Bzoj2946：[POI2000] 最长公共子串

题面

求多个串的最长公共子串

Sol

套路，拼在一起，二分答案+后缀数组判定

把大于答案的\(height\)分组，然后计算出一个组内是否有所有串的后缀

由于串只有\(5\)个开个桶就好了

# include <bits/stdc++.h>

# define IL inline

# define RG register

# define Fill(a, b) memset(a, b, sizeof(a))

using namespace std;

typedef long long ll;

const int _(100010);

int T, n, a[_], sa[_], rk[_], y[_], height[_], t[_], vis[_];

char s[_];

IL bool Cmp(RG int i, RG int j, RG int k){  return y[i] == y[j] && y[i + k] == y[j + k];  }

IL void Sort(){

	RG int m = 30;

	for(RG int i = 1; i <= n; ++i) ++t[rk[i] = a[i]];

	for(RG int i = 1; i <= m; ++i) t[i] += t[i - 1];

	for(RG int i = n; i; --i) sa[t[rk[i]]--] = i;

	for(RG int k = 1; k <= n; k <<= 1){

		RG int l = 0;

		for(RG int i = n - k + 1; i <= n; ++i) y[++l] = i;

		for(RG int i = 1; i <= n; ++i) if(sa[i] > k) y[++l] = sa[i] - k;

		for(RG int i = 0; i <= m; ++i) t[i] = 0;

		for(RG int i = 1; i <= n; ++i) ++t[rk[y[i]]];

		for(RG int i = 1; i <= m; ++i) t[i] += t[i - 1];

		for(RG int i = n; i; --i) sa[t[rk[y[i]]]--] = y[i];

		swap(rk, y); rk[sa[1]] = l = 1;

		for(RG int i = 2; i <= n; ++i) rk[sa[i]] = Cmp(sa[i - 1], sa[i], k) ? l : ++l;

		if(l >= n) break;

		m = l;

	}

	for(RG int i = 1, j = 0; i <= n; ++i){

		j = max(0, j - 1);

		while(a[j + i] == a[sa[rk[i] - 1] + j]) ++j;

		height[rk[i]] = j;

	}

}

IL bool Check(RG int x){

	RG int v[6], cnt = 0; Fill(v, 0);

	for(RG int i = 2; i <= n; ++i){

		if(height[i] >= x){

			if(vis[sa[i]] && !v[vis[sa[i]]]) ++cnt;

			++v[vis[sa[i]]];

		}

		else{

			Fill(v, 0); cnt = 0;

			if(vis[sa[i]] && !v[vis[sa[i]]]) ++cnt;

			++v[vis[sa[i]]];

		}

		if(cnt == T) return 1;

	}

	return 0;

}

int main(RG int argc, RG char* argv[]){

	scanf("%d", &T);

	for(RG int id = 1; id <= T; ++id){

		scanf(" %s", s); RG int l = strlen(s);

		for(RG int i = 0; i < l; ++i) a[++n] = s[i] - 'a' + 1, vis[n] = id;

		a[++n] = 27;

	}

	Sort();

	RG int l = 0, r = n, ans = 0;

	while(l <= r){

		RG int mid = (l + r) >> 1;

		if(Check(mid)) ans = mid, l = mid + 1;

		else r = mid - 1;

	}

	printf("%d\n", ans);

    return 0;

}