Given an n + n
bipartite graph, each edge has p
base i
,j
p
base i
,j
. The probability of occurrence of, find the probability that has a perfect match.
n ≤ 7,15s n ≤ 7,15s
Subtask: n ≤ 6,7s n ≤ 6,7s
Solution
Shape pressure, 2^n
Bit represents point X
2^n 2^n
. Whether there is a perfect match for the subsets, each time the Y-square point is enumerated, a solution to the X-square point is connected.
After the memorized search can be passed.
After running dfs
, I found that there are only 6e4
different states.
#include<bits/stdc++.h>
#define int long long
#define debug(...) fprintf(stderr,__VA_ARGS__)
#define DEBUG printf("Passing [%s] in LINE %d\n",__FUNCTION__,__LINE__)
#define Debug debug("Passing [%s] in LINE %d\n",__FUNCTION__,__LINE__)
#define all(x) x.begin(),x.end()
#define x first
#define y second
typedef long long ll;
typedef pair<int,int> pii;
const double eps=1e-8;
const double pi=acos(-1.0);
template<class T>int chkmin(T &a,T b){return a>b?a=b,1:0;}
template<class T>int chkmax(T &a,T b){return a<b?a=b,1:0;}
template<class T>T sqr(T a){return a*a;}
template<class T>T mmin(T a,T b){return a<b?a:b;}
template<class T>T mmax(T a,T b){return a>b?a:b;}
template<class T>T aabs(T a){return a<0?-a:a;}
template<class T>int dcmp(T a,T b){return a>b;}
template<int *a>int cmp_a(int x,int y){return a[x]<a[y];}
#define min mmin
#define max mmax
#define abs aabs
namespace io {
const int SIZE = (1 << 21) + 1;
char ibuf[SIZE], *iS, *iT, obuf[SIZE], *oS = obuf, *oT = oS + SIZE - 1, c, qu[55]; int f, qr;
// getchar
#define gc() (iS == iT ? (iT = (iS = ibuf) + fread (ibuf, 1, SIZE, stdin), (iS == iT ? EOF : *iS ++)) : *iS ++)
// print the remaining part
inline void flush () {
fwrite (obuf, 1, oS - obuf, stdout);
oS = obuf;
}
// putchar
inline void putc (char x) {
*oS ++ = x;
if (oS == oT) flush ();
}
// input a signed integer
inline bool read (signed &x) {
for (f = 1, c = gc(); c < '0' || c > '9'; c = gc()) if (c == '-') f = -1;else if(c==EOF)return 0;
for (x = 0; c <= '9' && c >= '0'; c = gc()) x = x * 10 + (c & 15); x *= f;
return 1;
}
inline bool read (long long &x) {
for (f = 1, c = gc(); c < '0' || c > '9'; c = gc()) if (c == '-') f = -1;else if(c==EOF)return 0;
for (x = 0; c <= '9' && c >= '0'; c = gc()) x = x * 10 + (c & 15); x *= f;
return 1;
}
inline bool read (char &x) {
x=gc();
return x!=EOF;
}
inline bool read(char *x){
while((*x=gc())=='\n' || *x==' '||*x=='\r')if(*x==EOF)return 0;
while(!(*x=='\n'||*x==' '||*x=='\r'||*x==EOF))*(++x)=gc();
*x=0;
return 1;
}
template<typename A,typename ...B>
inline bool read(A &x,B &...y){
return read(x)&&read(y...);
}
// print a signed integer
inline bool write (signed x) {
if (!x) putc ('0'); if (x < 0) putc ('-'), x = -x;
while (x) qu[++ qr] = x % 10 + '0', x /= 10;
while (qr) putc (qu[qr --]);
return 0;
}
inline bool write (long long x) {
if (!x) putc ('0'); if (x < 0) putc ('-'), x = -x;
while (x) qu[++ qr] = x % 10 + '0', x /= 10;
while (qr) putc (qu[qr --]);
return 0;
}
inline bool write (char x) {
putc(x);
return 0;
}
inline bool write(const char *x){
while(*x){putc(*x);++x;}
return 0;
}
inline bool write(char *x){
while(*x){putc(*x);++x;}
return 0;
}
template<typename A,typename ...B>
inline bool write(A x,B ...y){
return write(x)||write(y...);
}
//no need to call flush at the end manually!
struct Flusher_ {~Flusher_(){flush();}}io_flusher_;
}
using io :: read;
using io :: putc;
using io :: write;
int inf;
struct _init_{
_init_(){
memset(&inf,0x3f,sizeof(inf));
}
};
_init_ ___INIT__;
int a[11][11],n;
int f[11][131];
int is[131],i100;
const int p=1000000007;
int fpm(int a,int b){int c=1;for(;b;b>>=1,a=a*a%p)if(b&1)c=c*a%p;return c;}
struct lll{
unsigned long long a,b;
void set(int x){
if(x<64)a|=1ULL<<x;
else b|=1ULL<<(x-64);
}
void reset(){
a=b=0;
}
lll &operator |=(lll ano){this->a|=ano.a;this->b|=ano.b;return *this;}
bool operator ==(lll ano){return b==ano.b&&a==ano.a;}
bool operator [](int ano){return !!(ano<64?a&(1ULL<<ano):b&(1ULL<<(ano-64)));}
};
bool operator <(const lll &a,const lll &b){return a.b<b.b||(a.b==b.b && a.a<b.a);}
bool operator >(const lll &a,const lll &b){return a.b>b.b||(a.b==b.b && a.a>b.a);}
map<lll,int> st[8];
lll mx;
int dfs(int x,lll w){
if(x==n)return w==mx;
if(st[x].count(w))return st[x][w];
int ans=0;
lll ng;
lll s[n];
for(int j=0;j<n;++j)s[j].reset();
for(int i=0;i<1<<n;++i)if(w[i])for(int j=0;j<n;++j)if(!(i&(1<<j)))
s[j].set(i|(1<<j));
for(int i=0;i<1<<n;++i){
ng=w;
for(int j=0;j<n;++j)if(i&(1<<j))ng|=s[j];
ans=(ans+f[x][i]*dfs(x+1,ng))%p;
}
return st[x][w]=ans;
}
lll stt;
signed main(){
#ifdef QAQAutoMaton
freopen("E.in","r",stdin);
freopen("E.out","w",stdout);
#endif
read(n);
for(int i=0;i<1<<n;++i)mx.set(i);
stt.set(0);
i100=fpm(100,p-2);
for(int i=0;i<n;++i)for(int j=0;j<n;++j){
read(a[i][j]);
a[i][j]=a[i][j]*i100%p;
}
for(int i=0;i<n;++i)is[1<<i]=i;
for(int i=0;i<n;++i){
for(int j=0;j<1<<n;++j){
f[i][j]=1;
for(int k=0;k<n;++k)if(j&(1<<k))f[i][j]=f[i][j]*a[k][i]%p;
else f[i][j]=f[i][j]*(p+1-a[k][i])%p;
}
}
write(dfs(0,stt),'\n');
return 0;
}