NAIPC 2018

E. Prefix Free Code

大意: 给定$n$个串, 保证任意一个串都不是另一个串的前缀, 从中选出$k$个串可以拼成$inom{n}{k}k!$种串. 给定其中一个串, 求这个串的排名.

先用字典树处理一下, 从而转化为给定一个$n$元素中取$k$元素的排列, 求排名.

#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <cstring>
#include <bitset>
#include <functional>
#include <random>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '
'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<',';hr;})
using namespace std;
typedef long long ll;
const int P = 1e9+7;
const int N = 1e6+10;
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
int n,k,ch[N][26],T,val[N],a[N],tot,cnt,len;
char s[N];
void add(int &o, char *s) {
    if (!o) o = ++tot;
    if (*s) add(ch[o][*s-'a'],s+1);
    else val[o] = 1;
}
void dfs(int o) {
    if (!o) return;
    if (val[o]) return val[o] = ++cnt,void();
    REP(i,0,25) dfs(ch[o][i]);
}
void find(int o, char *s) {
    if (val[o]) a[++*a]=val[o];
    else ++len,find(ch[o][*s-'a'],s+1);
}

int c[N];
void add(int x, int v) {
    for (; x<=n; x+=x&-x) c[x]+=v;
}
int qry(int x) {
    int ret = 0;
    for (; x; x^=x&-x) ret+=c[x];
    return ret;
}

int main() {
    scanf("%d%d",&n,&k);
    REP(i,1,n) scanf("%s",s),add(T,s);
    dfs(T);
    scanf("%s",s+1);
    int m = strlen(s+1);
    REP(i,1,m) len=0,find(T,s+i),i+=len-1;
    int tot = 1;
    REP(i,n-k+1,n) tot = (ll)tot*i%P;
    REP(i,1,n) c[i] = i&-i;
    int ans = 1;
    REP(i,1,k) {
        tot = (ll)tot*inv(n-i+1)%P;
        ans = (ans+(ll)qry(a[i]-1)*tot)%P;
        add(a[i],-1);
    }
    printf("%d
",ans);
}

H. Recovery

大意: 给定每行每列和的奇偶性, 要求恢复一个$n imes m$的$01$矩阵, 若有多种方案输出$1$个数最多的, 还有多种的话输出字典序最小的.

先初始化全$1$, 那么就转化为要改变某些行某些列的奇偶性, 如果行列需要改动的个数之差为奇数则无解, 偶数时贪心构造一下.

#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <cstring>
#include <bitset>
#include <functional>
#include <random>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '
'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<',';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head



const int N = 55;
int n,m;
vector<int> va,vb;
char ans[N][N],a[N],b[N];

int main() {
    scanf("%s%s",a+1,b+1);
    n = strlen(a+1);
    m = strlen(b+1);
    REP(i,1,n) if (m%2==0&&a[i]=='1'||m%2==1&&a[i]=='0') va.pb(i);
    REP(i,1,m) if (n%2==0&&b[i]=='1'||n%2==1&&b[i]=='0') vb.pb(i);
    int sza = va.size(), szb = vb.size();
    if ((sza-szb)&1) return puts("-1"),0;
    REP(i,1,n) REP(j,1,m) ans[i][j]='1';
    if (szb>=sza) {
        REP(i,0,szb-sza-1) ans[1][vb[i]] = '0';
        int now = 0;
        REP(i,szb-sza,szb-1) ans[va[now++]][vb[i]] = '0';
    }
    else {
        REP(i,0,sza-szb-1) ans[va[i]][1] = '0';
        int now = 0;
        REP(i,sza-szb,sza-1) ans[va[i]][vb[now++]] = '0';
    }
    REP(i,1,n) puts(ans[i]+1);
}