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 <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 '\n'
#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 = 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] = i<mx?min(mx-i,z[i-l]):0;
while (i+z[i]<n&&s[z[i]]==s[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);
}
}
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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$i$, 集合$B$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 <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 '\n'
#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 = 2e3+10;
int n,m,a[N][N],deg[N],ans[N],vis[N];
bitset<N> 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]<a[i+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]);
}
}
}
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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 <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 '\n'
#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
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<<m)-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);
}
}
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