LightOJ 1079 - Just another Robbery （背包问题）

Just another Robbery
Time Limit:4000MS Memory Limit:32768KB 64bit IO Format:%lld & %llu
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Status
Description
As Harry Potter series is over, Harry has no job. Since he wants to make quick money, (he wants everything quick!) so he decided to rob banks. He wants to make a calculated risk, and grab as much money as possible. But his friends - Hermione and Ron have decided upon a tolerable probability P of getting caught. They feel that he is safe enough if the banks he robs together give a probability less than P.

Input
Input starts with an integer T (≤ 100), denoting the number of test cases.

Each case contains a real number P, the probability Harry needs to be below, and an integer N (0 < N ≤ 100), the number of banks he has plans for. Then follow N lines, where line j gives an integer Mj (0 < Mj ≤ 100) and a real number Pj . Bank j contains Mj millions, and the probability of getting caught from robbing it is Pj. A bank goes bankrupt if it is robbed, and you may assume that all probabilities are independent as the police have very low funds.

Output
For each case, print the case number and the maximum number of millions he can expect to get while the probability of getting caught is less than P.

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
Case 1: 2
Case 2: 4
Case 3: 6
Hint
For the first case, if he wants to rob bank 1 and 2, then the probability of getting caught is 0.02 + (1 - 0.02) * .03 = 0.0494 which is greater than the given probability (0.04). That’s why he has only option, just to rob rank 2.

Problem Setter: Jane Alam Jan
Developed and Maintained by
JANE ALAM JAN
Copyright © 2012
LightOJ, Jane Alam Jan

题意：
给你一些银行的存储金钱的数目及被抓的概率，若被抓总概率不超过p的话，问不被抓的条件下最多可以抢多少钱？

思路：
记住一个口号：没有背包创造背包也要上
这道题可以看出来和背包有关联，但是显然没有可以堆起来的背包，所以我们可以根据抢钱的总数构建一个背包，就变成了背包问题，最重要的是概率这个东西反面求比较简单

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;#define M 110double dp[M*M];//存的是被抓捕的概率，因为正面不好求所有反着来 int main(){    int t, n, sum, nm[M];    double p, np[M];    scanf("%d", &t);    for(int ca=1; ca<=t; ca++)    {        scanf("%lf%d", &p, &n);        sum = 0;        for(int i=1; i<=n; i++)        {            scanf("%d%lf", &nm[i], &np[i]);            sum += nm[i];//反着构建一个背包         }        memset(dp, 0, sizeof(dp));        int ans = 0;        dp[0] = 1;        for(int i=1; i<=n; i++)//转变成了背包问题        {            for(int j=sum; j>=nm[i]; j--)            {                dp[j] = max(dp[j], dp[j-nm[i]] * (1-np[i]));            }        }        for(int i=0; i<=sum; i++)        {            if((1-dp[i]) < p)//从小到大找到适合的            {                ans = i;            }        }        printf("Case %d: %d\n", ca, ans);    }    return 0;}