題目鏈接

BZOJ5315

題解

題目好嚇人= =點仙人掌 + 斯坦納樹

我們只需求出對於所有選點的方案的斯坦納樹邊長總和
\(n\)那麽大當然不能狀壓，但是考慮一下如果這是一棵樹，一個方案的貢獻就是連接這些點的所有邊
我們可以考慮計算每條邊的貢獻
一條邊在樹上有貢獻，當且僅當它兩端的樹都存在被選擇的點
那麽這條邊\((u,v)\)貢獻就是
\[(2^{siz[u]} - 1)(2^{siz[v] - 1})\]
其中\(siz[u]\)表示斷開這條邊後\(u\)一側的樹大小

如果放到仙人掌上呢？
對於割邊，和樹是一樣的
我們只需計算每個環的貢獻
考慮我們對於一個環，選擇了其中\(K\)個點所在外向樹，那麽就有連接\(K\)

直接轉移是\(O(n^4)\)的，常數很小數據很水可以跑過。。。

當然可以前綴和優化成\(O(n^3)\)

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
 

#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
using namespace std;
const int maxn = 205,maxm = 100005,INF = 1000000000,P = 1000000007;
inline int read(){
    int out = 0,flag = 1; char c = getchar();
    while (c < 48 || c > 57){if (c == ‘-‘) flag = -1; c = getchar();}
    while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
    return out * flag;
}
int h[maxn],ne = 1;
struct EDGE{int to,nxt;}ed[maxm];
inline void build(int u,int v){
    ed[++ne] = (EDGE){v,h[u]}; h[u] = ne;
    ed[++ne] = (EDGE){u,h[v]}; h[v] = ne;
}
int n,m;
inline int qpow(int a,int b){
    int re = 1;
    for (; b; b >>= 1,a = 1ll * a * a % P)
        if (b & 1) re = 1ll * re * a % P;
    return re;
}
int dfn[maxn],low[maxn],siz[maxn],fa[maxn],cnt,sum;
int v[maxn],K;
LL f[maxn][maxn][maxn],D[maxn][maxn][maxn],S[maxn][maxn][maxn],ans,bin[maxn];
void DP(int rt,int u){
    K = 0; int tot = 0;
    for (int i = u; i != rt; i = fa[i]){
        v[++K] = siz[i];
        siz[rt] += siz[i];
        tot += siz[i];
    }
    v[++K] = sum - tot;
    cls(f);
    for (int l = 1; l <= K; l++)
        for (int r = l; r <= K; r++)
            for (int k = 0; k <= r - l; k++){
                if (l == r){
                    if (k == 0) f[l][r][k] = bin[v[l]] - 1;
                    continue;
                }
                int d = 0,s = 0;
                for (int i = 0; i <= k; i++) d = (d + f[l][r - k][i]) % P;
                for (int i = r - k + 1; i < r; i++) s = (s + f[l][i][k]) % P;
                f[l][r][k] = (bin[v[r]] - 1) * (d + s) % P;
                ans = (ans + 1ll * (K - max(K - r + l,k)) * f[l][r][k] % P) % P;
            }
}
void dfs(int u){
    dfn[u] = low[u] = ++cnt; siz[u] = 1;
    Redge(u) if ((to = ed[k].to) != fa[u]){
        if (!dfn[to]){
            fa[to] = u; dfs(to);
            low[u] = min(low[u],low[to]);
        }
        else low[u] = min(low[u],dfn[to]);
        if (low[to] > dfn[u]){
            ans = (ans + 1ll * (bin[siz[to]] - 1) * (bin[sum - siz[to]] - 1) % P) % P;
            siz[u] += siz[to];
        }
    }
    Redge(u) if (fa[to = ed[k].to] != u && dfn[u] < dfn[to])
        DP(u,to);
}
int main(){
    bin[0] = 1; for (int i = 1; i <= 200; i++) bin[i] = bin[i - 1] * 2ll % P;
    n = read(); m = read();
    while (m--) build(read(),read());
    sum = n; dfs(1);
    ans = ans * qpow(bin[n],P - 2) % P;
    printf("%lld\n",ans);
    return 0;
}

BZOJ5315 [JSOI2018]防禦網絡 【仙人掌 + dp】