ASC 05 题解

E Graduated Lexicographical Ordering

数位dp

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
#include<vector>
#include<set>
#include<string>
#include<queue>
#include<complex>
#include<stack>
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()
#define gmax(a,b) a=max(a,b);
#define gmin(a,b) a=min(a,b);
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
void upd(ll &a,ll b){a+=b;}
#define MAXN (33)
char a[MAXN];
int n;
ll f[MAXN][10][2][10*18+10][2];
ll cnt[9*18+1];
bool vis[MAXN][10][2][10*18+10][2]={};
ll g[MAXN][10][2][10*18+10][2]={};
ll dfs(int i,int j,int k,int x,int y) {
    if (i==n) return g[i][j][k][x][y]=(x==0);
    if (vis[i][j][k][x][y]) {
        return g[i][j][k][x][y];
    }
    vis[i][j][k][x][y]=1;
    g[i][j][k][x][y]=0;
    Rep(t,10) if (t<=x){
        int nk=k|(t>0);
        int nx=x-t,ny=(y&&t==a[i+1]);
        if (y&&t>a[i+1]) continue;
        upd(g[i][j][k][x][y],dfs(i+1,t,nk,nx,ny));
    }
    return g[i][j][k][x][y];
}
char s[200];
void get_kth(ll W ,bool fl,ll w,ll sz ) {
    if (!fl) {
        w=1,sz=0;
        while(sz+cnt[w]<W) 
            sz+=cnt[w++];
        MEM(g) MEM(vis)
        dfs(0,0,0,w,1);
    }
    ll len=0,wsum=w,y=1,k=0;
    bool is_gre=0;
    while(1) {
        ++len;
        Rep(t,10) if (t<=W){
            if (t==0 &&!k) continue;
            s[len]=t+'0';
            ll ny=(y&&t==a[len]);
            ll nk=k|(t>0);
            ll pans=0;
            Fork(i,len,n) {
                if (i==len && (t>a[len] && y ||is_gre)) continue;
                pans+=g[i][t][nk][wsum-t][(i==len)&&ny];
            }
            if (sz+pans>=W) {
                if (t>a[len] && y) {
                    is_gre=1;
                } 
                y=ny,k=nk;

                break;
            }else sz+=pans;
        }

        s[len+1]=0;
        wsum-=s[len]-'0';
        if (sz+1==W&&!wsum) break;
        if (!wsum) ++sz;
    }
}
char b[100];
bool cmp() {
    int i=1;
//  cout<<s+1<<' '<<b+1<<endl;
    while(s[i]+b[i]) {
        if (s[i]^b[i]) return s[i]<b[i];
        ++i;
    }
    return 1;
}
void calc(ll k) {
    int len2=0;
    while(k) {
        b[++len2]='0'+k%10;
        k/=10LL;
    }
    b[len2+1]=0;
    reverse(b+1,b+1+len2);
    ll w=0;
    For(i,len2) b[i]-='0',w+=b[i];

    ll sz=0;
    For(i,w-1) sz+=cnt[i];
    MEM(g) MEM(vis)
    dfs(0,0,0,w,1);

    For(i,len2) b[i]+='0';
    ll L=sz+1,R=sz+cnt[w],ans=L;

    while(L<=R) {
        ll m=(L+R)/2;
//      cout<<m<<endl;
        get_kth(m,1,w,sz);
        if (cmp()) L=m+1,ans=m;
        else R=m-1;
    }
    cout<<ans<<endl;
}
int main()
{
    freopen("grlex.in","r",stdin);
    freopen("grlex.out","w",stdout);
    scanf("%s",a+1);
    ll ans=0;
    n=strlen(a+1);
    For(i,n) a[i]-='0';
    MEM(f)
    f[0][0][0][0][1]=1;
    Rep(i,n) {
        Rep(j,10) Rep(k,2) Rep(x,i*10+1) Rep(y,2) {
            if (f[i][j][k][x][y]) {
                Rep(t,10) {
                    int nk=k|(t>0);
                    int nx=x+t,ny=(y&&t==a[i+1]);
                    if (y&&t>a[i+1]) continue;
                    upd(f[i+1][t][nk][nx][ny],f[i][j][k][x][y]);
                }
            }
        }
    }
    Rep(j,10) Rep(k,2) Rep(x,n*10+1) Rep(y,2) {
        cnt[x]+=f[n][j][k][x][y];
    }
    ll k;
    cin>>k;

    calc(k);
//  For(k,200) {
//      get_kth(k,0,0,0);
//      cout<<s+1<<endl;
//  }
    get_kth(k,0,0,0);
    cout<<s+1<<endl;
//  Fork(k,2064,2174) {
//      get_kth(k,0,0,0);
//      cout<<s+1<<endl;
//  }
    return 0;
}

H Don’t Go Left

给一个自动机，然后模拟上面的操作，问磁带是否会回退？

根据题意写记忆化搜索

#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()
#define gmax(a,b) a=max(a,b);
#define gmin(a,b) a=min(a,b);
#define fr(x) freopen(x,"r",stdin)
#define fw(x) freopen(x,"w",stdout)
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 ONLINE_JUDGE
string Filename="left";
int f[120][120]={},n,m;
int a[120][120]={},c[120][120]={};
int vis[120][120][120]={};
bool dfs(int x,int p,int ed) { //p<=m+1 ' '
    if (x==n) return 1;
    else if (vis[x][p][ed]>=0) return vis[x][p][ed];
    else if (vis[x][p][ed]==-2) return 1;
    else {
//      cout<<x<<" "<<p<<' '<<ed<<endl;
        vis[x][p][ed]=-2;
        if (p==m+1) {
            For(j,m-1) {
                int v=(a[x][j]==1)?(m+1):c[x][j];
                if (a[x][j]==-1) return vis[x][p][ed]=0;
                if (!dfs(f[x][j],v,ed)) return vis[x][p][ed]=0;
            }
            int v=(a[x][m]==1)?(m):c[x][m];
            if (a[x][m]==-1) return vis[x][p][ed]=0;
            if (!dfs(f[x][m],v,1)) return vis[x][p][ed]=0;
            return vis[x][p][ed]=1;

        } else {
            if (a[x][p]==-1) return vis[x][p][ed]=0;
            return vis[x][p][ed]=dfs(f[x][p],(a[x][p]==1)?((ed)?m:(m+1) ):c[x][p],ed);
        }
    }
}

int main()
{
    fr("left.in");
    fw("left.out");
    MEMx(vis,-1)    
    n=read(),m=read();  
    For(i,n-1) {
        For(j,m) {
            f[i][j]=read();
            c[i][j]=read();
            a[i][j]=read();
        }
    }
    puts(dfs(1,m+1,0)? "YES":"NO");
    return 0;
}