Cactus CodeForces

大意: 给定无向图, 每个点最多属于一个简单环, 多组询问, 求给定起点终点, 有多少条简单路径.

先缩环, 然后假设两点树上路径经过$cnt$个环, 那么答案就为$2^{cnt}$.

要注意缩环建树时要加单向边.

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <queue>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define pb push_back
using namespace std;

template <class T> void rd(T &x){x=0;bool f=0;char c=getchar();while(c<'0'||c>'9'){if(c=='-')f=1;c=getchar();}while('0'<=c&&c<='9'){x=x*10+c-'0';c=getchar();}if(f)x=-x;}

const int N = 1e6+10, P = 1e9+7;
int n,m,dep[N],fa[N];
int s[N],vis[N],sz[N];
int son[N],top[N],v[N],fac[N];
vector<int> g[N],gg[N];
int Find(int x) {return s[x]?s[x]=Find(s[x]):x;}
void add(int x, int y) {
	x=Find(x),y=Find(y);
	if (x!=y) s[x]=y;
}
void get(int x, int y) {
    if (dep[x]<dep[y]) return;
	vis[y] = 1;
    for (; x!=y; x=fa[x]) add(x,y),vis[x]=1;
}
void dfs(int x, int f) {
    fa[x]=f,dep[x]=dep[f]+1;
    for (int y:g[x]) if (y!=f) {
        if (dep[y]) get(x,y);
        else dfs(y,x);
    }
}
void dfs2(int x, int f, int d) {
	sz[x]=1,fa[x]=f,dep[x]=d,v[x]=v[f]+vis[x];
	for (int y:gg[x]) if (y!=f) {
		dfs2(y,x,d+1),sz[x]+=sz[y];
		if (sz[y]>sz[son[x]]) son[x]=y;
	}
}
void dfs3(int x, int tf) {
	top[x] = tf;
	if (son[x]) dfs3(son[x],tf);
	for (int y:gg[x]) if (!top[y]) dfs3(y,y);
}
int lca(int x, int y) {
	while (top[x]!=top[y]) {
		if (dep[top[x]]<dep[top[y]]) swap(x,y);
		x = fa[top[x]];
	}
	return dep[x]<dep[y]?x:y;
}

int main() {
	fac[0]=1;
	REP(i,1,N-1) fac[i]=fac[i-1]*2%P;
	rd(n),rd(m);
	REP(i,1,m) {
		int u, v;
		rd(u),rd(v);
		if (u==v) continue;
		g[u].pb(v),g[v].pb(u);
	}
	dfs(1,0);
	int rt = 1;
	REP(i,1,n) if (vis[i]) rt = Find(i);
	REP(i,1,n) {
		for (int j:g[i]) {
			int u=Find(i),v=Find(j);
			if (u!=v) gg[u].pb(v);
		}
	}
	dfs2(rt,0,0),dfs3(rt,rt);
	scanf("%d", &m);
	while (m--) {
		int x, y;
		scanf("%d%d", &x, &y);
		x = Find(x), y = Find(y);
		int l = lca(x,y);
		printf("%d
", fac[v[x]+v[y]-v[l]-v[fa[l]]]);
	}
}