HDU 3366 Passage （概率DP）

Passage

Problem Description

Bill is a millionaire. But unfortunately he was trapped in a castle. There are only n passages to go out. For any passage i (1<=i<=n), Pi (0<=Pi<=1) denotes the probability that Bill will escape from this castle safely if he chose this passage. Qi (0<=Qi<=1-Pi) denotes the probability that there is a group of guards in this passage. And Bill should give them one million dollars and go back. Otherwise, he will be killed. The probability of this passage had a dead end is 1-Pi-Qi. In this case Bill has to go back. Whenever he came back, he can choose another passage.
We already know that Bill has M million dollars. Help Bill to find out the probability that he can escape from this castle if he chose the optimal strategy.

 

Input

The first line contains an integer T (T<=100) indicating the number of test cases.
The first line of each test case contains two integers n (1<=n<=1000) and M (0<=M<=10).
Then n lines follows, each line contains two float number Pi and Qi.

 

Output

For each test case, print the case number and the answer in a single line.
The answer should be rounded to five digits after the decimal point.
Follow the format of the sample output.

 

Sample Input

3 1 10 0.5 0 2 0 0.3 0.4 0.4 0.5 3 0 0.333 0.234 0.353 0.453 0.342 0.532

 

Sample Output

Case 1: 0.50000 Case 2: 0.43000 Case 3: 0.51458

 

Source

“光庭杯”第五届华中北区程序设计邀请赛 暨 WHU第八届程序设计竞赛

 

题目大意：

T组测试数据，一个人困在了城堡中，有n个通道，m百万money ，每个通道能直接逃出去的概率为 P[i] ，遇到士兵的概率为 q[i]，遇到士兵得给1百万money，否则会被杀掉，还有 1-p[i]-q[i] 的概率走不通，要回头。问在可以选择的情况下，逃出去的概率是多少？

解题思路：

首先，n个通道要选择哪个先走哪个后走呢？如果暴力是2^n，显然不可取。只需要贪心，选择逃生概率最大的通道，也就是 p[i]/q[i]最大的优先。

用 dp[i][j]记录 还剩j次机会，已经走到第i个通道能逃生的概率。

那么当前：

（1）遇到士兵，dp[i+1][j-1]+=dp[i][j]*q[i]

（2）走不通，dp[i+1][j]+=dp[i][j]*( 1-p[i]-q[i] )

（3）直接逃生，答案加上这个概率。

解题代码：

#include 
#include 
#include 
using namespace std;

const int maxn=1100;

struct route{
    double p,q;
    friend  bool operator < (route a,route b){
        return a.p/a.q>b.p/b.q;
    }
}r[maxn];

int n,m;
double dp[maxn][20];

void input(){
    scanf("%d%d",&n,&m);
    for(int i=0;i=0;j--){
            ans+=dp[i][j]*r[i].p;
            if(j-1>=0) dp[i+1][j-1]+=dp[i][j]*r[i].q;
            dp[i+1][j]+=dp[i][j]*(1.0-r[i].p-r[i].q);
        }
    }
    return ans;
}

int main(){
    int t;
    scanf("%d",&t);
    for(int i=1;i<=t;i++){
        input();
        printf("Case %d: %.5lf\n",i,solve());
    }
    return 0;
}

转载于:https://www.cnblogs.com/toyking/p/3893198.html