hdu2955 Robberies(背包问题)

Robberies

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 13043    Accepted Submission(s): 4827

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.


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

30.04 31 0.022 0.033 0.050.06 32 0.032 0.033 0.050.10 31 0.032 0.023 0.05

 

Sample Output

246

 

本题的题意为:一个贼想抢银行，然后给出被抓的概率的和 和测试样例的个数N，然后下面N行每行为每次偷的钱数和被抓的概率。

解题的思路:很明显这题考察的是01背包，然后我一开始本来想用贪心，结果想想必然是不能的，所以得用动态规划做，动态规划得有边界吧，边界是什么？当然是能偷的钱数啦，它是有封顶的哈，所以统计一下贼最多可以抢的钱，然后有了这个思路可以套用01背包的核心代码:

for i=1..N   forv=V..0        f[v]=max{f[v],f[v-c[i]]+w[i]};</span>

演变一下变为本题的核心代码:

for i=1..N   forv=sum..0        f[v]=max{f[v],f[v-c[i]]*w[i]}; //w[i]为成功率 ，c[i]为第i个的钱数</span>

下面为本题的完整代码：

#include<iostream>#include<cstring>#include<cstdio>#include<algorithm>using namespace std;const int maxn = 10000;double dp[maxn]; //成功率struct Map{    int value;    double rate;} p[maxn];/*bool cmp(struct Map a,struct Map b){    return (1-a.rate) * a.value > (1-b.rate) * b.value;}*/int main(){    int T,m,i,j,sum; //m为样例个数    double total;    //freopen("1","r",stdin);    cin>>T;    while(T--)    {        sum = 0;        cin>>total>>m;        for(i=0; i<m; i++)        {            cin>>p[i].value>>p[i].rate; //每次能偷的钱数和被抓的概率            sum += p[i].value; //总的能偷的钱数        }        for(i=0;i<=sum;i++)        {            dp[i] = 0;        }        dp[0]= 1; //因为没偷所以不会被抓，所以成功率为百分之百        //sort(p+1,p+m+1,cmp);        /*for(i=1;i<=m;i++)        {            cout<<p[i].value<<" "<<p[i].rate<<endl;        }*/        for(i=0;i<m;i++)        {            for(j=sum;j>=p[i].value;j--)            {                dp[j] = max(dp[j],dp[j-p[i].value]*(1-p[i].rate)); //拿了该件物品的价值乘以成功率                //cout<<dp[j]<<endl;            }        }        for(i=sum;i>=0;i--)        {            if(dp[i] - (1-total) > 0.000001) //从能偷的最大钱开始从大到小遍历，一旦发现不会被抓的数就输出，然后结束            {                cout<<i<<endl;                break;            }        }    }    return 0;}