CodeForces

( ext{Description})

传送门

( ext{Solution})

容易发现如果算出长度为 (L) 的公共前缀为 (x) 的 (A,B,C) 个数为 ((cnt_1,cnt_2,cnt_3))，那么长度为 (len) 小于 (L) 的公共前缀为 ”(x) 的前 (len) 位” 的 (A,B,C) 个数至少有 ((cnt_1,cnt_2,cnt_3))。

我们将 (h) 从大到小枚举，将新的满足条件的共同前缀合并。

比如有：

( ext{woshizhu})

( ext{woshixiannv})

本来在 (L>5) 时这两个串没有关联，当 (L=5) 时就可以合并。

( ext{Code})

#include <cstdio>

#define rep(i,_l,_r) for(register signed i=(_l),_end=(_r);i<=_end;++i)
#define fep(i,_l,_r) for(register signed i=(_l),_end=(_r);i>=_end;--i)
#define erep(i,u) for(signed i=head[u],v=to[i];i;i=nxt[i],v=to[i])
#define efep(i,u) for(signed i=Head[u],v=to[i];i;i=nxt[i],v=to[i])
#define print(x,y) write(x),putchar(y)

template <class T> inline T read(const T sample) {
    T x=0; int f=1; char s;
    while((s=getchar())>'9'||s<'0') if(s=='-') f=-1;
    while(s>='0'&&s<='9') x=(x<<1)+(x<<3)+(s^48),s=getchar();
    return x*f;
}
template <class T> inline void write(const T x) {
    if(x<0) return (void) (putchar('-'),write(-x));
    if(x>9) write(x/10);
    putchar(x%10^48);
}
template <class T> inline T Max(const T x,const T y) {if(x>y) return x; return y;}
template <class T> inline T Min(const T x,const T y) {if(x<y) return x; return y;}
template <class T> inline T fab(const T x) {return x>0?x:-x;}
template <class T> inline T gcd(const T x,const T y) {return y?gcd(y,x%y):x;}
template <class T> inline T lcm(const T x,const T y) {return x/gcd(x,y)*y;}
template <class T> inline T Swap(T &x,T &y) {x^=y^=x^=y;}

#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

const int mod=1e9+7,maxn=3e5+5;

int ans[maxn],ret,n,m=122,siz,minn,ori[maxn],op[maxn],x[maxn],y[maxn],sa[maxn],c[maxn],num,rk[maxn],h[maxn],f[maxn],dp[maxn][4],p[maxn];
char s[maxn];

void Read() {
	scanf("%s",s+1); n=siz=minn=strlen(s+1);
	rep(i,1,siz) ori[i]=s[i],op[i]=0;
	ori[++n]=1; op[n]=-1;
	scanf("%s",s+1); siz=strlen(s+1); minn=Min(minn,siz);
	rep(i,n+1,n+siz) ori[i]=s[i-n],op[i]=1;
	ori[n=n+siz+1]=2,op[n]=-1;
	scanf("%s",s+1); siz=strlen(s+1); minn=Min(minn,siz);
	rep(i,n+1,n+siz) ori[i]=s[i-n],op[i]=2;
	n+=siz;
}

void Suffix() {
	rep(i,1,n) ++c[x[i]=ori[i]];
	rep(i,2,m) c[i]+=c[i-1];
	rep(i,1,n) sa[c[x[i]]--]=i;
	for(int k=1;k<=n;k<<=1) {
		num=0;
		rep(i,n-k+1,n) y[++num]=i;
		rep(i,1,n) if(sa[i]>k) y[++num]=sa[i]-k;
		rep(i,1,m) c[i]=0;
		rep(i,1,n) ++c[x[i]];
		rep(i,2,m) c[i]+=c[i-1];
		fep(i,n,1) sa[c[x[y[i]]]--]=y[i],y[i]=0;
		swap(x,y);
		x[sa[1]]=num=1;
		rep(i,2,n)
			x[sa[i]]=(y[sa[i]]==y[sa[i-1]]&&y[sa[i]+k]==y[sa[i-1]+k])?num:++num;
		m=num;
	}
}

void LCP() {
	int k=0;
	rep(i,1,n) rk[sa[i]]=i;
	rep(i,1,n) {
		if(rk[i]==1) continue;
		if(k) --k;
		int j=sa[rk[i]-1];
		while(j+k<=n&&i+k<=n&&ori[j+k]==ori[i+k]) ++k;
		h[rk[i]]=k;
	}
}

void init() {
	rep(i,1,n) f[i]=p[i]=i;
}

bool cmp(int x,int y) {
	return h[x]>h[y];
}

int Find(int x) {return x==f[x]?x:f[x]=Find(f[x]);}

void Merge(int x,int y) {
	if(x==y) return;
	ret=(ret-1ll*dp[x][0]*dp[x][1]%mod*dp[x][2]%mod+mod)%mod; // 减去之前的贡献，不然会算重，因为我们是严格要求长度为 L
	ret=(ret-1ll*dp[y][0]*dp[y][1]%mod*dp[y][2]%mod+mod)%mod;
	rep(i,0,2) dp[x][i]=(dp[x][i]+dp[y][i])%mod;
	ret=(ret+1ll*dp[x][0]*dp[x][1]%mod*dp[x][2]%mod+mod)%mod;
	f[y]=x;
}

int main() {
	Read(); Suffix(); LCP();
	init(); sort(p+1,p+n+1,cmp);
	rep(i,1,n) if(~op[i]) dp[i][op[i]]=1;
	int j=1;
	fep(i,minn,1) {
		while(j<=n&&h[p[j]]>=i) {
			Merge(Find(sa[p[j]]),Find(sa[p[j]-1]));
			++j;
		}
		ans[i]=ret;
	}
	rep(i,1,minn) print(ans[i],' '); puts("");
	return 0;
}

( ext{Old Version})

感觉题目越来越不会做了怎么办(QWQ)。如今只有康题解辽。

你能想象这道题用并查集来优化吗？

(Solution)

枚举特别慢，我们就按顺序枚举。从大到小保证符合大的要求一定符合小的要求。

我们相当于做一个后缀和。每次加入新的满足的情况数。

巧妙的是并查集。因为无法确定下一次，所以只能用一坨的和来做。这就符合并查集的性质。

我的码风真好看。（自恋五秒钟）

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;

const int N = 3e5 + 8;
const ll mod = 1e9 + 7;

int ans[N], f[N], p[N], id[N], lg[N], len, rmq[N][20], n, m, h[N], tax[N], SA[N], tp[N], rk[N], a[N];
ll dp[N][3], tmp;

int read() {
	int x = 0, f = 1; char S;
	while((S = getchar()) > '9' || S < '0') {
		if(S == '-') f = -1;
		if(S == EOF) exit(0);
	}
	while(S <= '9' && S >= '0') {
		x = (x << 1) + (x << 3) + (S ^ 48);
		S = getchar();
	}
	return x * f;
}

bool cmp(const int x, const int y, const int d) {return tp[x] == tp[y] && tp[x + d] == tp[y + d];}

void Sort() {
    for(int i = 0; i <= m; ++ i) tax[i] = 0;
    for(int i = 1; i <= n; ++ i) ++ tax[rk[tp[i]]];
    for(int i = 1; i <= m; ++ i) tax[i] += tax[i - 1];
    for(int i = n; i >= 1; -- i) SA[tax[rk[tp[i]]] --] = tp[i];
}

void init() {
    char ch[N];
    scanf("%s", ch); int siz = strlen(ch); len = siz;
    for(int i = 0; i < siz; ++ i) a[++ n] = ch[i], id[n] = 1;
    a[++ n] = 1; id[n] = -1;
    scanf("%s", ch); siz = strlen(ch); len = min(len, siz);
    for(int i = 0; i < siz; ++ i) a[++ n] = ch[i], id[n] = 2;
    a[++ n] = 2; id[n] = -1;
    scanf("%s", ch); siz = strlen(ch); len = min(len, siz);
    for(int i = 0; i < siz; ++ i) a[++ n] = ch[i], id[n] = 3;
    a[++ n] = 3; id[n] = -1;
    m = 122;
}

void Suffix() {
    init();
    for(int i = 1; i <= n; ++ i) rk[i] = a[i], tp[i] = i;
    Sort();
    for(int w = 1, p = 1, i; p < n; m = p, w <<= 1) {
        for(p = 0, i = n - w + 1; i <= n; ++ i) tp[++ p] = i;
        for(i = 1; i <= n; ++ i) if(SA[i] > w) tp[++ p] = SA[i] - w;
        Sort(); swap(rk, tp); rk[SA[1]] = p = 1;
        for(i = 2; i <= n; ++ i) rk[SA[i]] = cmp(SA[i], SA[i - 1], w) ? p : ++ p;
    }
    int j, k = 0;
    for(int i = 1; i <= n; h[rk[i ++]] = k)
        for(k = k ? k - 1 : k, j = SA[rk[i] - 1]; a[i + k] == a[j + k]; ++ k);
    for(int i = 2; i <= n; ++ i) lg[i] = lg[i >> 1] + 1;
    for(int i = 1; i <= n; ++ i) rmq[i][0] = h[i];
    for(int j = 1; (1 << j) <= n; ++ j)
        for(int i = 1; i + (1 << j) - 1 <= n; ++ i)
            rmq[i][j] = min(rmq[i][j - 1], rmq[i + (1 << j - 1)][j - 1]);
}

int lcp(const int x, const int y) {
    int l = rk[x], r = rk[y];
    if(l == r) return n - SA[l] + 1;
    if(l > r) swap(l, r);
    int t = lg[r - l];
    return min(rmq[l + 1][t], rmq[r - (1 << t) + 1][t]);
}

void makeSet() {
    for(int i = 1; i <= n; ++ i) p[i] = i, f[i] = i;
}

int findSet(const int x) {
    return x == f[x] ? x : f[x] = findSet(f[x]);
}

void sub(ll &x, const int y) {
    x = (x - y + mod) % mod;
}

void unionSet(const int x, const int y) {
    sub(tmp, dp[x][0] * dp[x][1] % mod * dp[x][2] % mod);
    sub(tmp, dp[y][0] * dp[y][1] % mod * dp[y][2] % mod);
    for(int i = 0; i < 3; ++ i) (dp[x][i] += dp[y][i]) %= mod;
    (tmp += dp[x][0] * dp[x][1] % mod * dp[x][2] % mod) %= mod;
    f[y] = x;
}

bool Cmp(const int x, const int y) {
    return h[x] > h[y];
}

int main() {
    Suffix();
    makeSet();
    sort(p + 1, p + n + 1, Cmp);
    for(int i = 1; i <= n; ++ i)
        if(id[i] >= 1)
            dp[i][id[i] - 1] = 1;
    int j = 1;
    for(int i = len; i >= 1; -- i) {
        while(j <= n && h[p[j]] >= i) {
            unionSet(findSet(SA[p[j] - 1]), findSet(SA[p[j]]));
            ++ j;
        }
        ans[i] = tmp;
    }
    for(int i = 1; i <= len; ++ i) printf("%d ", ans[i]);
    putchar('
');
    return 0;
}