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QOJ

IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#590062#6837. AC Automatoncyj888TL 3ms36816kbC++115.0kb2024-09-25 21:21:092024-09-25 21:21:11

Judging History

你现在查看的是最新测评结果

  • [2024-09-25 21:21:11]
  • 评测
  • 测评结果:TL
  • 用时:3ms
  • 内存:36816kb
  • [2024-09-25 21:21:09]
  • 提交

answer

#include <bits/stdc++.h>
#define pb push_back 
#define ott(i,l,r) for (int i = l; i <= r; i ++)
#define tto(i,l,r) for (int i = r; i >= l; i --)
using namespace std;
using ll = long long;
inline int read () {
	int x = 0; bool f = 0; char c = getchar ();
	while (!isdigit (c)) f |= (c == '-'), c = getchar ();
	while (isdigit (c)) x = (x << 3) + (x << 1) + (c ^ 48), c = getchar ();
	if (f) x = -x; return x;
}
const int N = 3e5 + 110;
int n, m, q, idx, B;
ll res;
vector <int> e[N], E[N];
int up[N], dn[N];
int dfn[N], dep[N], fa[N];
int lg[N << 1], f[N << 1][20];
char ch[N];
inline void dfs0 (int u) {
	f[dfn[u] = ++ idx][0] = u, dep[u] = dep[fa[u]] + 1;
	for (int v : e[u]) {
		dfs0 (v);
		f[++ idx][0] = u;
	}
	return ;
}
inline int lca (int x, int y) {
	x = dfn[x], y = dfn[y];
	if (x > y) swap (x, y);
	int z = lg[y - x + 1], f1 = f[x][z], f2 = f[y - (1 << z) + 1][z];
	return dep[f1] < dep[f2] ? f1 : f2;
}
struct OPT {
	int x; char c;
} a[N];
struct BLOCK {
	int ac, now, sum;//ac = 'A'count
	int s[1201];
	inline void ins (int v) {
		if (v >= now) ++ sum;
		if (-m <= v && v <= m) ++ s[v + m];
		return ;
	}
	inline void upd (int i) {
		if (ch[i] == 'A') ++ ac;
		if (ch[i] == '?') ins (dn[i] - up[i]);
		return ;
	}
	inline void add (int v) {
		if (v == 1) {
			sum -= s[(now ++) + m];
			res -= 1ll * sum;
		}
		if (v == -1) {
			res += 1ll * sum;
			sum += s[(-- now) + m];
		}
		return ;
	}
	void clear () {
		ac = now = sum = 0;
		ott (i, 0, m + m) s[i] = 0;
		return ;
	}
} b[N];
int id[N], val[N], FA[N];
inline bool cmp (const int &x, const int &y) {
	return dfn[x] < dfn[y];
}
inline void push (int u, int z) {
	b[z].upd (u);
	for (int v : e[u]) push (v, z);
	return ;
}
int Z, OTH;
inline void find (int x) {
	if (id[fa[x]]) {
		val[Z = id[x] = ++ idx] = x;
		return ;
	}
	find (fa[x]);
	b[Z].upd (fa[x]);
	for (int v : e[fa[x]]) {
		if (v == x) continue;
		if (!OTH) val[OTH = id[v] = ++ idx] = v;
		push (v, OTH);
	}
	return ;
}
inline void red (int u) {
	dn[u] = 0;
	for (int v : e[u]) {
		up[v] = up[u] + (ch[u] != 'C');
		red (v);
		dn[u] += dn[v] + (ch[v] != 'A');
	}
	if (ch[u] == 'A') res += 1ll * dn[u];
	if (ch[u] == '?') res += 1ll * max (0, dn[u] - up[u]);
	return ;
}
inline void cup (int u, int V) {
	for (int v : E[u]) {
		if (val[v] < 0) {
			int tv = -val[v];
			if (ch[tv] == '?') res -= 1ll * max (0, dn[tv] - up[tv]);
			up[tv] += V;
			if (ch[tv] == '?') res += 1ll * max (0, dn[tv] - up[tv]);
		}
		else {
			b[v].add (V);
		}
		cup (v, V);
	}
	return ;
}
inline void cdn (int u, int V) {
	if (!u) return ;
	if (val[u] < 0) {
		int tu = -val[u];
		if (ch[tu] == '?') res -= 1ll * max (0, dn[tu] - up[tu]);
		dn[tu] += V; if (ch[tu] == 'A') res += 1ll * V;
		if (ch[tu] == '?') res += 1ll * max (0, dn[tu] - up[tu]);
	}
	else {
		b[u].add (-V), res += 1ll * b[u].ac * V;
	}
	cdn (FA[u], V);
	return ;
}
inline void sol () {
    ott (i, 1, m) val[++ idx] = a[i].x;
    sort (val + 1, val + 1 + idx), idx = unique (val + 1, val + 1 + idx) - val - 1;
    sort (val + 1, val + 1 + idx, cmp);
    ott (i, 2, idx) val[idx + i - 1] = lca (val[i - 1], val[i]); val[idx += idx] = 1;
    sort (val + 1, val + 1 + idx), idx = unique (val + 1, val + 1 + idx) - val - 1;
    ott (i, 1, idx) {
    	id[val[i]] = i;
    	val[i] = -val[i];//区分会修改的点
	}
    ott (i, 2, idx) {
    	if (val[i] > 0) break;
    	if (!id[fa[-val[i]]]) {
    		OTH = Z = 0, find (-val[i]);
    		if (OTH) E[FA[OTH] = i].pb (OTH);
    		E[FA[Z] = id[fa[val[Z]]]].pb (Z), E[FA[i] = Z].pb (i);
		}
		else E[FA[i] = id[fa[-val[i]]]].pb (i);
	}
	ott (i, 1, idx) {
		if (val[i] > 0) break;
		OTH = 0; for (int v : e[-val[i]]) {
			if (id[v]) continue;
			if (!OTH) {
				val[OTH = id[v] = ++ idx] = v;
				E[FA[OTH] = i].pb (OTH);
			}
			push (v, OTH);
		}
	}
	ott (i, 1, m) {
		int x = a[i].x; char c = a[i].c;
		if (ch[x] != 'C') cup (id[x], -1); if (ch[x] != 'A') cdn (FA[id[x]], -1);
		if (ch[x] == 'A') res -= 1ll * dn[x];
		if (ch[x] == '?') res -= 1ll * max (0, dn[x] - up[x]);
		ch[x] = c;
		if (ch[x] == 'A') res += 1ll * dn[x];
		if (ch[x] == '?') res += 1ll * max (0, dn[x] - up[x]);
		if (ch[x] != 'C') cup (id[x], 1); if (ch[x] != 'A') cdn (FA[id[x]], 1);
		printf ("%lld\n", res);
	}
	ott (i, 1, idx) {
		E[i].clear (), FA[i] = 0;
		if (val[i] < 0) val[i] = -val[i];
		else b[i].clear ();
		id[val[i]] = 0, val[i] = 0;
	}
	m = idx = 0, res = 0; red (1);
	return ;
}
int main () {
	//freopen ("data.in", "r", stdin); freopen ("res.out", "w", stdout);
	n = read (), q = read (), scanf ("%s", ch + 1); ott (i, 2, n) e[fa[i] = read ()].pb (i);
	B = 600; dfs0 (1);
	ott (j, 1, 19) {
		ott (i, 1, idx - (1 << j) + 1) {
			int f1 = f[i][j - 1], f2 = f[i + (1 << j - 1)][j - 1];
			f[i][j] = dep[f1] < dep[f2] ? f1 : f2;
		}
	}
	lg[0] = -1;	ott (i, 1, idx) lg[i] = lg[i >> 1] + 1;
	idx = 0, red (1); ott (i, 1, q) {
		int x = read (); char c = getchar ();
		a[++ m] = {x, c};
		if (i % B == 0 || i == q) sol ();
	}
	return 0;
}

Details

Tip: Click on the bar to expand more detailed information

Test #1:

score: 100
Accepted
time: 0ms
memory: 33344kb

input:

5 3
AC??C
1 2 2 1
1 ?
4 A
2 ?

output:

4
3
3

result:

ok 3 lines

Test #2:

score: 0
Accepted
time: 3ms
memory: 36816kb

input:

1000 1000
AC?ACCCCC?CCA??CCCCC?A?AC?C??C?ACCCACC?C?CCC?CACACA???????C?????CC?C?AAAAACCA??AAAACACACA???AACAC?CCC?CAC?AAAACCAC???ACAA??A??A?CA?A?ACACACCAA?ACACA?AC??AC??ACAAACCC??CAA?A???CC?AA??AC???A?CCCC??CACCAACC?AA?C?AAACCCA?AAA??C?CC??CACCACAACCA?AACCC?A?CCC?CC???AA?CACCCAAC??CAA??C?AA??CA???CAAA...

output:

2344
2345
2342
2342
2768
2768
2772
2772
2772
2772
2772
2772
2772
2767
2767
2767
2767
2764
2766
2766
2769
2765
2761
2764
2767
2772
2772
2772
2772
2772
2772
2777
2777
2777
2777
2774
2771
2774
2782
2778
2778
2772
2768
2772
2772
2772
2772
2772
2774
2774
2778
2781
2781
2779
2782
2784
2787
2782
2786
2788
...

result:

ok 1000 lines

Test #3:

score: -100
Time Limit Exceeded

input:

300000 300000
AAA?CA?AA?AC?A?CCA?AACCAAA???CA?ACCAACCCCAACAAA?CCAAAC?A?C??CC?C?C?CCCA?CAA?ACA??C?C?AC??CA??ACA?AA???CACAAA?CACCCCCCC?A?AAAAAC?AACCA????CCC?C?AAACCCAA?C???CCCC?AAACAAA???A?CAAC??A??A??CCCC??AA?C??ACA?AACAAA????CAA???AAAAACC?C?CCA?CCAA?AAC?CC?CA?A??CC??CCAC??C??????AAC?AA?AA?AAC?C??AAC...

output:

14995917235
14995917235
14996064601
14996083631
14995980103
14995925797
14995925797
14995925797
14995967213
14995967213
14995967213
14995876211
14995774037
14995774037
14995774037
14995876791
14995866113
14995756158
14995647554
14995647554
14995560537
14995560537
14995583619
14995583619
14995583619
...

result: