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QOJ

IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#187935#7509. 01treebulijiojiodibuliduo#TL 13ms8916kbC++174.3kb2023-09-25 06:57:122023-09-25 06:57:12

Judging History

你现在查看的是最新测评结果

  • [2023-09-25 06:57:12]
  • 评测
  • 测评结果:TL
  • 用时:13ms
  • 内存:8916kb
  • [2023-09-25 06:57:12]
  • 提交

answer

#include <bits/stdc++.h>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define pb push_back
#define eb emplace_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef vector<int> VI;
typedef basic_string<int> BI;
typedef long long ll;
typedef pair<int,int> PII;
typedef double db;
mt19937 mrand(1); 
const ll mod=1000000007;
int rnd(int x) { return mrand() % x;}
ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}
// head

template<int MOD, int RT> struct mint {
	static const int mod = MOD;
	static constexpr mint rt() { return RT; } // primitive root for FFT
	int v; explicit operator int() const { return v; } // explicit -> don't silently convert to int
	mint():v(0) {}
	mint(ll _v) { v = int((-MOD < _v && _v < MOD) ? _v : _v % MOD);
		if (v < 0) v += MOD; }
	bool operator==(const mint& o) const {
		return v == o.v; }
	friend bool operator!=(const mint& a, const mint& b) { 
		return !(a == b); }
	friend bool operator<(const mint& a, const mint& b) { 
		return a.v < b.v; }
   
	mint& operator+=(const mint& o) { 
		if ((v += o.v) >= MOD) v -= MOD; 
		return *this; }
	mint& operator-=(const mint& o) { 
		if ((v -= o.v) < 0) v += MOD; 
		return *this; }
	mint& operator*=(const mint& o) { 
		v = int((ll)v*o.v%MOD); return *this; }
	mint& operator/=(const mint& o) { return (*this) *= inv(o); }
	friend mint pow(mint a, ll p) {
		mint ans = 1; assert(p >= 0);
		for (; p; p /= 2, a *= a) if (p&1) ans *= a;
		return ans; }
	friend mint inv(const mint& a) { assert(a.v != 0); 
		return pow(a,MOD-2); }
		
	mint operator-() const { return mint(-v); }
	mint& operator++() { return *this += 1; }
	mint& operator--() { return *this -= 1; }
	friend mint operator+(mint a, const mint& b) { return a += b; }
	friend mint operator-(mint a, const mint& b) { return a -= b; }
	friend mint operator*(mint a, const mint& b) { return a *= b; }
	friend mint operator/(mint a, const mint& b) { return a /= b; }
};

const int MOD=1000000007; 
using mi = mint<MOD,5>; // 5 is primitive root for both common mods

namespace simp {
	vector<mi> fac,ifac,invn;
	void check(int x) {
		if (fac.empty()) {
			fac={mi(1),mi(1)};
			ifac={mi(1),mi(1)};
			invn={mi(0),mi(1)};
		}
		while (SZ(fac)<=x) {
			int n=SZ(fac),m=SZ(fac)*2;
			fac.resize(m);
			ifac.resize(m);
			invn.resize(m);
			for (int i=n;i<m;i++) {
				fac[i]=fac[i-1]*mi(i);
				invn[i]=mi(MOD-MOD/i)*invn[MOD%i];
				ifac[i]=ifac[i-1]*invn[i];
			}
		}
	}
	mi gfac(int x) {
		check(x); return fac[x];
	}
	mi ginv(int x) {
		check(x); return invn[x];
	}
	mi gifac(int x) {
		check(x); return ifac[x];
	}
	mi binom(int n,int m) {
		if (m < 0 || m > n) return mi(0);
		return gfac(n)*gifac(m)*gifac(n - m);
	}
}

const int N=101000;
VI e[N];
int par[N],c0[N],cm[N],d0[N],dm[N],a[N],b[N];
int a0,am,b0,bm,n;
char s1[N],s2[N];
mi ans;
void dfs0(int u,int f) {
	for (auto v:e[u]) if (v!=f) {
		par[v]=par[u]^1;
		dfs0(v,u);
	}
}
void dfs(int u,int f) {
	c0[u]=a[u]==0,cm[u]=a[u]==-1;
	d0[u]=b[u]==0,dm[u]=b[u]==-1;
	for (auto v:e[u]) if (v!=f) {
		dfs(v,u);
		c0[u]+=c0[v];
		cm[u]+=cm[v];
		d0[u]+=d0[v];
		dm[u]+=dm[v];
	}
	int c1=c0[u],c2=cm[u],c3=a0-c1,c4=am-c2;
	int d1=d0[u],d2=dm[u],d3=b0-d1,d4=bm-d2;
	rep(i1,0,c2+1) rep(i2,0,c4+1) rep(j1,0,d2+1) rep(j2,0,d4+1) {
		if (c1+c3+i1+i2==d1+d3+j1+j2) {
			ans+=simp::binom(c2,i1)*simp::binom(d2,j1)*simp::binom(c4,i2)
				*simp::binom(d4,j2)*mi(abs((c1+i1)-(d1+j1)));
		}
	}
}
void solve() {
	scanf("%d",&n);
	rep(i,1,n+1) e[i].clear();
	rep(i,1,n) {
		int u,v;
		scanf("%d%d",&u,&v);
		e[u].pb(v);
		e[v].pb(u);
	}
	dfs0(1,0);
	scanf("%s",s1+1);
	rep(i,1,n+1) {
		a[i]=s1[i]=='?'?-1:(s1[i]-'0');
		if (a[i]!=-1) a[i]^=par[i];
	}
	scanf("%s",s2+1);
	rep(i,1,n+1) {
		b[i]=s2[i]=='?'?-1:(s2[i]-'0');
		if (b[i]!=-1) b[i]^=par[i];
	}
	a0=0,am=0;
	b0=0,bm=0;
	rep(i,1,n+1) {
		a0+=(a[i]==0),am+=(a[i]==-1);
		b0+=(b[i]==0),bm+=(b[i]==-1);
	}
	ans=0;
	dfs(1,0);
	printf("%d\n",(int)ans);
}

int _;
int main() {
	for (scanf("%d",&_);_;_--) {
		solve();
	}
}

Details

Tip: Click on the bar to expand more detailed information

Test #1:

score: 100
Accepted
time: 0ms
memory: 8916kb

input:

3
2
1 2
00
11
3
1 2
2 3
???
???
3
1 2
2 3
??1
0?0

output:

1
16
1

result:

ok 3 number(s): "1 16 1"

Test #2:

score: 0
Accepted
time: 13ms
memory: 7292kb

input:

1000
23
1 2
1 3
1 4
2 5
5 6
4 7
3 8
4 9
8 10
8 11
8 12
1 13
7 14
10 15
7 16
7 17
5 18
18 19
12 20
9 21
21 22
6 23
00?10?0000??1?00111?010
011??1?10?01?110?0??101
6
1 2
1 3
1 4
4 5
3 6
000?10
1???01
25
1 2
2 3
2 4
4 5
5 6
2 7
4 8
5 9
7 10
8 11
11 12
5 13
11 14
3 15
6 16
14 17
1 18
4 19
6 20
4 21
5 22...

output:

53545
24
11724
2228
162
14
711
28
550
1680
52
2
13
988
9480
2342
626416
0
71780
1
88
39207
19448
4
37395
9602
3253496
38057200
1066
3
303732
1608
281022
11718
170
78
15
1219376
29364
9092
7
0
2
7014
1942448
170371
75845
167217
16554
64
904
564290
14822
1127090
1758504
1227646
14833316
14954550
36055...

result:

ok 1000 numbers

Test #3:

score: -100
Time Limit Exceeded

input:

1
3000
1 2
2 3
2 4
1 5
3 6
2 7
5 8
8 9
9 10
10 11
2 12
11 13
7 14
11 15
7 16
13 17
8 18
1 19
11 20
10 21
18 22
7 23
7 24
15 25
23 26
24 27
14 28
15 29
25 30
16 31
6 32
10 33
3 34
30 35
16 36
9 37
36 38
28 39
26 40
33 41
1 42
11 43
20 44
23 45
14 46
31 47
41 48
11 49
48 50
45 51
6 52
10 53
32 54
38 5...

output:


result: