Portal-->bzoj4894

Solution

  这题的话其实,一句话题意就是求。。外向树(方向是根往叶子)。。

  然后关于有向图的生成树计数的话,求外向树就是将度数矩阵改成入度,内向树就是改成出度

​  然后其他的一样的

  

  代码大概长这个样子

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int N=310,MOD=1e9+7;
int a[N][N],mp[N][N],d[N][N];
int n,m;
int solve(int n); int main(){
#ifndef ONLINE_JUDGE
freopen("a.in","r",stdin);
#endif
char ch;
scanf("%d\n",&n);
for (int i=1;i<=n;++i){
for (int j=1;j<=n;++j){
scanf("%c",&ch);
mp[i][j]=ch-'0';
if (mp[i][j]) ++d[j][j];
}
scanf("\n");
}
for (int i=1;i<=n;++i)
for (int j=1;j<=n;++j)
a[i][j]=(d[i][j]-mp[i][j]+MOD)%MOD;
printf("%d\n",solve(n));
} int solve(int n){
int id,mark=1;
int tmp;
for (int i=2;i<=n;++i){
id=i;
for (int j=i+1;j<=n;++j)
if (a[j][i]){id=j;break;}
if (id!=i){
mark=-mark;
for (int j=2;j<=n;++j) swap(a[id][j],a[i][j]);
}
for (int j=i+1;j<=n;++j){
while (a[j][i]){
tmp=a[j][i]/a[i][i];
for (int k=2;k<=n;++k)
a[j][k]=(1LL*a[j][k]+MOD-1LL*tmp*a[i][k]%MOD)%MOD;
if (a[j][i]==0) break;
mark=-mark;
for (int k=2;k<=n;++k)
swap(a[j][k],a[i][k]);
}
}
}
int ret=mark;
for (int i=2;i<=n;++i)
ret=1LL*ret*a[i][i]%MOD;
return (ret+MOD)%MOD;
}
05-11 16:11