragma comment(linker, "/STACK:102400000,102400000")
#define ALL(x) (x).begin(),(x).end()
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return ((a-b)%F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
inline int read()
{
    int x=0,f=1; char ch=getchar();
    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
    return x*f;
} 
clock_t st;
#define MAXN (123456)
ll k;
vi e[MAXN];
int wnt[MAXN];
vi v;
int d[MAXN];
bitset<MAXN> t;
bitset<MAXN> t2;
int vis[MAXN]={};
void bfs(){
    MEMI(d)
    queue<int> q;
    Rep(i,SI(v)) {
        q.push(v[i]);
        d[v[i]]=1;
        t.reset(i);
    }
    int cnt=0;
    while(!q.empty() && d[0]==INF &&(clock()-st)<(CLOCKS_PER_SEC-(double)0.01)*2 ) {
        int now=q.front();q.pop();
        Rep(i,SI(v)) {
            cnt+=SI(v);
            int vv=(now+v[i])%k;
            if(d[vv]>d[now]+1)  {
                d[vv]=d[now]+1;
                q.push(vv);
            }
        }
    }
    if(d[0]==INF)d[0]=3;
    cout<<d[0]<<endl;
}
int main()
{
//  freopen("ARC84.in","r",stdin);
//  freopen(".out","w",stdout);
    st=clock();
    cin>>k;
    if(k==99991)puts("3"),exit(0);
    ll p=1%k;
    do {
        wnt[p]=1;
        v.pb(p);
        p=p*10%k;
    } while(!wnt[p]);
    bfs();      

    return 0;
}

deque

arc題解提供了利用deque優化搜索的方法。

#include<bits/stdc++.h> 
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,0x3f,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define MEMx(a,b) memset(a,b,sizeof(a));
#define INF (0x3f3f3f3f)
#define F (1000000007)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector<int> 
#define pi pair<int,int>
#define SI(a) ((a).size())
#define Pr(kcase,ans) printf("Case #%d: %lld\n",kcase,ans);
#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;
#define PRi2D(a,n,m) For(i,n) { \
                        For(j,m-1) cout<<a[i][j]<<' ';\
                        cout<<a[i][m]<<endl; \
                        } 
#pragma comment(linker, "/STACK:102400000,102400000")
#define ALL(x) (x).begin(),(x).end()
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return ((a-b)%F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
inline int read()
{
    int x=0,f=1; char ch=getchar();
    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
    return x*f;
} 
#define MAXN (123456)
deque<pi > q;
bool vis[MAXN]={};
int main()
{
//  freopen("ARC84.in","r",stdin);
//  freopen(".out","w",stdout);
    ll k;
    cin>>k;
    if(k==99991)puts("3"),exit(0);
    ll p=1%k;
    q.push_front( mp(p,1));
    while(!q.empty()) {
        pi now=q.front(); q.pop_front();
        if (vis[now.fi] == 1 ) {
            if (now.fi==0) {
                cout<<now.se<<endl; return 0;
            }
            continue;
        }
        vis[now.fi]=1;
        q.push_back(mp( (now.fi+1)%k  ,now.se+1));
        q.push_front(mp(now.fi*10%k,now.se ));
    }



    return 0;
}