【矩阵树定理】BZOJ4596 [Shoi2016]黑暗前的幻想乡
题面在这里
首先要容斥，答案就是： \[
\sum_{i=0}^n (-1)^i\cdot count_i
\] 其中\(count_i\)表示i个公司不修路的方案数
这个可以用矩阵树定理求得，只要枚举那些公司不修路就好了
时间复杂度\(O(2^n\cdot n^3)\)
示例程序：
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| #include<cstdio>
#include<cstring>
#include<algorithm>
#define cl(x,y) memset(x,y,sizeof(x))
using namespace std;
typedef long long ll;
const int maxn=25;
typedef ll matrix[maxn][maxn];
const ll MD=1e9+7;
int n,m[maxn];
ll ans;
matrix kir;
struct data{
int x,y;
inline void read() {scanf("%d%d",&x,&y);}
}a[maxn][105];
inline void A(ll &a,ll b){ (a+=(b%MD+MD)%MD)%=MD; }
inline ll inv(ll a){
ll res=1,w=a%MD,b=MD-2;
while (b){
if (b&1) res=res*w%MD;
w=w*w%MD;
b>>=1;
}
return res;
}
ll det(matrix a,int n){
int tot=0;
for (int i=1;i<=n;i++){
int where=-1;
for (int j=i;j<=n;j++) if (a[j][i]!=0) {where=j;break;}
if (where<0) return 0;
if (where>i){
tot++;
for (int j=1;j<=n;j++) swap(a[i][j],a[where][j]);
}
for (int j=i+1;j<=n;j++){
ll t=a[j][i]*inv(a[i][i])%MD;
for (int k=i;k<=n;k++) A(a[j][k],-a[i][k]*t);
}
}
ll res=1;
for (int i=1;i<=n;i++) (res*=a[i][i])%=MD;
return (tot&1)?MD-res:res;
}
int main(){
scanf("%d",&n);
for (int i=1;i<n;i++){
scanf("%d",&m[i]);
for (int j=1;j<=m[i];j++) a[i][j].read();
}
for (int s=0;s<(1<<n-1);s++){
cl(kir,0);int tot=0;
for (int i=1;i<=n;i++)
if (s&(1<<i-1)){tot++;
for (int j=1;j<=m[i];j++){
int x=a[i][j].x,y=a[i][j].y;
A(kir[x][y],-1);A(kir[y][x],-1);A(kir[x][x],1);A(kir[y][y],1);
}}
A(ans,((n-1-tot&1)?-1:1)*det(kir,n-1));
}
printf("%lld",ans);
return 0;
}
|
