【LOJ】#2027. 「SHOI2016」黑暗前的幻想乡

题解

我一开始写的最小表示法写的插头dp，愉快地TLE成60分

然后我觉得我就去看正解了！
发现是容斥 + 矩阵树定理

矩阵树定理对于有重边的图只要邻接矩阵的边数设置a[u][v]表示u,v之间有几条边就好

我们枚举哪些公司不用，然后用矩阵树求一下生成几棵树，复杂度\(2^{n - 1}(n - 1)^3\)

代码

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <vector>
#include <set>
#include <cmath>
#define enter putchar('\n')
#define space putchar(' ')
//#define ivorysi
#define pb push_back
#define MAXN 200005
#define mo 974711
#define pii pair<int,int>
#define mp make_pair
#define fi first
#define se second
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 - '0' + c;
    c = getchar();
    }
    res = 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 = 1000000007;
int N,cnt[(1 << 17) + 5],D[18][18],len[25];
pii E[25][505];
int lowbit(int x) {
    return x & (-x);
}
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;
}
void Init() {
    read(N);
    int M,u,v;
    for(int i = 1 ; i < N ; ++i) {
    read(M);len[i] = M;
    for(int j = 1 ; j <= M ; ++j) {
        read(u);read(v);
        E[i][j] = mp(u,v);
    }
    }
}
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;
}
int Guass() {
    int res = 1;
    for(int i = 2 ; i <= N ; ++i) {
    int l = i;
    if(!D[l][i]) {
        for(int j = i + 1 ; j <= N ; ++j) {
        if(D[j][i]) {l = j;break;}
        }
    }
    if(!D[l][i]) return 0;
    if(l != i) {
        res = -res;
        for(int j = i ; j <= N ; ++j) swap(D[l][j],D[i][j]);
    }
    for(int j = i + 1; j <= N ; ++j) {
        int t = mul(D[j][i],fpow(D[i][i],MOD - 2));
        for(int k = i ; k <= N ; ++k) {
        D[j][k] = inc(D[j][k],MOD - mul(D[i][k],t));
        }
    }
    }
    if(res == -1) res = MOD - 1;
    for(int i = 2 ; i <= N ; ++i) {
    res = mul(res,D[i][i]);
    }
    return res;
}
int calc(int S) {
    memset(D,0,sizeof(D));
    for(int i = 1 ; i <= N - 1; ++i) {
    if(S >> (i - 1) & 1) {
        for(int j = 1 ; j <= len[i] ; ++j) {
        D[E[i][j].fi][E[i][j].se] -= 1;
        D[E[i][j].se][E[i][j].fi] -= 1;
        D[E[i][j].fi][E[i][j].fi]++;
        D[E[i][j].se][E[i][j].se]++;
        }
    }
    }
    for(int i = 2 ; i <= N; ++i) {
    if(!D[i][i]) return 0;
    for(int j = 2 ; j <= N ; ++j) {
        D[i][j] = inc(D[i][j],MOD);
    }
    }
    return Guass();
}
void Solve() {
    int ans = 0;
    for(int S = 0 ; S < (1 << (N - 1)) ; ++S) {
    if(S) cnt[S] = cnt[S - lowbit(S)] + 1;
    if((N - 1 - cnt[S]) & 1) ans = inc(ans,MOD - calc(S));
    else ans = inc(ans,calc(S));
    }
    out(ans);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Init();
    Solve();
}