2019牛客国庆集训派对day1

C. Distinct Substrings

大意: 给定串$s$, 字符集$m$, 对于每个字符$c$, 求$s$末尾添加字符$c$后本质不同子串增加多少.

exkmp求出每个前缀与后缀匹配的最大长度, 统计一下贡献即可

#include 
#include 
#include 
#include 
#include 
#include <set>
#include 
#include 
#include <string>
#include 
#include 
#include 
#include 
#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 '\n'
#define DB(a) ({REP(__i,1,n) cout<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 = 1e6+50;
int n,m,a[N],z[N],po[N],h[N],b[N];

void init(int *s, int *z, int n) {
    int mx=0,l=0;
    REP(i,1,n-1) {
        z[i] = i0;
        while (i+z[i]z[i];
        if (i+z[i]>mx) mx=i+z[i],l=i;
    }
}

int main() {
    po[0] = 1;
    REP(i,1,N-1) po[i] = po[i-1]*3ll%P;
    while (~scanf("%d%d", &n, &m)) {
        REP(i,1,m) h[i] = -1;
        REP(i,1,n) { 
            scanf("%d",a+i);
            h[a[i]] = 0;
        }
        REP(i,1,n) b[i]=a[n-i+1];
        init(b+1,z+1,n);
        REP(i,1,n-1) h[a[i+1]] = max(h[a[i+1]],z[n-i+1]);
        ll ans = 0;
        REP(i,1,m) ans ^= (ll)(n-h[i])*po[i]%P;
        printf("%lld\n", ans);
    }
}
View Code

  

G. 字典序

大意: 给定$nm$矩阵$C$, 求将所有列重排, 使得每一行的字典序都不超过下一行, 输出字典序最小方案.

好题. 假设只有两行, 假设集合$A$为$C_{1,i}C_{2,i}$的$i$.

那么必须满足$B$中所有元素必须全排在某一个$A$中元素后面.

$n$的情况就相当于给了$n-1$个这样的限制. 维护一个堆, 拓排一下即可.

#include 
#include 
#include 
#include 
#include 
#include <set>
#include 
#include 
#include <string>
#include 
#include 
#include 
#include 
#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 '\n'
#define DB(a) ({REP(__i,1,n) cout<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 = 2e3+10;
int n,m,a[N][N],deg[N],ans[N],vis[N];
bitset A[N];
priority_queue<int,vector<int>,greater<int> > q;
int main() {
    while (~scanf("%d%d", &n, &m)) {
        REP(i,1,n) REP(j,1,m) scanf("%d",a[i]+j);
        REP(i,1,n) vis[i] = 0;
        REP(i,1,m) { 
            deg[i] = 0;
            REP(j,1,n) A[i][j]=0;
        }
        REP(i,1,n-1) REP(j,1,m) {
            if (a[i][j]>a[i+1][j]) ++deg[j];
            if (a[i][j]1][j]) A[j][i] = 1;
        }
        while (q.size()) q.pop();
        REP(i,1,m) if (!deg[i]) q.push(i);
        int ok = 1;
        REP(i,1,m) {
            if (q.empty()) {ok = 0;break;}
            ans[i] = q.top(); q.pop();
            REP(j,1,n-1) if (A[ans[i]][j]&&!vis[j]) {
                vis[j] = 1;
                REP(x,1,m) if (a[j][x]>a[j+1][x]) {
                    if (!--deg[x]) q.push(x);
                }
            }
        }
        if (!ok) puts("-1");
        else { 
            REP(i,1,m) printf("%d%c", ans[i]," \n"[i==m]);
        }
    }
}
View Code

 

J. Parity of Tuples (Easy)

大意: 给定$n$个$m$维向量$v_i=(a_{i,1},...,a_{i,m})$

定义$count(x)$表示对于$x$合法的向量数.

一个向量$v_i$对于$x$合法, 要满足对于所有的$j$, $a_{i,j}\wedge x$二进制有奇数个$1$.

求$\sum\limits_{x=0}^{2^k-1}count(x)\cdot 3^{x}$

关键是要发现每行贡献是独立的, 求出每行贡献再加一下即可.

求出$dp_{x,S}$表示考虑到第$x$位, 满足条件的$j$的状态为$S$的贡献.

答案就为$dp_{k-1,2^{m}-1}$

#include 
#include 
#include 
#include 
#include 
#include <set>
#include 
#include 
#include <string>
#include 
#include 
#include 
#include 
#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 '\n'
#define DB(a) ({REP(__i,1,n) cout<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


int n,m,k,a[100],po[100];
int dp[2][1<<10],nxt[1<<10];
void add(int &a, ll b) {a=(a+b)%P;}
int calc() {
    int cur = 0, mx = (1<1;
    REP(i,0,mx) dp[cur][i] = 0;
    int sta = 0;
    REP(i,0,m-1) if (a[i]&1) sta ^= 1<<i;
    dp[cur][0] = 1;
    add(dp[cur][sta],3);
    REP(i,1,k-1) {
        cur ^= 1;
        //第i位填0
        memcpy(dp[cur],dp[!cur],sizeof dp[0]);
        int sta = 0;
        REP(j,0,m-1) if (a[j]>>i&1) sta ^= 1<<j;
        REP(s,0,mx) {
            //第i位填1
            add(dp[cur][s^sta],(ll)dp[!cur][s]*po[i]);
        }
    }
    return dp[cur][mx];
}

int main() {
    int t = 1;
    REP(i,0,50) { 
        po[i] = qpow(3,t);
        t = (ll)t*2%(P-1);
    }
    while (~scanf("%d%d%d",&n,&m,&k)) {
        int ans = 0;
        REP(i,1,n) { 
            REP(j,0,m-1) scanf("%d",a+j);
            ans = (ans+calc())%P;
        }
        printf("%d\n", ans);
    }
}
View Code

 

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