Fantasia 【HDU - 5739】【广义圆方树】

题目链接

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  这道题的题意在这里就不展开了，因为这次的英文比较的好读，也没有生词，好懂一些些hh。

  然后，这道题的关键点，肯定就是来看这个点是不是一个割点，也就是割去这个点之后就将原来的联通块展开成两个以上联通块，把原来的联通性给断开了，于是，基于这个，我想到的解法就是广义圆方树了。

  很明显的，广义圆方树会形成新的方点，我们给新的方点赋值为1，肯定就不会影响到答案了，于是真正影响答案的就只有圆点了，于是就很好解了，一个点，肯定是圆点，它能拆分成的贡献，就一定是它的根结点到它父亲的上面的这部分，还有就是下面的几个方点会被割开，于是就会又有这些贡献，然而，边的数量是有限的，于是这道题的复杂度就是O(M)级别的了。

  为什么是O(M)？因为每个点都要查它分割出来的子树所以肯定是直接跑边。一个边最多两个结点，所以每条边最多也就是跑两遍。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <bitset>
//#include <unordered_map>
//#include <unordered_set>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define INF 0x3f3f3f3f3f3f3f3f
#define eps 1e-8
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
#define MP(a, b) make_pair(a, b)
#define MAX_3(a, b, c) max(a, max(b, c))
using namespace std;
typedef unsigned long long ull;
typedef unsigned int uit;
typedef long long ll;
const ll mod = 1e9 + 7;
inline ll fast_mi(ll a, ll b = mod - 2LL)
{
    ll ans = 1;
    while(b) { if(b & 1) ans = ans * a % mod; a = a * a % mod; b >>= 1LL; }
    return ans;
}
const int maxN = 2e5 + 7, maxM = 4e5 + 7;
int N, M;
ll w[maxN];
struct Graph
{
    int head[maxN], cnt;
    struct Eddge
    {
        int nex, to;
        Eddge(int a=-1, int b=0):nex(a), to(b) {}
    }edge[maxM];
    inline void addEddge(int u, int v)
    {
        edge[cnt] = Eddge(head[u], v);
        head[u] = cnt++;
    }
    inline void _add(int u, int v) { addEddge(u, v); addEddge(v, u); }
    inline void clear()
    {
        cnt = 0;
        for(int i=1; i<=N; i++) head[i] = -1;
    }
} Old, Now;
vector<int> rt;
ll pro[maxN], sum;
struct Tarjan_Algorlth
{
    int dfn[maxN], low[maxN], tot, Stap[maxN], Stop, Belong[maxN], Bcnt, top[maxN], root;
    void Tarjan(int u, int fa)
    {
        dfn[u] = low[u] = ++tot; top[u] = root; pro[root] = pro[root] * w[u] % mod;
        Stap[++Stop] = u;
        for(int i=Old.head[u], v, p; ~i; i=Old.edge[i].nex)
        {
            v = Old.edge[i].to;
            if(v == fa) continue;
            if(!dfn[v])
            {
                Tarjan(v, u);
                low[u] = min(low[u], low[v]);
                if(low[v] >= dfn[u])
                {
                    Bcnt++; Now.head[N + Bcnt] = -1; w[N + Bcnt] = 1;
                    do
                    {
                        p = Stap[Stop--];
                        Belong[p] = Bcnt;
                        Now._add(p, N + Bcnt);
                    } while(p ^ v);
                    Now._add(u, N + Bcnt);
                }
            }
            else low[u] = min(low[u], dfn[v]);
        }
    }
    inline void clear()
    {
        tot = Bcnt = Stop = 0;
        for(int i=1; i<=N; i++) dfn[i] = 0;
    }
} Tj;
ll siz[maxN];
int fa[maxN];
void dfs(int u, int father)
{
    fa[u] = father;
    siz[u] = w[u];
    for(int i=Now.head[u], v; ~i; i=Now.edge[i].nex)
    {
        v = Now.edge[i].to;
        if(v == father) continue;
        dfs(v, u);
        siz[u] = siz[u] * siz[v] % mod;
    }
}
inline void init()
{
    Old.clear(); Now.clear(); Tj.clear();
    rt.clear(); sum = 0;
}
int main()
{
    int T; scanf("%d", &T);
    while(T--)
    {
        scanf("%d%d", &N, &M);
        init();
        for(int i=1; i<=N; i++) scanf("%lld", &w[i]);
        for(int i=1, u, v; i<=M; i++)
        {
            scanf("%d%d", &u, &v);
            Old._add(u, v);
        }
        int len = 0;
        for(int i=1; i<=N; i++) if(!Tj.dfn[i])
        {
            rt.push_back(i); len++;
            pro[i] = 1; Tj.root = i;
            Tj.Tarjan(i, 0);
            sum = (sum + pro[i]) % mod;
        }
        for(int i=0; i<len; i++) dfs(rt[i], 0);
        ll ans = 0, tmp; int gen;
        for(ll u=1; u<=N; u++)
        {
            tmp = 0;
            gen = Tj.top[u];
            ans = (ans + (sum - siz[gen] + mod) % mod * u % mod) % mod;
            if(u ^ gen)
            {
                ans = (ans + u * siz[gen] % mod * fast_mi(siz[u]) % mod) % mod;
            }
            for(int i=Now.head[u], v; ~i; i=Now.edge[i].nex)
            {
                v = Now.edge[i].to;
                if(v == fa[u]) continue;
                ans = (ans + u * siz[v] % mod) % mod;
            }
        }
        printf("%lld\n", ans);
    }
    return 0;
}