BZOJ 4698: [SDOI2008]Sandy的卡片

题面

有权限号的去看吧
Luogu

Sol

差分后就是求多个串的最长公共子串
套路啊
拼在一起用不同字符隔开,后缀数组,二分答案,分块height,开桶记录即可
我把差分值离散了

# 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 _(2000000);

IL ll Read(){
    RG char c = getchar(); RG ll x = 0, z = 1;
    for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
    for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
    return x * z;
}

int T, n, a[_], sa[_], rk[_], tmp[_], height[_], t[_], ans, mx, vis[_], cnt[1010], Index;
int o[_], num[1010][1010], len, l[_];
 
IL bool Cmp(RG int i, RG int j, RG int k){  return tmp[i] == tmp[j] && tmp[i + k] == tmp[j + k];  }

IL void Suffix_Sort(){
    RG int m = mx;
    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) tmp[++l] = i;
        for(RG int i = 1; i <= n; ++i) if(sa[i] > k) tmp[++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[tmp[i]]];
        for(RG int i = 1; i <= m; ++i) t[i] += t[i - 1];
        for(RG int i = n; i; --i) sa[t[rk[tmp[i]]]--] = tmp[i];
        swap(rk, tmp); 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, h = 0; i <= n; ++i){
		if(h) --h;
        while(a[i + h] == a[sa[rk[i] - 1] + h]) ++h;
		height[rk[i]] = h;
    }
}

IL bool Check(RG int x){
	RG int tot = (vis[sa[1]] != 0); cnt[vis[sa[1]]] = ++Index;
	for(RG int i = 2; i <= n; ++i){
		if(height[i] >= x){
			if(cnt[vis[sa[i]]] != Index) tot += (vis[sa[i]] != 0), cnt[vis[sa[i]]] = Index;
		}
		else cnt[vis[sa[i]]] = ++Index, tot = (vis[sa[i]] != 0);
		if(tot == T) return 1;
	}
	return 0;
}

int main(RG int argc, RG char* argv[]){
	T = Read();
	for(RG int i = 1; i <= T; ++i){
		l[i] = Read();
		for(RG int j = 1; j <= l[i]; ++j) num[i][j] = Read();
	}
	for(RG int i = 1; i <= T; ++i)
		for(RG int j = 1; j <= l[i]; ++j) o[++len] = num[i][j] = num[i][j + 1] - num[i][j];
	sort(o + 1, o + len + 1); len = unique(o + 1, o + len + 1) - o - 1;
	mx = len;
	for(RG int i = 1; i <= T; ++i){
		for(RG int j = 1; j <= l[i]; ++j){
			num[i][j] = lower_bound(o + 1, o + len + 1, num[i][j]) - o;
			a[++n] = num[i][j]; vis[n] = i;
		}
		a[++n] = ++mx;
	}
	Suffix_Sort();
	Check(1);
	RG int l = 0, r = n;
	while(l <= r){
		RG int mid = (l + r) >> 1;
		if(Check(mid)) ans = mid + 1, l = mid + 1;
		else r = mid - 1;
	}
	printf("%d\n", ans);
    return 0;
}

posted @ 2018-01-25 10:36  Cyhlnj  阅读(91)  评论(0编辑  收藏  举报