杭电2955(01背包)
Robberies
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 11119 Accepted Submission(s):
4121
Problem Description
The aspiring Roy the Robber has seen a lot of American
movies, and knows that the bad guys usually gets caught in the end, often
because they become too greedy. He has decided to work in the lucrative business
of bank robbery only for a short while, before retiring to a comfortable job at
a university.
![](http://acm.hdu.edu.cn/data/images/con211-1010-1.jpg)
For a few months now, Roy has been assessing the security of various banks and the amount of cash they hold. He wants to make a calculated risk, and grab as much money as possible.
His mother, Ola, has decided upon a tolerable probability of getting caught. She feels that he is safe enough if the banks he robs together give a probability less than this.
![](http://acm.hdu.edu.cn/data/images/con211-1010-1.jpg)
For a few months now, Roy has been assessing the security of various banks and the amount of cash they hold. He wants to make a calculated risk, and grab as much money as possible.
His mother, Ola, has decided upon a tolerable probability of getting caught. She feels that he is safe enough if the banks he robs together give a probability less than this.
Input
The first line of input gives T, the number of cases.
For each scenario, the first line of input gives a floating point number P, the
probability Roy needs to be below, and an integer N, the number of banks he has
plans for. Then follow N lines, where line j gives an integer Mj and a floating
point number Pj .
Bank j contains Mj millions, and the probability of getting caught from robbing it is Pj .
Bank j contains Mj millions, and the probability of getting caught from robbing it is Pj .
Output
For each test case, output a line with the maximum
number of millions he can expect to get while the probability of getting caught
is less than the limit set.
Notes and Constraints
0 < T <= 100
0.0 <= P <= 1.0
0 < N <= 100
0 < Mj <= 100
0.0 <= Pj <= 1.0
A bank goes bankrupt if it is robbed, and you may assume that all probabilities are independent as the police have very low funds.
Notes and Constraints
0 < T <= 100
0.0 <= P <= 1.0
0 < N <= 100
0 < Mj <= 100
0.0 <= Pj <= 1.0
A bank goes bankrupt if it is robbed, and you may assume that all probabilities are independent as the police have very low funds.
Sample Input
3
0.04 3
1 0.02
2 0.03
3 0.05
0.06 3
2 0.03
2 0.03
3 0.05
0.10 3
1 0.03
2 0.02
3 0.05
Sample Output
2
4
6
思路:
01背包,但是如果把概率当做背包容量的话,得乘以1000000,
必然超时,所以应该把钱数当做容量。。。
另外,概率求值,化成成功逃跑,两个相乘。
代码:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include<stdio.h> 2 #include<stdlib.h> 3 #include<string.h> 4 5 double dp[10005]; 6 7 double getmax(double x, double y){ 8 return x > y ? x : y; 9 } 10 11 int main(){ 12 int t, i, j, n, money[105],zmoney; 13 double zp, p[105]; 14 scanf("%d", &t); 15 while(t --){ 16 scanf("%lf %d", &zp, &n); 17 zmoney = 0; 18 for(i = 0; i < n; i ++){ 19 scanf("%d %lf", &money[i], &p[i]); 20 p[i] = 1 - p[i]; 21 zmoney += money[i]; 22 } 23 memset(dp, 0, sizeof(dp)); 24 dp[0] = 1; 25 for(i = 0; i < n; i ++){ 26 for(j = zmoney; j >= money[i]; j --){ 27 dp[j] = getmax(dp[j], p[i] * dp[j - money[i]]); 28 } 29 } 30 for(i = zmoney; i >= 0; i --){ 31 if(dp[i] >= (1 - zp)){ 32 printf("%d\n", i); 33 break; 34 } 35 } 36 } 37 return 0; 38 }