Passage
总提交: 40 测试通过: 20
描述
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.
输入
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.
输出
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.
样例输入
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
样例输出
Case 1: 0.50000
Case 2: 0.43000
Case 3: 0.51458
题目来源
“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
如果有两总选择 P1, Q1 and P2, Q2
先选第一条路被抓的概率为 ret1 = (1 - p1 - Q1) * Q2 + Q1
先选第二条路被抓的概率为 ret2 = (1 - p2 - Q2) * Q1 + Q2
设P1 / Q1 > P2 / Q2 => P1 * Q2 > P2 * Q1
ret1 - ret2 = P2 * Q1 - P1 * Q2 < 0
所以优先选择第一条路
先按Pi / Qi 从大到小排序
dp[i][j] 表示到第 i 条路, 消耗 j million dollars的概率
逃跑的概率 ret += dp[i-1][j] * Pi ( j <= k )
dp[i][j] = dp[i-1][j] * (1.0 - Pi - Qi) + dp[i-1][j-1] * Qi
1 #include <stdio.h> 2 #include <iostream> 3 #include <algorithm> 4 using namespace std; 5 6 struct Node{ 7 double Pi, Qi; 8 }; 9 10 bool cmp(Node a, Node b){ 11 return a.Pi / a.Qi > b.Pi / b.Qi; 12 } 13 14 Node path[1010]; 15 double dp[1010][15]; 16 double deal(int n, int k){ 17 memset(dp, 0, sizeof(dp)); 18 dp[0][0] = 1; 19 double ret = 0; 20 for(int i = 1; i <= n; i++){ 21 22 for(int j = k; j >= 1; j--){ 23 ret += dp[i-1][j] * path[i].Pi; 24 dp[i][j] += dp[i-1][j] * (1.0 - path[i].Pi - path[i].Qi); 25 dp[i][j] += dp[i-1][j-1] * path[i].Qi; 26 } 27 ret += dp[i-1][0] * path[i].Pi; 28 dp[i][0] = dp[i-1][0] * (1.0 - path[i].Pi - path[i].Qi); 29 } 30 return ret; 31 } 32 33 int main() 34 { 35 int T; 36 scanf("%d", &T); 37 for(int cas = 1; cas <= T; cas++){ 38 int n, k; 39 scanf("%d %d", &n, &k); 40 for(int i = 1; i <= n; i++){ 41 scanf("%lf %lf", &path[i].Pi, &path[i].Qi); 42 } 43 sort(path+1, path+n+1, cmp); 44 printf("Case %d: %.5lf ", cas, deal(n, k)); 45 } 46 return 0; 47 }