CF1229E

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。

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
/*
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;
}
# DP, 搜索

评论

Your browser is out-of-date!

Update your browser to view this website correctly. Update my browser now

×