ZOJ-3640 Help Me Escape 概率DP

　　题目链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3640

　　题意：Cain被困在一个洞穴里，洞穴有n个出口，每个出口有一个难度值C[i]，Cain有一个初始的战斗值f。现在Cain随机选择一个出口，如果f大于出后的难度，那么Cain将会花floor( (1+sqrt(5))/2*C[i]*C[i] )天出去，否则Cain的f将会增加C[i]并且消耗掉一天时间，然后重新尝试。求Cain逃出洞穴天数的期望值。。。

　　首先用BFS把Cain所有的f值的可能情况求出来，然后就可以列出方程，这里因为期望都是由大的f推过来，所以f直接从大到小递推消元就可以了。。

//STATUS:C++_AC_730MS_1436KB
#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=40010;
const int INF=0x3f3f3f3f;
//const LL MOD=1000000007,STA=8000010;
//const LL LNF=1LL<<55;
const double EPS=1e-9;
const double OO=1e50;
const int dx[8]={-1,-1,0,1,1,1,0,-1};
const int dy[8]={0,1,1,1,0,-1,-1,-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 E
,d
;
int w
,id
,c
,vis
;
int n,f,cnt;

void bfs(int s)
{
int i,x;
queue<int> q;
q.push(s);
mem(vis,0);
vis[s]=1;
while(!q.empty()){
x=q.front();q.pop();
for(i=1;i<=n;i++){
if(x>c[i] || vis[x+c[i]])continue;
w[cnt++]=x+c[i];
vis[x+c[i]]=1;
q.push(x+c[i]);
}
}
}

int main(){
//   freopen("in.txt","r",stdin);
int i,j;
double t=(1+sqrt(5.0))/2,p;
while(~scanf("%d%d",&n,&f))
{
p=1.0/n;
for(i=1;i<=n;i++){
scanf("%d",&c[i]);
d[i]=floor(t*c[i]*c[i]);
}

w[cnt=0]=f;cnt++;
bfs(f);
sort(w,w+cnt);
for(i=0;i<cnt;i++)id[w[i]]=i;
for(i=cnt-1;i>=0;i--){
E[i]=0;
for(j=1;j<=n;j++){
if(w[i]>c[j])E[i]+=p*d[j];
else E[i]+=p*(E[id[w[i]+c[j]]]+1);
}
}

printf("%.3lf\n",E[0]);
}
return 0;
}