POJ 3744 Scout YYF I [概率DP]

Description

YYF is a couragous scout. Now he is on a dangerous mission which is to penetrate into the enemy's base. After overcoming a series difficulties, YYF is now at the start of enemy's famous "mine road". This is a very long road, on which there are numbers of mines. At first, YYF is at step one. For each step after that, YYF will walk one step with a probability of p, or jump two step with a probality of 1- p. Here is the task, given the place of each mine, please calculate the probality that YYF can go through the "mine road" safely.

Input

The input contains many test cases ended with EOF. 
Each test case contains two lines. 
The First line of each test case is N (1 ≤ N ≤ 10) and p (0.25 ≤ p ≤ 0.75) seperated by a single blank, standing for the number of mines and the probability to walk one step. 
The Second line of each test case is N integer standing for the place of N mines. Each integer is in the range of [1, 100000000].

Output

For each test case, output the probabilty in a single line with the precision to 7 digits after the decimal point.

題意：

一條路上每個點標號爲1,2,3....，你一開始站在1往正方向出發， 有P的概率走到下一格，（1-P）的概率走到下一格的下一格，現在路上可能會有一些地雷，給出N個地雷的座標，問你不踩地雷走過這條路的概率。

範圍：

地雷數小於等於10，座標小於10憶

解法：

OXOOOOX

若O表示可以走的，X表示有地雷，那麼走過這段路可以分成三個部分

第一部分：座標1->座標3 這一階段達成的概率是1-p

第二部分：座標3->座標6

第三部分：座標6->座標8 這一階段達成的概率是1-p

可以發現，只要求出第二部分的概率即可，這一部分的概率可以DP

記F[i]爲走過長度爲i的沒有地雷的路程的概率，那麼公式爲f[i]=f[i-1]*p+f[i-2]*(1.0-p);

但是因爲I可能很大，需要考慮對它進一步分析，最後發現i趨於無窮時，F[I]是有極限的，實際過程中這個極限很快就會達到。

所以處理出大約前1000個F[I]，大於1000的取F[1000]，然後按階段處理即可。

代碼:

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<math.h>
#include<iostream>
#include<stdlib.h>
#include<set>
#include<map>
#include<queue>
#include<vector>
#include<bitset>
#pragma comment(linker, "/STACK:1024000000,1024000000")
template <class T>
bool scanff(T &ret){ //Faster Input
    char c; int sgn; T bit=0.1;
    if(c=getchar(),c==EOF) return 0;
    while(c!='-'&&c!='.'&&(c<'0'||c>'9')) c=getchar();
    sgn=(c=='-')?-1:1;
    ret=(c=='-')?0:(c-'0');
    while(c=getchar(),c>='0'&&c<='9') ret=ret*10+(c-'0');
    if(c==' '||c=='\n'){ ret*=sgn; return 1; }
    while(c=getchar(),c>='0'&&c<='9') ret+=(c-'0')*bit,bit/=10;
    ret*=sgn;
    return 1;
}
#define inf 1073741823
#define llinf 4611686018427387903LL
#define PI acos(-1.0)
#define lth (th<<1)
#define rth (th<<1|1)
#define rep(i,a,b) for(int i=int(a);i<=int(b);i++)
#define drep(i,a,b) for(int i=int(a);i>=int(b);i--)
#define gson(i,root) for(int i=ptx[root];~i;i=ed[i].next)
#define tdata int testnum;scanff(testnum);for(int cas=1;cas<=testnum;cas++)
#define mem(x,val) memset(x,val,sizeof(x))
#define mkp(a,b) make_pair(a,b)
#define findx(x) lower_bound(b+1,b+1+bn,x)-b
#define pb(x) push_back(x)
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;

int n,a[1111];
double p,f[1111];
int main(){
    while(scanf("%d%lf",&n,&p)!=EOF){
        mem(f,0);
        f[1]=1.0;
        rep(i,2,1000){
           f[i]=f[i-1]*p+f[i-2]*(1.0-p);
        }
        double ans=1.0;
        rep(i,1,n)scanff(a[i]);
        sort(a+1,a+1+n);
        int an=unique(a+1,a+1+n)-a-1;
        a[0]=0;
        rep(i,1,an){
            int x=a[i];
            int dis=min(a[i]-a[i-1]-1,1000);
            ans=ans*f[dis]*(1.0-p);
        }
        printf("%.7f\n",ans);
    }
    return 0;
}