hdu 2955 Robberies(概率01背包,反向思维)

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
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
概率01背包，反向思维，把钱当做背包，把概率当做物品

#include<cstdio>
#include<cstring>
#include<stdlib.h>
#include<fstream>
#include<ctype.h>
#include<math.h>
#include<stack>
#include<queue>
#include<map>
#include<set>
#include<vector>
#include<string>
#include<iostream>
#include<algorithm>
#include<utility>
#include<iomanip>
#include<time.h>
#include<iostream>
#define lowbit(x) (x&-x)
#define abs(x) ((x)>0?(x):-(x))
using namespace std;
typedef long long ll;
const double Pi = acos(-1.0);
const int N = 1e6+10, M = 1e3+20, mod = 1e9+7, inf = 2e9+10;
const double e=2.718281828459 ;
const double esp=1e-9;
int t,n;
double P;
double dp[10100];//获得i钱最大不被抓的概率
struct node
{
    int m;
    double p;
} a[110];
int main()
{
    // freopen("in.txt","r",stdin);    freopen("out.txt","w",stdout);
    std::ios::sync_with_stdio(false);
    std::cin.tie(0);
    cin>>t;
    while(t--)
    {

        memset(dp,0,sizeof(dp));
        dp[0]=1;//没偷钱的概率等于1
        cin>>P>>n;
        int maxn=0;
        for(int i=0; i<n; i++)
        {
            int x; double y;
            cin>>x>>y;//y是被抓的概率
            maxn+=x;      a[i].m=x;     a[i].p=y;
        }
        for(int i=0; i<n; i++)
        {
            for(int j=maxn; j>=a[i].m; j--)
            {
                dp[j]=max(dp[j],dp[j-a[i].m]*(1-a[i].p));
            }
        }
        int ans=0;
        for(int i=maxn; i>=0; i--)
        {
            if(dp[i]>=1-P)//最大不被抓的概率大于最小不被抓的概率
            {
                ans=i;  break;
            }
        }
        cout<<ans<<endl;
    }
    return 0;
}