【LOJ】#2496. 「AHOI / HNOI2018」毒瘤

题面

还有这么诚实的出题人!

我们最多影响20个点,然后把这20个点的虚树建出来,并且枚举每个点的选举状态,如果一个点选或不选可以通过改(dp[u][0] = 0)(dp[u][1] = 0)完成

状态应该不多,因为每条边只有三种选的情况,上限是(3^{m - n + 1})
然后我们考虑递推出(dp[u][0])(dp[u][1])在更新它虚树上的父亲的方案数,是可以用(k_0 * dp[u][0] + k_1 * dp[u][1])表示的,这个可以递推出来

然后就剩别把代码敲错了= =

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define enter putchar('
')
#define space putchar(' ')
#define MAXN 100005
#define eps 1e-8
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;char c = getchar();T f = 1;
    while(c < '0' || c > '9') {
	if(c == '-') f = -1;
	c = getchar();
    }
    while(c >= '0' && c <= '9') {
	res = res * 10 + c - '0';
	c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) {
	out(x / 10);
    }
    putchar('0' + x % 10);
}
const int MOD = 998244353;
int N,M;
struct node {
    int to,next;
}E[MAXN * 2];
int head[MAXN],sumE,belong[MAXN],dfn[MAXN],idx,aux[MAXN],faAux[MAXN],acnt,tot,fa[MAXN][20],dep[MAXN];
int sta[MAXN],top,dp[MAXN][2],g[MAXN][2][2],val[MAXN][2],id[MAXN],Ncnt,qa[MAXN][2];
pii p[MAXN];
int inc(int a,int b) {
    return a + b >= MOD ? a + b - MOD : a + b; 
}
int mul(int a,int b) {
    return 1LL * a * b % MOD;
}
int fpow(int x,int c) {
    int res = 1,t = x;
    while(c) {
	if(c & 1) res = mul(res,t);
	t = mul(t,t);
	c >>= 1;
    }
    return res;
}
void update(int &x,int y) {
    x = inc(x,y);
}
int getbl(int x) {
    return belong[x] == x ? x : belong[x] = getbl(belong[x]);
}
void add(int u,int v) {
    E[++sumE].to = v;
    E[sumE].next = head[u];
    head[u] = sumE;
}
bool cmp(int a,int b) {
    return dfn[a] < dfn[b];
}
int dfs(int u) {
    dfn[u] = ++idx;
    dep[u] = dep[fa[u][0]] + 1;
    dp[u][0] = dp[u][1] = 1;
    for(int i = head[u] ; i ; i = E[i].next) {
	int v = E[i].to;
	if(v != fa[u][0]) {
	    fa[v][0] = u;
	    dfs(v);
	    dp[u][0] = mul(dp[u][0],inc(dp[v][1],dp[v][0]));
	    dp[u][1] = mul(dp[u][1],dp[v][0]);
	}
    }
}
int lca(int u,int v) {
    if(dep[u] < dep[v]) swap(u,v);
    int l = 18;
    while(dep[u] > dep[v]) {
	if(dep[fa[u][l]] >= dep[v]) u = fa[u][l];
	--l;
    }
    if(u == v) return u;
    l = 18;
    while(fa[u][0] != fa[v][0]) {
	if(fa[u][l] != fa[v][l]) {
	    u = fa[u][l];
	    v = fa[v][l];
	}
	--l;
    }
    return fa[u][0];
}
void build_AuxTree() {
    int t = acnt;
    top = 0;
    for(int i = 1 ; i <= acnt ; ++i) {
	if(!top) sta[++top] = aux[i];
	else {
	    int f = lca(aux[i],sta[top]);
	    while(top >= 1 && dep[sta[top]] > dep[f]) {
		if(top == 1 || dep[sta[top - 1]] <= dep[f]) {
		    faAux[sta[top]] = f;
		}
		--top;
	    }
	    if(sta[top] != f) {
		faAux[f] = sta[top];
		sta[++top] = f;
		aux[++t] = f;
	    }
	    sta[++top] = aux[i];
	    faAux[aux[i]] = f;
	}
    }
    acnt = t;
    sort(aux + 1,aux + acnt + 1,cmp);
    if(aux[1] != 1) {faAux[aux[1]] = 1;aux[++acnt] = 1;}
    sort(aux + 1,aux + acnt + 1,cmp);
    for(int i = 1 ; i <= acnt ; ++i) val[aux[i]][0] = dp[aux[i]][0],val[aux[i]][1] = dp[aux[i]][1];
    for(int i = acnt ; i >= 2 ; --i) {
	int u = aux[i],f = faAux[u];
        memset(g[u],0,sizeof(g[u]));
	g[u][0][0] = 1;g[u][1][1] = 1;
	while(fa[u][0] != f) {
	    int t = fa[u][0];
	    val[t][0] = mul(dp[t][0],fpow(inc(dp[u][0],dp[u][1]),MOD - 2));
	    val[t][1] = mul(dp[t][1],fpow(dp[u][0],MOD - 2));
	    memset(g[t],0,sizeof(g[t]));
	    g[t][0][0] = mul( inc( g[u][1][0] , g[u][0][0] ) , val[t][0]);
	    g[t][0][1] = mul( inc( g[u][1][1] , g[u][0][1] ) , val[t][0]);
	    g[t][1][0] = mul( g[u][0][0] , val[t][1]);
	    g[t][1][1] = mul( g[u][0][1] , val[t][1]);	    
	    u = t;
	}
	g[f][0][0] = inc( g[u][1][0] , g[u][0][0] );
	g[f][0][1] = inc( g[u][1][1] , g[u][0][1] );
	g[f][1][1] = g[u][0][1];
	g[f][1][0] = g[u][0][0];
	memcpy(g[aux[i]],g[f],sizeof(g[f]));
	val[f][0] = mul(val[f][0],fpow(inc(dp[u][0],dp[u][1]),MOD - 2));
	val[f][1] = mul(val[f][1],fpow(dp[u][0],MOD - 2));
    }
}
void Init() {
    read(N);read(M);
    for(int i = 1 ; i <= N ; ++i) belong[i] = i;
    int u,v;
    for(int i = 1 ; i <= M ; ++i) {
	read(u);read(v);
	if(getbl(u) != getbl(v)) {
	    belong[getbl(u)] = getbl(v);
	    add(u,v);add(v,u);
	}
	else {
	    p[++tot] = mp(u,v);
	}
    }
    dfs(1);
    if(M == N - 1) {
	out(inc(dp[1][0],dp[1][1]));enter;
	exit(0);
    }
    for(int j = 1 ; j <= 18 ; ++j) {
	for(int i = 1 ; i <= N ; ++i) {
	    fa[i][j] = fa[fa[i][j - 1]][j - 1];
	}
    }
    for(int i = 1 ; i <= tot ; ++i) {
	aux[++acnt] = p[i].fi;aux[++acnt] = p[i].se;
    }
    sort(aux + 1,aux + acnt + 1);
    
    acnt = unique(aux + 1,aux + acnt + 1) - aux - 1;
    for(int i = 1 ; i <= acnt ; ++i) {
	id[aux[i]] = ++Ncnt;
    }
    sort(aux + 1,aux + acnt + 1,cmp);
    build_AuxTree();
}
int Calc(int S) {
    for(int i = 1 ; i <= acnt ; ++i) qa[aux[i]][0] = qa[aux[i]][1] = 1;
    for(int i = acnt ; i >= 1 ; --i) {
	int u = aux[i];
	qa[u][0] = mul( qa[u][0] , val[u][0] );
	qa[u][1] = mul( qa[u][1] , val[u][1] );
	if(id[u]) {
	    qa[u][(S >> (id[u] - 1) & 1) ^ 1] = 0;
	}
	if(u == 1) break;
	qa[faAux[u]][0] = mul(inc( mul( g[u][0][1] , qa[u][1] ) , mul( g[u][0][0] , qa[u][0]) )  , qa[faAux[u]][0]);
	qa[faAux[u]][1] = mul(inc( mul( g[u][1][1] , qa[u][1] ) , mul( g[u][1][0] , qa[u][0]) )  , qa[faAux[u]][1]);
    }
    return inc(qa[1][0],qa[1][1]);
}
void Solve() {
    int ans = 0;
    for(int S = 0 ; S < (1 << Ncnt) ; ++S) {
	bool flag = 1;
	for(int i = 1 ; i <= tot ; ++i) {
	    if((S >> (id[p[i].fi] - 1) & 1) && (S >> (id[p[i].se] - 1) & 1)) {
		flag = 0;
		break;
	    }
	}
	if(!flag) continue;
	else update(ans,Calc(S));
    }
    out(ans);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Init();
    Solve();
    return 0;
}
原文地址:https://www.cnblogs.com/ivorysi/p/9970569.html