HDU——2955Robberies

Posted 2020-12-29 lwsh123k

Robberies
Time Limit: 1000msMemory Limit: 32768KB This problem will be judged on HDU. Original ID: 2955
64-bit integer IO format: %I64d Java class name: Main
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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.

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 .
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.
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

题意：Roy想要抢劫银行，每家银行多有一定的金额和被抓到的概率，知道Roy被抓的最大概率P，求Roy在被抓的情况下，抢劫最多。

分析：被抓概率可以转换成安全概率，Roy的安全概率大于1-P时都是安全的。抢劫的金额为0时，肯定是安全的，所以d[0]=1;其他金额初始为最危险的所以概率全为0；

题目中给定价值和被抓几率，但是被抓几率不可以用乘积来组合计算，举个例子，比如第一个银行3%被抓几率，第二个5%被抓几率，那么乘起来会变成0.15%，抢的越多，被抓几率却越小了，显然不对，因此要转换成不被抓几率，上述例子则变为第一家97%不被抓，第二家95%不被抓，乘起来就是92.15%，抢的越多，不被抓的几率越来越小即被抓几率越来越大，这样才是符合常理的。

#include<cstdio>
#include<string.h>
#include<algorithm>
using namespace std;
double f[10100],p[10100];
int w[500]; //存储钱
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n;
        double p1;
        scanf("%lf%d",&p1,&n);
        int sum = 0;
        for(int i = 1; i <= n; i++)
        {
            scanf("%d%lf",&w[i],&p[i]);
            sum += w[i];
        }
        memset(f,0,sizeof f);
        f[0] = 1;   //抢0元，逃跑率为1
        for(int i = 1; i <= n; i++)
            for(int j = sum; j >= w[i]; j--)
            f[j] = max(f[j],f[j-w[i]]*(1-p[i]));
        for(int i = sum; ;i--)
            if(f[i] > (1-p1))
            {
                printf("%d
",i);
                break;
            }
    }
    return 0;
}