题意

给定一张n+n的二分图,每条边有$p_{i,j}$的概率出现,求存在完美匹配的概率。

$n\le 7,15s$

Subtask: $n\le 6,7s$

题解

状压,$2^n$位表示X点中$2^n$个子集是否存在完美匹配,每次枚举Y方点中一个对X方点连边方案。

记忆化搜索之后可以过。

gcz: dfs跑一遍发现不同状态只有6e4个。

/*
Author: QAQ Automaton
Lang: C++
Prog: E.cpp
Mail: lk@qaq-am.com
Blog: https://www.qaq-am.com/
*/
#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 
using namespace std;
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;
}

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