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ID题目提交者结果用时内存语言文件大小提交时间测评时间
#583294#6837. AC Automatoncyj888WA 18ms32672kbC++114.4kb2024-09-22 19:26:342024-09-22 19:26:36

Judging History

你现在查看的是最新测评结果

  • [2024-09-22 19:26:36]
  • 评测
  • 测评结果:WA
  • 用时:18ms
  • 内存:32672kb
  • [2024-09-22 19:26:34]
  • 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;
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, res;
vector <int> e[N], E[N];
int up[N], dn[N];
int fa[N][20], dfn[N], dep[N];
char s[N];
void dfs0 (int u) {
	dfn[u] = ++ idx, dep[u] = dep[fa[u][0]] + 1;
	ott (j, 1, 18) fa[u][j] = fa[fa[u][j - 1]][j - 1];
	for (int v : e[u]) dfs0 (v);
	return ;
}
int lca (int x, int y) {
	if (dep[x] < dep[y]) swap (x, y);
	tto (j, 0, 18) if (dep[fa[x][j]] >= dep[y]) x = fa[x][j];
	if (x == y) return x;
	tto (j, 0, 18) if (fa[x][j] != fa[y][j]) x = fa[x][j], y = fa[y][j];
	return fa[x][0];
}
struct OPT {
	int x; char c;
} a[N];
struct BLOCK {
	int ac, now, sum;//ac = 'A'count
	int s[1100];
	void ins (int v) {
		if (v >= now) ++ sum;
		if (-m <= v && v <= m) ++ s[v + m];
		return ;
	}
	void add (int v) {
		if (v == 1) {
			sum -= s[(now ++) + m];
			res -= sum;
		}
		if (v == -1) {
			res += sum;
			sum += s[(-- now) + m];
		}
		return ;
	}
	void clear () {
		ac = now = sum = 0;
		ott (i, 0, m + m) s[i] = 0;
		return ;
	}
} b[N];
int t[N], FA[N], bl[N], oth[N];
bool cmp (int x, int y) {
	return dfn[x] < dfn[y];
}
void fuck (int u, int z) {
	bl[u] = z;
	for (int v : e[u]) fuck (v, z);
	return ;
}
void cpr (int u) {
	int OTH = 0;
	for (int v : e[u]) {
		if (bl[v]) cpr (v);
		else if (bl[u] < 0) {
			if (!OTH) OTH = t[++ t[0]] = v;
			fuck (v, OTH);
		}
		else {
			if (!oth[bl[u]]) oth[bl[u]] = t[++ t[0]] = v;
			fuck (v, oth[bl[u]]);
		}
		
	}
	return ;
}
int find (int u) {
	return bl[u] = (bl[fa[u][0]] ? u : find (fa[u][0]));
}
void red (int u) {
	dn[u] = 0;
	for (int v : e[u]) {
		up[v] = up[u] + (s[u] != 'C');
		red (v);
		dn[u] += dn[v] + (s[v] != 'A');
	}
	if (s[u] == 'A') res += dn[u];
	if (s[u] == '?') res += max (0, dn[u] - up[u]);
	return ;
}
void cup (int u, int val) {
	for (int v : E[u]) {
		if (bl[v] < 0) {
			if (s[v] == '?') res -= max (0, dn[v] - up[v]);
			up[v] += val;
			if (s[v] == '?') res += max (0, dn[v] - up[v]);
		}
		else {
			b[v].add (val);
		}
		cup (v, val);
	}
	return ;
}
void cdn (int u, int val) {
	if (!u) return ;
	if (bl[u] < 0) {
		if (s[u] == '?') res -= max (0, dn[u] - up[u]);
		dn[u] += val; if (s[u] == 'A') res += val;
		if (s[u] == '?') res += max (0, dn[u] - up[u]);
	}
	else {
		b[u].add (-val), res += b[u].ac * val;
	}
	cdn (FA[u], val);
	return ;
}

void sol () {
    ott (i, 1, m) t[++ t[0]] = a[i].x;//ott (i, 1, n) t[++ t[0]] = i; 
    sort (t + 1, t + 1 + t[0]), t[0] = unique (t + 1, t + 1 + t[0]) - t - 1;
    sort (t + 1, t + 1 + t[0], cmp);
    ott (i, 2, t[0]) t[t[0] + i - 1] = lca (t[i - 1], t[i]); assert (t[0]); t[t[0] += t[0]] = 1;
    sort (t + 1, t + 1 + t[0]), t[0] = unique (t + 1, t + 1 + t[0]) - t - 1;
    ott (i, 1, t[0]) bl[t[i]] = -t[i];//区分会修改的点
    ott (i, 2, t[0]) {
    	int now = fa[t[i]][0];
    	if (!bl[now]) t[++ t[0]] = find (now);
	}
	cpr (1), sort (t + 1, t + 1 + t[0], cmp);
	ott (i, 1, n) {
		assert (bl[i]);
		if (bl[i] < 0) continue;
		if (s[i] == 'A') ++ b[bl[i]].ac;
		if (s[i] == '?') b[bl[i]].ins (dn[i] - up[i]);
	}
	ott (i, 2, t[0]) E[FA[t[i]] = lca (t[i - 1], t[i])].pb (t[i]);
	ott (i, 1, m) {
		int x = a[i].x; char c = a[i].c;
		if (s[x] != 'C') cup (x, -1); if (s[x] != 'A') cdn (FA[x], -1);
		if (s[x] == 'A') res -= dn[x];
		if (s[x] == '?') res -= max (0, dn[x] - up[x]);
		s[x] = c;
		if (s[x] == 'A') res += dn[x];
		if (s[x] == '?') res += max (0, dn[x] - up[x]);
		if (s[x] != 'C') cup (x, 1); if (s[x] != 'A') cdn (FA[x], 1);
		printf ("%d\n", res);
	}
	ott (i, 1, t[0]) {
		FA[t[i]] = 0, E[t[i]].clear ();
		if (bl[t[i]] == t[i]) b[t[i]].clear (), oth[t[i]] = 0;
		t[i] = 0;
	}
	ott (i, 1, n) bl[i] = 0;
	t[0] = m = 0, res = 0;
	return ;
}
int main () {
	n = read (), q = read (), cin >> s + 1; ott (i, 2, n) e[fa[i][0] = read ()].pb (i);
	B = 1;
	dfs0 (1), red (1); ott (i, 1, q) {
		int x = read (); char c = getchar ();
		a[++ m] = {x, c};
		if (i % B == 0 || i == q) {
			sol (), red (1);
		}
	}
	return 0;
}
/*
10 5
??????????
1 1 2 2 3 3 4 4 5
4 A
10 A
4 C
10 C
1 C


*/

詳細信息

Test #1:

score: 100
Accepted
time: 0ms
memory: 28644kb

input:

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

output:

4
3
3

result:

ok 3 lines

Test #2:

score: -100
Wrong Answer
time: 18ms
memory: 32672kb

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:

2342
2342
2345
2340
2760
2768
2767
2772
2772
2772
2772
2772
2772
2772
2767
2767
2767
2764
2766
2766
2766
2765
2761
2764
2764
2772
2772
2772
2772
2772
2772
2772
2777
2777
2777
2779
2771
2774
2774
2783
2775
2772
2772
2768
2769
2772
2772
2769
2774
2774
2774
2777
2781
2779
2779
2779
2783
2787
2782
2788
...

result:

wrong answer 1st lines differ - expected: '2344', found: '2342'