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CodeForces 1286E Fedya the Potter Strikes Back

作者:互联网

洛谷传送门

CF 传送门

KMP 好题。

思路

合法的子区间其实就是原串的 \(\mathrm{border}\),考虑维护 \(\mathrm{border}\) 的集合。每次加入一个字符,就保留原来合法的 \(\mathrm{border}\) 并加入新的合法 \(\mathrm{border}\)(如果 \(s_1 = s_i\))。

重点在于如何删掉不合法的 \(\mathrm{border}\)。设 \(fa_{i,j}\) 表示 \([1,pos+1]\) 能被 \([i-pos+1,i]\) 和字符 \(j\) 拼接成的 \(pos\) 最大值,每次可以 \(O(26)\) 处理。枚举 \(i - 1\) 后接上的 \(\ne s_i\) 的字符,暴力跳 \(fa\) 删前缀即可。用一个 map 维护每个 \(w_i\) 出现的次数,则删完之后把 map 中 \(> w_i\) 的权值全部修改为 \(w_i\) 即可。

线段树维护区间 \(w_i\) 最小值,时间复杂度为 \(O(n \log n + 26n)\)。

代码

code
/*

p_b_p_b txdy
AThousandMoon txdy
AThousandSuns txdy
hxy txdy

*/

#include <bits/stdc++.h>
#define pb push_back
#define fst first
#define scd second

using namespace std;
typedef long long ll;
typedef pair<ll, ll> pii;

struct bignum {
	const int P = 10000;
	
	int a[10], n;
	
	void add(ll x) {
		int pos = 0;
		while (x) {
			a[pos] += x % P;
			x /= P;
			++pos;
		}
		for (int i = 0; i < 9; ++i) {
			a[i + 1] += a[i] / P;
			a[i] %= P;
		}
		n = 9;
		while (n && !a[n]) {
			--n;
		}
	}
	
	ll mod(ll p) {
		ll res = 0;
		for (int i = n; ~i; --i) {
			res = (res * P + a[i]) % p;
		}
		return res;
	}
	
	void write() {
		printf("%d", a[n]);
		for (int i = n - 1; ~i; --i) {
			printf("%04d", a[i]);
		}
		putchar('\n');
	}
} ans;

const int maxn = 600100;

int n, a[maxn], b[maxn], tree[maxn << 2], fail[maxn];
int fa[maxn][26];
map<ll, ll> mp;

void pushup(int x) {
	tree[x] = min(tree[x << 1], tree[x << 1 | 1]);
}

void update(int rt, int l, int r, int x, int y) {
	if (l == r) {
		tree[rt] = y;
		return;
	}
	int mid = (l + r) >> 1;
	if (x <= mid) {
		update(rt << 1, l, mid, x, y);
	} else {
		update(rt << 1 | 1, mid + 1, r, x, y);
	}
	pushup(rt);
}

int query(int rt, int l, int r, int ql, int qr) {
	if (ql <= l && r <= qr) {
		return tree[rt];
	}
	int mid = (l + r) >> 1, res = 2e9;
	if (ql <= mid) {
		res = min(res, query(rt << 1, l, mid, ql, qr));
	}
	if (qr > mid) {
		res = min(res, query(rt << 1 | 1, mid + 1, r, ql, qr));
	}
	return res;
}

void solve() {
	char op[9];
	scanf("%d%s%d", &n, op, &b[1]);
	a[1] = op[0] - 'a';
	update(1, 1, n, 1, b[1]);
	ans.add(b[1]);
	ans.write();
	ll cur = 0;
	for (int i = 2, j = 0; i <= n; ++i) {
		scanf("%s%d", op, &b[i]);
		a[i] = (op[0] - 'a' + ans.mod(26)) % 26;
		b[i] ^= ans.mod(1LL << 30);
		update(1, 1, n, i, b[i]);
		ans.add(query(1, 1, n, 1, i));
		if (a[1] == a[i]) {
			cur += b[i];
			++mp[b[i]];
		}
		while (j && a[i] != a[j + 1]) {
			j = fail[j];
		}
		if (a[i] == a[j + 1]) {
			++j;
		}
		fail[i] = j;
		for (int k = 0; k < 26; ++k) {
			fa[i][k] = fa[fail[i]][k];
		}
		fa[i][a[fail[i] + 1]] = fail[i];
		for (int k = 0; k < 26; ++k) {
			if (a[i] == k) {
				continue;
			}
			for (int u = fa[i - 1][k]; u; u = fa[u][k]) {
				int mn = query(1, 1, n, i - u, i - 1);
				cur -= mn;
				--mp[mn];
			}
		}
		ll cnt = 0;
		vector<ll> tv;
		for (auto it = mp.upper_bound(b[i]); it != mp.end(); ++it) {
			tv.pb(it->fst);
			cnt += it->scd;
			cur -= it->fst * it->scd;
		}
		for (ll x : tv) {
			mp.erase(x);
		}
		cur += cnt * b[i];
		mp[b[i]] += cnt;
		ans.add(cur);
		ans.write();
	}
}

int main() {
	int T = 1;
	// scanf("%d", &T);
	while (T--) {
		solve();
	}
	return 0;
}

标签:int,res,Fedya,Potter,Back,pos,mp,border,ll
来源: https://www.cnblogs.com/zltzlt-blog/p/16437093.html