Codeforces 231E - Cactus

题意：给定n（2<=n<=100000）个点的无向图，一个点最多只在一个simple cycle中，定义simple path为u->v经历的路径不同即不同，

         k（1<=k<=100000）组询问u v，问u->v的simple路径的个数。

题解：缩点+倍增lca

Sure原创，转载请注明出处

#include <iostream>
#include <cstdio>
#include <memory.h>
#define MIN(a , b) ((a) < (b) ? (a) : (b))
using namespace std;
const int mod = 1000000007;
const int maxn = 100002;
const int fix = 16;
struct node
{
    int v;
    int next;
}edge[maxn << 1];
int head[maxn],dfn[maxn],low[maxn],belong[maxn],cnt[maxn],cycle[maxn];
int s[maxn],dis[maxn],dep[maxn],fa[maxn][fix],U[maxn],V[maxn];
bool vis[maxn],instack[maxn];
int m,n,q,idx,tmpdfn,tot,top;

void swap(int &a,int &b)
{
    int tmp = a;
    a = b;
    b = tmp;
    return;
}

void init()
{
    memset(head,-1,sizeof(head));
    memset(cnt,0,sizeof(cnt));
    memset(fa,0,sizeof(fa));
    memset(dis,0,sizeof(dis));
    memset(cycle,0,sizeof(cycle));
    memset(vis,false,sizeof(vis));
    memset(instack,false,sizeof(instack));
    idx = tmpdfn = tot = 0;
    return;
}

void addedge(int u,int v)
{
    edge[idx].v = v;
    edge[idx].next = head[u];
    head[u] = idx++;

    edge[idx].v = u;
    edge[idx].next = head[v];
    head[v] = idx++;
    return;
}

void read()
{
    for(int i=0;i<m;i++)
    {
        scanf("%d %d",&U[i],&V[i]);
        addedge(U[i],V[i]);
    }
    return;
}

void tarjan(int st,int pre)
{
    dfn[st] = low[st] = tmpdfn++;
    vis[st] = instack[st] = true;
    s[top++] = st;
    for(int i=head[st];i != -1;i=edge[i].next)
    {
        if(edge[i].v == pre) continue;
        if(vis[edge[i].v] == false)
        {
            tarjan(edge[i].v , st);
            low[st] = MIN(low[st] , low[edge[i].v]);
        }
        else if(instack[edge[i].v])
        {
            low[st] = MIN(low[st] , dfn[edge[i].v]);
        }
    }
    if(low[st] == dfn[st])
    {
        int u;
        tot++;
        do
        {
            u = s[--top];
            instack[u] = false;
            belong[u] = tot;
            cnt[tot]++;
        }while(u != st);
        if(cnt[tot] > 1) cycle[tot] = 1;
    }
    return;
}

void make()
{
    for(int i=1;i<=n;i++)
    {
        if(vis[i] == false)
        {
            top = 0;
            tarjan(i , -1);
        }
    }
    memset(head,-1,sizeof(head));
    idx = 0;
    for(int i=0;i<m;i++)
    {
        int u = belong[U[i]];
        int v = belong[V[i]];
        if(u != v) addedge(u , v);
    }
    return;
}

void dfs(int st,int pre,int d)
{
    dep[st] = d;
    for(int i=head[st];i != -1;i=edge[i].next)
    {
        if(edge[i].v == pre) continue;
        fa[edge[i].v][0] = st;
        for(int j=1;j<fix;j++)
        {
            fa[edge[i].v][j] = fa[fa[edge[i].v][j-1]][j-1];
        }
        dis[edge[i].v] = dis[st] + cycle[edge[i].v];
        dfs(edge[i].v , st , d + 1);
    }
    return;
}

int lca(int u,int v)
{
    if(dep[u] > dep[v]) swap(u , v);
    if(dep[u] < dep[v])
    {
        int del = dep[v] - dep[u];
        for(int i=0;i<fix;i++)
        {
            if(del & (1 << i))
            {
                v = fa[v][i];
            }
        }
    }
    if(u != v)
    {
        for(int i=fix-1;i>=0;i--)
        {
            if(fa[u][i] != fa[v][i])
            {
                u = fa[u][i];
                v = fa[v][i];
            }
        }
        u = fa[u][0];
        v = fa[v][0];
    }
    return u;
}

int quick_pow(int x)
{
    __int64 res = 1LL,tmp = 2 * 1LL;
    while(x)
    {
        if(x & 1)
        {
            res *= tmp;
            res %= mod;
        }
        tmp *= tmp;
        tmp %= mod;
        x >>= 1;
    }
    return (int)res;
}

void solve()
{
    dis[belong[1]] = cycle[belong[1]];
    dfs(belong[1] , -1 , 0);
    scanf("%d",&q);
    int u,v;
    while(q--)
    {
        scanf("%d %d",&u,&v);
        u = belong[u];
        v = belong[v];
        int l = lca(u , v);
        int c = dis[u] + dis[v] - 2 * dis[l] + cycle[l];
        printf("%d\n",quick_pow(c));
    }
    return;
}

int main()
{
    while(~scanf("%d %d",&n,&m))
    {
        init();
        read();
        make();
        solve();
    }
    return 0;
}