HDU-4336 Card Collector 概率DP

　　题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4336

　　题意：买食品收集n个卡片，每个卡片的概率分别是pi，且Σp[i]<=1，求收集n个卡片需要买的食品数的期望。

　　压缩DP：把每个食品用二进制表示，0和1分别表示没有卡片和已经收集到此卡片的期望，则

　　　　 f[s]=(1-Σp[i])*f[s]+Σp[j]*f[s]+Σp[k]*f[s|(1<<k)]  

　　　　　　s表示状态，i表示所有卡片编号，j表示s状态中已经有的卡片编号，k表示s状态中没有的卡片编号

　　->  Σp[i]*f[s]=Σp[i]*f[s|(1<<i)] 

　　或者容斥原理做：

　　压缩DP:

//STATUS:C++_AC_281MS_7128KB
#include <functional>
#include <algorithm>
#include <iostream>
//#include <ext/rope>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <numeric>
#include <cstring>
#include <cassert>
#include <cstdio>
#include <string>
#include <vector>
#include <bitset>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <list>
#include <set>
#include <map>
using namespace std;
//#pragma comment(linker,"/STACK:102400000,102400000")
//using namespace __gnu_cxx;
//define
#define pii pair<int,int>
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define PI acos(-1.0)
//typedef
typedef __int64 LL;
typedef unsigned __int64 ULL;
//const
const int N=(1<<20)+10;
const int INF=0x3f3f3f3f;
const int MOD= 1000000007,STA=8000010;
const LL LNF=1LL<<55;
const double EPS=1e-9;
const double OO=1e30;
const int dx[4]={-1,0,1,0};
const int dy[4]={0,1,0,-1};
const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
//Daily Use ...
inline int sign(double x){return (x>EPS)-(x<-EPS);}
template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
template<class T> inline T Min(T a,T b){return a<b?a:b;}
template<class T> inline T Max(T a,T b){return a>b?a:b;}
template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
//End

double p[23],f[N];
int n;

int main(){
 //   freopen("in.txt","r",stdin);
    int i,j,up;
    double s;
    while(~scanf("%d",&n))
    {
        for(i=0;i<n;i++){
            scanf("%lf",&p[i]);
        }
        up=(1<<n)-1;
        f[up]=0;
        for(i=up-1;i>=0;i--){
            f[i]=1;s=0;
            for(j=0;j<n;j++){
                if(i&(1<<j))continue;
                f[i]+=p[j]*f[i|(1<<j)];
                s+=p[j];
            }
            f[i]/=s;
        }

        printf("%lf
",f[0]);
    }
    return 0;
}

　　容斥原理：

//STATUS:C++_AC_203MS_244KB
#include <functional>
#include <algorithm>
#include <iostream>
//#include <ext/rope>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <numeric>
#include <cstring>
#include <cassert>
#include <cstdio>
#include <string>
#include <vector>
#include <bitset>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <list>
#include <set>
#include <map>
using namespace std;
//#pragma comment(linker,"/STACK:102400000,102400000")
//using namespace __gnu_cxx;
//define
#define pii pair<int,int>
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define PI acos(-1.0)
//typedef
typedef __int64 LL;
typedef unsigned __int64 ULL;
//const
const int N=(1<<20)+10;
const int INF=0x3f3f3f3f;
const int MOD= 1000000007,STA=8000010;
const LL LNF=1LL<<55;
const double EPS=1e-9;
const double OO=1e30;
const int dx[4]={-1,0,1,0};
const int dy[4]={0,1,0,-1};
const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
//Daily Use ...
inline int sign(double x){return (x>EPS)-(x<-EPS);}
template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
template<class T> inline T Min(T a,T b){return a<b?a:b;}
template<class T> inline T Max(T a,T b){return a>b?a:b;}
template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
//End

double p[23];
int n;

int main(){
 //   freopen("in.txt","r",stdin);
    int i,j,up,cnt;
    double ans,s;
    while(~scanf("%d",&n))
    {
        for(i=0;i<n;i++){
            scanf("%lf",&p[i]);
        }
        up=(1<<n)-1;ans=0;
        for(i=1;i<=up;i++){
            s=0;
            for(j=cnt=0;j<n;j++){
                if(i&(1<<j)){
                    cnt++;
                    s+=p[j];
                }
            }
            if(cnt&1)ans+=1/s;
            else ans-=1/s;
        }

        printf("%lf
",ans);
    }
    return 0;
}